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Question 908084: find the value of r so that the line that passes through each pair of points has the given slope. (6,r), (3,3), m=2
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39792)  (Show Source):
Answer by MathTherapy(10806)  (Show Source):
Question 445381: What are the solutions to
6x+3y+z=19
x-3y+2z=21
17x-2y+3z=86
Found 2 solutions by MathTherapy, ikleyn:
Answer by MathTherapy(10806) (Show Source):
Answer by ikleyn(53748)  (Show Source):
You can put this solution on YOUR website! .
What are the solutions to
6x+3y+z=19
x-3y+2z=21
17x-2y+3z=86
~~~~~~~~~~~~~~~~~~~~~
The solution in the post by @mananth x = 4.27, y = -9.99, z = 3.37 is incorrect.
To check, I substituted these values into first equation and got the value of the left side -0.98,
different from 19.
This is just enough to refute the solution by @mananth.
If you want to get a precise solution, go to website https:\\www.reshish.com/
Find there free of charge solver and use it.
The solver can provide step-by-step procedure with all necessary detailed explanations.
Question 1165604: Consider the following functions T : R → R . Show that each is a linear transformation and determine for each the matrix A such that T (x) = A x
a) x1. x1+2x2 +3x3
T x2. = 2x2-3x1+x3
¶ x3
b) x1. 7x1+2x2 +x3
T x2. = 3x1-11x2+2x3
¶ x3
Answer by CPhill(2189)  (Show Source):
You can put this solution on YOUR website! This question asks you to demonstrate that the given transformations $T: \mathbb{R}^3 \to \mathbb{R}^3$ are linear and to find the standard matrix $A$ for each, such that $T(\mathbf{x}) = A\mathbf{x}$.
## General Definition of a Linear Transformation
A transformation $T: V \to W$ is linear if, for all vectors $\mathbf{u}, \mathbf{v}$ in $V$ and all scalars $c$ in $\mathbb{R}$:
1. **Additivity:** $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$
2. **Homogeneity:** $T(c\mathbf{u}) = cT(\mathbf{u})$
---
## Part (a)
The transformation is defined by:
$$T \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} x_1 + 2x_2 + 3x_3 \\ -3x_1 + 2x_2 + x_3 \\ 0 \end{pmatrix}$$
*(Note: I assumed the third component is $0$ as the input has three components but the output only shows two distinct linear expressions, and a $3 \times 3$ matrix is expected for $T: \mathbb{R}^3 \to \mathbb{R}^3$. I'll proceed by assuming the third component is a constant like 0 or $x_1$ from the format)*
**Based on the layout, it is most likely that the output is meant to be a $3 \times 1$ column vector. Given the formatting ambiguity, I will use the coefficients as presented, and assume the third component is 0 for the linear combination interpretation.**
Let's re-write the output as a $3 \times 1$ vector where the structure dictates the coefficients:
$$T \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} x_1 + 2x_2 + 3x_3 \\ -3x_1 + 2x_2 + x_3 \\ 0 \end{pmatrix}$$
### A. Finding the Standard Matrix $A$
The standard matrix $A$ is formed by applying $T$ to the standard basis vectors $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$:
$$A = [T(\mathbf{e}_1) \mid T(\mathbf{e}_2) \mid T(\mathbf{e}_3)]$$
1. $$T\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1(1) + 2(0) + 3(0) \\ -3(1) + 2(0) + 1(0) \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}$$
2. $$T\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1(0) + 2(1) + 3(0) \\ -3(0) + 2(1) + 1(0) \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}$$
3. $$T\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1(0) + 2(0) + 3(1) \\ -3(0) + 2(0) + 1(1) \\ 0 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix}$$
The standard matrix $A$ is:
$$\mathbf{A = \begin{pmatrix} 1 & 2 & 3 \\ -3 & 2 & 1 \\ 0 & 0 & 0 \end{pmatrix}}$$
### B. Proving Linearity
Since the components of $T(\mathbf{x})$ are **linear homogeneous polynomials** (polynomials where every term has degree one) in the variables $x_1, x_2, x_3$ and there is no constant term, $T$ is automatically a linear transformation.
---
## Part (b)
The transformation is defined by:
$$T \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 7x_1 + 2x_2 + x_3 \\ 3x_1 - 11x_2 + 2x_3 \end{pmatrix}$$
*(Note: In this case, $T: \mathbb{R}^3 \to \mathbb{R}^2$, as the output vector has 2 components.)*
### A. Finding the Standard Matrix $A$
The standard matrix $A$ is a $2 \times 3$ matrix formed by applying $T$ to the standard basis vectors $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$:
$$A = [T(\mathbf{e}_1) \mid T(\mathbf{e}_2) \mid T(\mathbf{e}_3)]$$
1. $$T\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 7(1) + 2(0) + 1(0) \\ 3(1) - 11(0) + 2(0) \end{pmatrix} = \begin{pmatrix} 7 \\ 3 \end{pmatrix}$$
2. $$T\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 7(0) + 2(1) + 1(0) \\ 3(0) - 11(1) + 2(0) \end{pmatrix} = \begin{pmatrix} 2 \\ -11 \end{pmatrix}$$
3. $$T\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 7(0) + 2(0) + 1(1) \\ 3(0) - 11(0) + 2(1) \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$$
The standard matrix $A$ is:
$$\mathbf{A = \begin{pmatrix} 7 & 2 & 1 \\ 3 & -11 & 2 \end{pmatrix}}$$
### B. Proving Linearity
Since the components of $T(\mathbf{x})$ are again **linear homogeneous polynomials** in $x_1, x_2, x_3$ (no terms with degree other than one and no constant terms), $T$ is a linear transformation.
**Formal Proof for Additivity (Homogeneity is similar):**
Let $\mathbf{u} = \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}$.
$$T(\mathbf{u} + \mathbf{v}) = T \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \\ u_3 + v_3 \end{pmatrix}$$
$$T(\mathbf{u} + \mathbf{v}) = \begin{pmatrix} 7(u_1 + v_1) + 2(u_2 + v_2) + (u_3 + v_3) \\ 3(u_1 + v_1) - 11(u_2 + v_2) + 2(u_3 + v_3) \end{pmatrix}$$
Rearranging the terms based on $\mathbf{u}$ and $\mathbf{v}$:
$$T(\mathbf{u} + \mathbf{v}) = \begin{pmatrix} (7u_1 + 2u_2 + u_3) + (7v_1 + 2v_2 + v_3) \\ (3u_1 - 11u_2 + 2u_3) + (3v_1 - 11v_2 + 2v_3) \end{pmatrix}$$
$$T(\mathbf{u} + \mathbf{v}) = \begin{pmatrix} 7u_1 + 2u_2 + u_3 \\ 3u_1 - 11u_2 + 2u_3 \end{pmatrix} + \begin{pmatrix} 7v_1 + 2v_2 + v_3 \\ 3v_1 - 11v_2 + 2v_3 \end{pmatrix} = T(\mathbf{u}) + T(\mathbf{v})$$
Thus, $T$ is linear.
Question 1165858: Let matrix3x2 A = 1, -1 , 2, 0, 3, -4. Determine whether multiplication by A is a one-to-one transformation.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! To determine whether multiplication by a matrix $A$ is a **one-to-one transformation**, we need to analyze the properties of the linear transformation $T(\mathbf{x}) = A\mathbf{x}$.
A linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is **one-to-one (injective)** if and only if **$T(\mathbf{x}) = \mathbf{0}$ has only the trivial solution** ($\mathbf{x} = \mathbf{0}$). Equivalently, this means the **null space (kernel)** of $A$ contains only the zero vector, or that the matrix $A$ has **linearly independent columns**.
## 1. Analyze the Matrix and Transformation
The given matrix $A$ is a $3 \times 2$ matrix (3 rows, 2 columns):
$$A = \begin{pmatrix} 1 & -1 \\ 2 & 0 \\ 3 & -4 \end{pmatrix}$$
The transformation $T$ maps vectors from $\mathbb{R}^2$ to $\mathbb{R}^3$, i.e., $T: \mathbb{R}^2 \to \mathbb{R}^3$.
$$T \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ 2 & 0 \\ 3 & -4 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$$
## 2. Check for One-to-One Property
We check the condition by solving the homogeneous equation $A\mathbf{x} = \mathbf{0}$:
$$\begin{pmatrix} 1 & -1 \\ 2 & 0 \\ 3 & -4 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$
This corresponds to the system of equations:
1. $x_1 - x_2 = 0$
2. $2x_1 + 0x_2 = 0$
3. $3x_1 - 4x_2 = 0$
From equation (2):
$$2x_1 = 0 \implies x_1 = 0$$
Substitute $x_1 = 0$ into equation (1):
$$0 - x_2 = 0 \implies x_2 = 0$$
Substitute $x_1 = 0$ and $x_2 = 0$ into equation (3) to verify consistency:
$$3(0) - 4(0) = 0 \implies 0 = 0$$
Since the only solution to $A\mathbf{x} = \mathbf{0}$ is the **trivial solution** ($\mathbf{x} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$), the columns of $A$ are **linearly independent**.
---
## Conclusion
Yes, multiplication by matrix $A$ **is a one-to-one transformation**.
Question 1165864: Kuttler, pg. 279, Ex. 5.2.8) Consider the following functions T : R3→ R2 . Show that each is a linear transformation and determine for each the matrix A such that
T (x) = A x.
T . (matrix3x1(x1, x2, x3) )= Matrix2x1(x1+2x2+3x3, 2x2−3x1+x3)
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! This problem asks you to confirm that a given function $T: \mathbb{R}^3 \to \mathbb{R}^2$ is a **linear transformation** and then find its **standard matrix** $A$.
The given transformation is:
$$T \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} x_1 + 2x_2 + 3x_3 \\ -3x_1 + 2x_2 + x_3 \end{pmatrix}$$
---
## 1. Showing $T$ is a Linear Transformation
A function $T$ is a linear transformation if it satisfies two properties for any vectors $\mathbf{u}$ and $\mathbf{v}$ in the domain ($\mathbb{R}^3$) and any scalar $c$ in $\mathbb{R}$:
1. **Additivity:** $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$
2. **Homogeneity:** $T(c\mathbf{u}) = cT(\mathbf{u})$
Let $\mathbf{u} = \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}$.
### 1.1. Additivity
$$\mathbf{u} + \mathbf{v} = \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \\ u_3 + v_3 \end{pmatrix}$$
$$\begin{aligned} T(\mathbf{u} + \mathbf{v}) &= T \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \\ u_3 + v_3 \end{pmatrix} \\ &= \begin{pmatrix} (u_1+v_1) + 2(u_2+v_2) + 3(u_3+v_3) \\ -3(u_1+v_1) + 2(u_2+v_2) + (u_3+v_3) \end{pmatrix} \\ &= \begin{pmatrix} (u_1 + 2u_2 + 3u_3) + (v_1 + 2v_2 + 3v_3) \\ (-3u_1 + 2u_2 + u_3) + (-3v_1 + 2v_2 + v_3) \end{pmatrix} \\ &= \begin{pmatrix} u_1 + 2u_2 + 3u_3 \\ -3u_1 + 2u_2 + u_3 \end{pmatrix} + \begin{pmatrix} v_1 + 2v_2 + 3v_3 \\ -3v_1 + 2v_2 + v_3 \end{pmatrix} \\ &= T(\mathbf{u}) + T(\mathbf{v}) \end{aligned}$$
**Additivity is satisfied.**
### 1.2. Homogeneity
$$c\mathbf{u} = \begin{pmatrix} cu_1 \\ cu_2 \\ cu_3 \end{pmatrix}$$
$$\begin{aligned} T(c\mathbf{u}) &= T \begin{pmatrix} cu_1 \\ cu_2 \\ cu_3 \end{pmatrix} \\ &= \begin{pmatrix} (cu_1) + 2(cu_2) + 3(cu_3) \\ -3(cu_1) + 2(cu_2) + (cu_3) \end{pmatrix} \\ &= \begin{pmatrix} c(u_1 + 2u_2 + 3u_3) \\ c(-3u_1 + 2u_2 + u_3) \end{pmatrix} \\ &= c \begin{pmatrix} u_1 + 2u_2 + 3u_3 \\ -3u_1 + 2u_2 + u_3 \end{pmatrix} \\ &= cT(\mathbf{u}) \end{aligned}$$
**Homogeneity is satisfied.**
Since both properties hold, $T$ is a **linear transformation**. (In general, any transformation defined by a system of linear equations with no constant terms is a linear transformation.)
---
## 2. Determining the Standard Matrix $A$
The standard matrix $A$ of a linear transformation $T$ is the matrix whose columns are the images of the **standard basis vectors** of the domain ($\mathbb{R}^3$):
$$A = \begin{bmatrix} T(\mathbf{e}_1) & T(\mathbf{e}_2) & T(\mathbf{e}_3) \end{bmatrix}$$
where $\mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, $\mathbf{e}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$, and $\mathbf{e}_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.
1. **Image of $\mathbf{e}_1$:**
$$T \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} (1) + 2(0) + 3(0) \\ -3(1) + 2(0) + (0) \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \end{pmatrix}$$
2. **Image of $\mathbf{e}_2$:**
$$T \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} (0) + 2(1) + 3(0) \\ -3(0) + 2(1) + (0) \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \end{pmatrix}$$
3. **Image of $\mathbf{e}_3$:**
$$T \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} (0) + 2(0) + 3(1) \\ -3(0) + 2(0) + (1) \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$$
### The Matrix $A$
Assembling the column vectors:
$$A = \begin{pmatrix} 1 & 2 & 3 \\ -3 & 2 & 1 \end{pmatrix}$$
We can verify that $T(\mathbf{x}) = A\mathbf{x}$:
$$A\mathbf{x} = \begin{pmatrix} 1 & 2 & 3 \\ -3 & 2 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1x_1 + 2x_2 + 3x_3 \\ -3x_1 + 2x_2 + 1x_3 \end{pmatrix} = T(\mathbf{x})$$
The standard matrix is:
$$A = \begin{pmatrix} 1 & 2 & 3 \\ -3 & 2 & 1 \end{pmatrix}$$
Question 1166341: Let W be the set of all sequences which converge to 0. Is W a subspace of the vector space of convergent sequences? You must justify your answer.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! Yes, **W**, the set of all sequences which converge to $0$, is a **subspace** of the vector space of convergent sequences, which we can call $V$.
The vector space $V$ is the set of all sequences of real numbers, $\{a_n\}_{n=1}^\infty$, such that $\lim_{n \to \infty} a_n = L$, where $L$ is a finite real number.
The subset $W$ is defined as:
$$W = \left\{ \{a_n\} \in V \ \bigg| \ \lim_{n \to \infty} a_n = 0 \right\}$$
To prove that $W$ is a subspace of $V$, we must verify the three subspace conditions:
---
## 1. The Zero Vector Condition (Contains the Zero Sequence)
The zero vector in $V$ is the **zero sequence**, $\mathbf{z} = \{z_n\}$, where $z_n = 0$ for all $n \ge 1$.
We check if the zero sequence satisfies the condition for $W$:
$$\lim_{n \to \infty} z_n = \lim_{n \to \infty} 0 = 0$$
Since the limit of the zero sequence is $0$, the zero sequence is in $W$.
Thus, $W$ is non-empty.
---
## 2. Closure under Vector Addition ➕
Let $\mathbf{a} = \{a_n\}$ and $\mathbf{b} = \{b_n\}$ be two arbitrary sequences in $W$. This means:
* $\lim_{n \to \infty} a_n = 0$
* $\lim_{n \to \infty} b_n = 0$
We need to check if their sum, $\mathbf{a} + \mathbf{b} = \{a_n + b_n\}$, is also in $W$. We use the **Limit Sum Law**:
$$\lim_{n \to \infty} (a_n + b_n) = \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n$$
Substitute the conditions for $\mathbf{a}$ and $\mathbf{b}$:
$$\lim_{n \to \infty} (a_n + b_n) = 0 + 0 = 0$$
Since the sum of the sequences converges to $0$, the sequence $\mathbf{a} + \mathbf{b}$ is in $W$.
Thus, $W$ is closed under sequence addition.
---
## 3. Closure under Scalar Multiplication ✖️
Let $\mathbf{a} = \{a_n\}$ be an arbitrary sequence in $W$ (so $\lim_{n \to \infty} a_n = 0$), and let $c$ be an arbitrary scalar (a real number).
We need to check if the scalar multiple, $c\mathbf{a} = \{c \cdot a_n\}$, is also in $W$. We use the **Limit Constant Multiple Law**:
$$\lim_{n \to \infty} (c \cdot a_n) = c \cdot \lim_{n \to \infty} a_n$$
Substitute the condition for $\mathbf{a}$:
$$\lim_{n \to \infty} (c \cdot a_n) = c \cdot (0) = 0$$
Since the scalar multiple of the sequence converges to $0$, the sequence $c\mathbf{a}$ is in $W$.
Thus, $W$ is closed under scalar multiplication.
***
Since $W$ satisfies all three conditions, it is a **subspace** of the vector space of convergent sequences, $V$.
Would you like to analyze a different set of sequences, such as the set of sequences that converge to $1$, to see if it is a subspace?
Question 1166343: 6 Let V = C[a, b] be the vector space of continuous functions on interval
[a, b]. Let W be the subset of functions in V such that:
(integralsign) from(a,b) f(x) dx = 0.
Is W a subspace of V ? You must justify your answer.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! Yes, **W** is a **subspace** of $\mathbf{V}$, the vector space of continuous functions on the interval $[a, b]$.
The vector space $V$ is $C[a, b]$, the set of all continuous real-valued functions on the interval $[a, b]$. The subset $W$ is defined as:
$$W = \left\{ f(x) \in V \ \bigg| \ \int_a^b f(x) \ dx = 0 \right\}$$
To prove that $W$ is a subspace of $V$, we must verify the three conditions:
---
## 1. The Zero Vector Condition (Contains the Zero Function)
The zero vector in $V$ is the **zero function**, $\mathbf{z}(x) = 0$, for all $x \in [a, b]$.
We must check if the zero function satisfies the condition for $W$:
$$\int_a^b \mathbf{z}(x) \ dx = \int_a^b 0 \ dx = 0$$
Since the integral of the zero function is $0$, the zero function is in $W$.
Thus, $W$ is non-empty.
---
## 2. Closure under Vector Addition ➕
Let $f(x)$ and $g(x)$ be two arbitrary functions in $W$. This means:
* $\int_a^b f(x) \ dx = 0$
* $\int_a^b g(x) \ dx = 0$
We need to check if their sum, $(f+g)(x) = f(x) + g(x)$, is also in $W$. We check the integral of the sum:
$$\int_a^b (f(x) + g(x)) \ dx = \int_a^b f(x) \ dx + \int_a^b g(x) \ dx \quad \text{(Property of Integrals)}$$
Substitute the conditions for $f$ and $g$:
$$\int_a^b (f(x) + g(x)) \ dx = 0 + 0 = 0$$
Since the integral of the sum is $0$, the function $(f+g)(x)$ is in $W$.
Thus, $W$ is closed under function addition.
---
## 3. Closure under Scalar Multiplication ✖️
Let $f(x)$ be an arbitrary function in $W$ (so $\int_a^b f(x) \ dx = 0$), and let $c$ be an arbitrary scalar (a real number).
We need to check if the scalar multiple, $(cf)(x) = c \cdot f(x)$, is also in $W$. We check the integral of the scalar multiple:
$$\int_a^b c \cdot f(x) \ dx = c \cdot \int_a^b f(x) \ dx \quad \text{(Property of Integrals)}$$
Substitute the condition for $f$:
$$\int_a^b c \cdot f(x) \ dx = c \cdot (0) = 0$$
Since the integral of the scalar multiple is $0$, the function $(cf)(x)$ is in $W$.
Thus, $W$ is closed under scalar multiplication.
***
Since $W$ satisfies all three conditions, it is a **subspace** of the vector space $V$.
Would you like to analyze a different set of continuous functions to see if it forms a subspace?
Question 1166345: Yes, **W**, the set of skew-symmetric $n \times n$ matrices, is a **subspace** of $\mathbf{V}$, the vector space of all $n \times n$ matrices.
To prove that $W$ is a subspace of $V$, we must verify the three conditions for a subset to be a subspace:
1. **The Zero Vector Condition:** The zero vector of $V$ must be in $W$.
2. **Closure under Vector Addition:** If $\mathbf{A}$ and $\mathbf{B}$ are in $W$, then their sum $(\mathbf{A} + \mathbf{B})$ must also be in $W$.
3. **Closure under Scalar Multiplication:** If $\mathbf{A}$ is in $W$ and $c$ is any scalar, then the scalar multiple $(c\mathbf{A})$ must also be in $W$.
Recall that a matrix $\mathbf{A}$ is **skew-symmetric** if and only if $\mathbf{A}^T = -\mathbf{A}$.
***
### 1. The Zero Vector Condition
The zero vector in $V$ is the $n \times n$ **zero matrix**, $\mathbf{Z}$, where every entry is $0$.
We check if $\mathbf{Z}$ is skew-symmetric:
$$\mathbf{Z}^T = \mathbf{Z}$$
$$-\mathbf{Z} = -\mathbf{Z} \text{ (which is still } \mathbf{Z})$$
Since $\mathbf{Z}^T = \mathbf{Z} = -\mathbf{Z}$, the condition $\mathbf{Z}^T = -\mathbf{Z}$ holds.
Thus, the zero matrix $\mathbf{Z}$ is skew-symmetric, and $\mathbf{Z} \in W$.
### 2. Closure under Vector Addition
Let $\mathbf{A}$ and $\mathbf{B}$ be two arbitrary matrices in $W$. This means:
* $\mathbf{A}^T = -\mathbf{A}$
* $\mathbf{B}^T = -\mathbf{B}$
We need to check if their sum $(\mathbf{A} + \mathbf{B})$ is also skew-symmetric. We check the transpose of the sum:
$$(\mathbf{A} + \mathbf{B})^T = \mathbf{A}^T + \mathbf{B}^T \quad (\text{Property of Transpose})$$
Substitute the skew-symmetric conditions:
$$(\mathbf{A} + \mathbf{B})^T = (-\mathbf{A}) + (-\mathbf{B})$$
$$(\mathbf{A} + \mathbf{B})^T = -(\mathbf{A} + \mathbf{B})$$
Since the transpose of $(\mathbf{A} + \mathbf{B})$ is equal to $-(\mathbf{A} + \mathbf{B})$, the sum $(\mathbf{A} + \mathbf{B})$ is skew-symmetric.
Thus, $W$ is closed under matrix addition.
### 3. Closure under Scalar Multiplication
Let $\mathbf{A}$ be an arbitrary matrix in $W$ (so $\mathbf{A}^T = -\mathbf{A}$), and let $c$ be an arbitrary scalar.
We need to check if the scalar multiple $(c\mathbf{A})$ is also skew-symmetric. We check the transpose of the multiple:
$$(c\mathbf{A})^T = c(\mathbf{A}^T) \quad (\text{Property of Transpose})$$
Substitute the skew-symmetric condition for $\mathbf{A}$:
$$(c\mathbf{A})^T = c(-\mathbf{A})$$
$$(c\mathbf{A})^T = -(c\mathbf{A})$$
Since the transpose of $(c\mathbf{A})$ is equal to $-(c\mathbf{A})$, the scalar multiple $(c\mathbf{A})$ is skew-symmetric.
Thus, $W$ is closed under scalar multiplication.
***
Since $W$ satisfies all three conditions, it is a **subspace** of the vector space $V$.
---
Would you like to know the **dimension** of the subspace $W$ of skew-symmetric $n \times n$ matrices?
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! Yes, **W**, the set of skew-symmetric $n \times n$ matrices, is a **subspace** of $\mathbf{V}$, the vector space of all $n \times n$ matrices.
To prove that $W$ is a subspace of $V$, we must verify the three conditions for a subset to be a subspace:
1. **The Zero Vector Condition:** The zero vector of $V$ must be in $W$.
2. **Closure under Vector Addition:** If $\mathbf{A}$ and $\mathbf{B}$ are in $W$, then their sum $(\mathbf{A} + \mathbf{B})$ must also be in $W$.
3. **Closure under Scalar Multiplication:** If $\mathbf{A}$ is in $W$ and $c$ is any scalar, then the scalar multiple $(c\mathbf{A})$ must also be in $W$.
Recall that a matrix $\mathbf{A}$ is **skew-symmetric** if and only if $\mathbf{A}^T = -\mathbf{A}$.
***
### 1. The Zero Vector Condition
The zero vector in $V$ is the $n \times n$ **zero matrix**, $\mathbf{Z}$, where every entry is $0$.
We check if $\mathbf{Z}$ is skew-symmetric:
$$\mathbf{Z}^T = \mathbf{Z}$$
$$-\mathbf{Z} = -\mathbf{Z} \text{ (which is still } \mathbf{Z})$$
Since $\mathbf{Z}^T = \mathbf{Z} = -\mathbf{Z}$, the condition $\mathbf{Z}^T = -\mathbf{Z}$ holds.
Thus, the zero matrix $\mathbf{Z}$ is skew-symmetric, and $\mathbf{Z} \in W$.
### 2. Closure under Vector Addition
Let $\mathbf{A}$ and $\mathbf{B}$ be two arbitrary matrices in $W$. This means:
* $\mathbf{A}^T = -\mathbf{A}$
* $\mathbf{B}^T = -\mathbf{B}$
We need to check if their sum $(\mathbf{A} + \mathbf{B})$ is also skew-symmetric. We check the transpose of the sum:
$$(\mathbf{A} + \mathbf{B})^T = \mathbf{A}^T + \mathbf{B}^T \quad (\text{Property of Transpose})$$
Substitute the skew-symmetric conditions:
$$(\mathbf{A} + \mathbf{B})^T = (-\mathbf{A}) + (-\mathbf{B})$$
$$(\mathbf{A} + \mathbf{B})^T = -(\mathbf{A} + \mathbf{B})$$
Since the transpose of $(\mathbf{A} + \mathbf{B})$ is equal to $-(\mathbf{A} + \mathbf{B})$, the sum $(\mathbf{A} + \mathbf{B})$ is skew-symmetric.
Thus, $W$ is closed under matrix addition.
### 3. Closure under Scalar Multiplication
Let $\mathbf{A}$ be an arbitrary matrix in $W$ (so $\mathbf{A}^T = -\mathbf{A}$), and let $c$ be an arbitrary scalar.
We need to check if the scalar multiple $(c\mathbf{A})$ is also skew-symmetric. We check the transpose of the multiple:
$$(c\mathbf{A})^T = c(\mathbf{A}^T) \quad (\text{Property of Transpose})$$
Substitute the skew-symmetric condition for $\mathbf{A}$:
$$(c\mathbf{A})^T = c(-\mathbf{A})$$
$$(c\mathbf{A})^T = -(c\mathbf{A})$$
Since the transpose of $(c\mathbf{A})$ is equal to $-(c\mathbf{A})$, the scalar multiple $(c\mathbf{A})$ is skew-symmetric.
Thus, $W$ is closed under scalar multiplication.
***
Since $W$ satisfies all three conditions, it is a **subspace** of the vector space $V$.
---
Would you like to know the **dimension** of the subspace $W$ of skew-symmetric $n \times n$ matrices?
Question 1166667: : Let V be the vector space of all polynomials defined on the real number line. (12 points) Suppose that T : V → V is the transformation T(p(x)) = xp′(x) where the prime denotes
derivative.
(a) Show T is a linear transformation. (b) Show T is not one-to-one.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! That's a great problem in linear algebra! Here's the demonstration that $T$ is a linear transformation but not one-to-one.
Let $V$ be the vector space of all polynomials defined on the real number line, and let the transformation $T: V \to V$ be defined by $T(p(x)) = x p'(x)$.
---
## (a) Show T is a Linear Transformation
To show that $T$ is a **linear transformation**, we must verify two properties for any two polynomials $p(x)$ and $q(x)$ in $V$, and any scalar $c \in \mathbb{R}$:
### 1. Additivity: $T(p(x) + q(x)) = T(p(x)) + T(q(x))$
Start with the left side (LHS):
$$T(p(x) + q(x))$$
By the definition of $T$:
$$T(p(x) + q(x)) = x \cdot \frac{d}{dx} (p(x) + q(x))$$
Using the linearity of the derivative ($\frac{d}{dx}(f+g) = \frac{df}{dx} + \frac{dg}{dx}$):
$$T(p(x) + q(x)) = x (p'(x) + q'(x))$$
Distributing $x$:
$$T(p(x) + q(x)) = x p'(x) + x q'(x)$$
Now, substitute back the definition of $T$:
$$T(p(x) + q(x)) = T(p(x)) + T(q(x))$$
Since LHS = RHS, **additivity holds**.
### 2. Homogeneity: $T(c \cdot p(x)) = c \cdot T(p(x))$
Start with the left side (LHS):
$$T(c \cdot p(x))$$
By the definition of $T$:
$$T(c \cdot p(x)) = x \cdot \frac{d}{dx} (c \cdot p(x))$$
Using the property of the derivative ($\frac{d}{dx}(c \cdot f) = c \cdot \frac{df}{dx}$):
$$T(c \cdot p(x)) = x (c \cdot p'(x))$$
Rearranging the terms (moving the scalar $c$ to the front):
$$T(c \cdot p(x)) = c (x \cdot p'(x))$$
Now, substitute back the definition of $T$:
$$T(c \cdot p(x)) = c \cdot T(p(x))$$
Since LHS = RHS, **homogeneity holds**.
Since $T$ satisfies both additivity and homogeneity, **$T$ is a linear transformation.**
---
## (b) Show T is Not One-to-One
A linear transformation $T$ is **not one-to-one** (or injective) if and only if its **Kernel** (or null space), $\text{Ker}(T)$, contains more than just the zero vector.
The Kernel is defined as the set of all vectors $p(x) \in V$ such that $T(p(x)) = 0$ (the zero polynomial).
$$\text{Ker}(T) = \{ p(x) \in V \mid T(p(x)) = 0 \}$$
We need to find if there is a non-zero polynomial $p(x)$ such that:
$$T(p(x)) = x p'(x) = 0$$
If $x p'(x) = 0$ for all $x \in \mathbb{R}$, this implies that the derivative $p'(x)$ must be the zero polynomial:
$$p'(x) = 0$$
If the derivative of a polynomial is zero, the polynomial itself must be a **constant polynomial**:
$$p(x) = c$$
where $c$ is any real constant.
The polynomials $p(x) = 1$, $p(x) = 5$, or $p(x) = -10$ are all non-zero vectors in $V$ (the space of all polynomials).
For example, let $p(x) = 5$.
$$p'(x) = 0$$
$$T(p(x)) = T(5) = x \cdot (0) = 0$$
Since $\text{Ker}(T)$ contains all non-zero constant polynomials (i.e., $\text{Ker}(T) = \{p(x) = c \mid c \in \mathbb{R}\}$), the **Kernel is non-trivial** (it is not just the zero vector).
Therefore, **$T$ is not one-to-one.**
Question 736146: A basketball player scored 18 times during one game. He scored a total of 30 points, which consists of 1 point free throws and 2 point baskets. How many one point free throws did he make? How many 2 point baskets did he make?
Found 4 solutions by mccravyedwin, greenestamps, Edwin McCravy, ikleyn:
Answer by mccravyedwin(421)  (Show Source):
You can put this solution on YOUR website!
I knew in advance Greenestamps would come up with the way to solve it the same
way as the 2-legged and 4-legged animal problems, by first thinking how it would
be if they were all one or the other. Then there would either be too few or too
many legs and how many of the other type animal you would need to have to
increase or decrease the number of legs to get the given number of legs. I've
also seen bicycle/tricycle and 3 and 4-legged stool problems that could be done
the same way.
Greenestamps used the "too many points" way. I'll use the "too-few points way.
(1) If all 18 shots were 1-point free throws, (a really strange basketball
game lol), the point total would be 18*1 = 18.
(2) The actual point total, 30, is 30-18=12 more than that.
(3) So 12 of those 18 shots must be increased by 1 point to increase the
total points from 18 to 30.
ANSWERS: 12 2-point shots; so 18-12 = 6 1-point shots.
Edwin
Answer by greenestamps(13327)  (Show Source):
You can put this solution on YOUR website!
You have received two responses showing different ways of solving the problem. Briefly summarized, here are those two methods.
First tutor (set up the problem using only one variable)....
x = # of 1-point shot, so (18-x) = # of 2-point shots.
The point total is 30, so x(1)+2(18-x) = 30.
Solve that equation to find the answer.
Second tutor (set up the problem using two variables and two equations and solve using substitution)....
x = # of 1-point shots; y = # of 2-points shots.
The total number of shots is 18, so x+y = 18.
The point total is 30, so x(1)+2(y) = 30.
Solve the first equation for x: x = 18-y.
Substitute "18-y" for "x" in the second equation, giving x(1)+2(18-x) = 30.
At that point the equation to solve is the same as in the solution from the first tutor, so solving the problem from there is the same.
A different method for solving the problem -- which a student should also know -- is using two variables and two equations and solving the pair of equations using elimination.
x+y = 18; x+2y = 30
Subtract the first equation from the second, eliminating x: y = 12
Substitute y=12 in either equation to find x = 6
ANSWER: x = 6 1-point shots and y = 12 2-point shots
Finally, while a student should understand and be able to use either substitution or elimination to solve the problem, it should be noted that solving the problem informally, using logical reasoning and simple arithmetic instead of formal algebra, is excellent mental exercise.
For this problem, the calculations are exactly those used in the above solution using elimination. Here is how the thinking can go:
(1) If all 18 shots were 2-point shots, the point total would be 18*2 = 36.
(2) The actual point total, 30, is 6 points less than that.
(3) Since each 1-point shot is worth 1 point less than each 2-point shot, the number of 1-point shots must have been 6.
ANSWERS: 6 1-point shot; so 18-6 = 12 2-point shots
Answer by Edwin McCravy(20077)  (Show Source):
Answer by ikleyn(53748) (Show Source):
You can put this solution on YOUR website! .
A basketball player scored 18 times during one game. He scored a total of 30 points, which consists
of 1 point free throws and 2 point baskets.
How many one point free throws did he make? How many 2 point baskets did he make?
~~~~~~~~~~~~~~~~~~~~~~~~~~~
There are different methods to solve this problem.
I will show here one of these methods, which is a basic.
It means that every student solving such problem/problems MUST know it.
Probably, other tutors will come and add their versions - but this one is that every student must know.
Let x be the number of 1-point throws.
Then the number of 2-point throws is (18-x).
1-point throws contribute 1*x = x points to the total counting.
2-point throws contribute 2*(18-x) points to the total counting.
Now you are prepared to write an equation for the total points
x + 2*(18-x) = 30 points total.
Simplify it and find 'x'
x + 36 - 2x = 30,
x - 2x = 30 - 36,
-x = -6,
x = 6.
Thus the number of the 1-point throws is 6.
Hence, the number of the 2-point throws is 18-6 = 12.
ANSWER. 6 one-point throws and 12 two-point throws.
Solved, and now you know the method.
Question 629298: Find the quadratic polynomial whose graph goes through the points (−1,5), (0,5), and (2,29).
f(x)= _x2+_ x+_ <<< that form
I got to the step 4a+2b+0=29
that was like my 6/7th step. I am getting stuck from there. Can i get any help please?
Found 3 solutions by Edwin McCravy, n2, ikleyn:
Answer by Edwin McCravy(20077) (Show Source):
You can put this solution on YOUR website!
On your TI-84
press STAT
press ENTER
Under L1 enter the x-coordinates and under L2 enter the y-coordinates.
That is, make the screen look like this under L1 and L2
L1 | L2
-1 | 5
0 | 5
2 | 29
Press STAT then right arrow once to highlight CALC
Press 5 to choose this--> 5: QuadReg
Press ENTER five times
You read
y=ax2+bx+c
a=4
b=4
c=5
The answer is f(x) = 4x2 + 4x + 5
Edwin
Answer by n2(79) (Show Source):
You can put this solution on YOUR website! .
Find the quadratic polynomial whose graph goes through the points (−1,5), (0,5), and (2,29).
f(x)= _x2+_ x+_ <<< that form
I got to the step 4a+2b+0=29
that was like my 6/7th step. I am getting stuck from there. Can i get any help please?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Notice that the y-coordinate is the same, '5', for both points (-1,5) and (0,5).
It means that if you consider another quadratic function, g(x) = f(x) - 5, instead of f(x),
this new quadratic function will have x-intercepts at x= -1 and x= 0.
In turn, it means that g(x) = ax*(x+1) with some real coefficient 'a',
so f(x) = ax(x+1) + 5.
Great (!) So, to determine f(x), we only need to find single unknown coefficient 'a'.
For it, we use the condition that the parabola passes through the point (2,29).
It gives us this equation
f(2) = 29, or a*2*(2+1) + 5 = 29, or 2*3*a = 29 - 5, or 6a = 24.
Thus we find a = 24/6 = 4.
So, our quadratic function is f(x) = 4x(x+1) + 5, or f(x) = 4x^2 + 4x + 5.
At this point, the problem is solved completely, and we obtained the solution
with minimal computations and maximal fun.
This trick works for many other similar problem, where
two points of the three given points have the same y-coordinate.
So, this method is worth to learn ( ! )
Answer by ikleyn(53748) (Show Source):
You can put this solution on YOUR website! .
Find the quadratic polynomial whose graph goes through the points (−1,5), (0,5), and (2,29).
f(x)= _x2+_ x+_ <<< that form
I got to the step 4a+2b+0=29
that was like my 6/7th step. I am getting stuck from there. Can i get any help please?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This task, as it is solved in the post by @Theo, seems routine and does not create any inspiration.
But it depends on how to look at it.
I will show you another solution from different perspective that will make it seem beautiful like a sparkling diamond.
Notice that the y-coordinate is the same, '5', for both points (-1,5) and (0,5).
It means that if you consider another quadratic function, g(x) = f(x) - 5, instead of f(x),
this new quadratic function will have x-intercepts at x= -1 and x= 0.
In turn, it means that g(x) = ax*(x+1) with some real coefficient 'a',
so f(x) = ax(x+1) + 5.
Great (!) So, to determine f(x), we only need to find single unknown coefficient 'a'.
For it, we use the condition that the parabola passes through the point (2,29).
It gives us this equation
f(2) = 29, or a*2*(2+1) + 5 = 29, or 2*3*a = 29 - 5, or 6a = 24.
Thus we find a = 24/6 = 4.
So, our quadratic function is f(x) = 4x(x+1) + 5, or f(x) = 4x^2 + 4x + 5.
At this point, the problem is solved completely, and we obtained the solution
with minimal computations and maximal fun.
This trick works for many other similar problem, where
two points of the three given points have the same y-coordinate.
So, this method is worth to learn ( ! )
Question 1167408: Giovanni invested N$90 000 in three different account at the beginning of 2018 according to the
following table.
2018 yield
Saving 6%
Unit trust 7%
32 days 5%
If he invested the same amount in the unit trust as well as in the 32 days accounts and his 2018 yield
for the year from the saving and the 32 days amounted to N$400, how much did he invest in each
account? Use Gaussian elimination.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! This problem can be solved by setting up a system of linear equations and using **Gaussian elimination** to find the investment amounts.
Let:
* $S$ be the amount invested in the **Saving** account.
* $U$ be the amount invested in the **Unit Trust** account.
* $T$ be the amount invested in the **32 Days** account.
## 1. Set Up the System of Equations
We derive three equations from the problem description:
**Equation 1: Total Investment**
The total amount invested is N\$90,000.
$$S + U + T = 90000$$
**Equation 2: Equal Investment**
The amount invested in the Unit Trust ($U$) is the same as the amount invested in the 32 Days account ($T$).
$$U = T \quad \implies \quad 0S + U - T = 0$$
**Equation 3: Specific Yield**
The 2018 yield (interest earned) from the Saving (6%) and the 32 Days (5%) accounts amounted to N\$400.
$$0.06S + 0.05T = 400$$
To work with integers, multiply the equation by 100:
$$6S + 5T = 40000$$
Since $U$ is not in this equation, we include a $0U$ term:
$$6S + 0U + 5T = 40000$$
The system of linear equations is:
1. $S + U + T = 90000$
2. $0S + U - T = 0$
3. $6S + 0U + 5T = 40000$
## 2. Gaussian Elimination
We form the augmented matrix and use row operations to achieve row echelon form.
$$\begin{pmatrix} 1 & 1 & 1 & | & 90000 \\ 0 & 1 & -1 & | & 0 \\ 6 & 0 & 5 & | & 40000 \end{pmatrix}$$
**Step 1: Eliminate $S$ from Row 3**
$$R_3 \leftarrow R_3 - 6R_1$$
$$\begin{aligned} R_3: & \quad 6 & 0 & 5 & | & 40000 \\ -6R_1: & \quad -6 & -6 & -6 & | & -540000 \\ \text{New } R_3: & \quad 0 & -6 & -1 & | & -500000 \end{aligned}$$
The new matrix is:
$$\begin{pmatrix} 1 & 1 & 1 & | & 90000 \\ 0 & 1 & -1 & | & 0 \\ 0 & -6 & -1 & | & -500000 \end{pmatrix}$$
**Step 2: Eliminate $U$ from Row 3**
$$R_3 \leftarrow R_3 + 6R_2$$
$$\begin{aligned} R_3: & \quad 0 & -6 & -1 & | & -500000 \\ 6R_2: & \quad 0 & 6 & -6 & | & 0 \\ \text{New } R_3: & \quad 0 & 0 & -7 & | & -500000 \end{aligned}$$
The matrix is now in row echelon form:
$$\begin{pmatrix} 1 & 1 & 1 & | & 90000 \\ 0 & 1 & -1 & | & 0 \\ 0 & 0 & -7 & | & -500000 \end{pmatrix}$$
## 3. Back-Substitution
The row echelon form corresponds to the following simplified system:
1. $S + U + T = 90000$
2. $U - T = 0$
3. $-7T = -500000$
**From Equation 3 (32 Days Account, $T$):**
$$-7T = -500000$$
$$T = \frac{500000}{7} \approx 71428.5714$$
**From Equation 2 (Unit Trust Account, $U$):**
Since $U - T = 0$, we have $U = T$.
$$U = \frac{500000}{7}$$
**From Equation 1 (Saving Account, $S$):**
$$S + U + T = 90000$$
$$S + \frac{500000}{7} + \frac{500000}{7} = 90000$$
$$S + \frac{1000000}{7} = 90000$$
$$S = 90000 - \frac{1000000}{7}$$
To simplify, use a common denominator:
$$S = \frac{90000 \times 7 - 1000000}{7}$$
$$S = \frac{630000 - 1000000}{7}$$
$$S = \frac{-370000}{7}$$
The result for $S$ is negative, which is impossible for an investment amount. This indicates that **no solution exists** for the given problem parameters.
***
## Conclusion
Based on the required constraints and Gaussian elimination, the resulting investment amount for the Saving account ($S$) is negative: $S = -\frac{370,000}{7}$.
**This means that there is no amount Giovanni could invest in the three accounts that satisfies all the conditions simultaneously.**
### Breakdown of the Contradiction:
The contradiction arises because the required yield is too small for the large investment.
1. **Total Investment:** N\$90,000.
2. **Equal Investments:** $U = T$. This means $S + 2T = 90000$.
3. **Maximum Yield on S & T:** The maximum possible interest from the $S$ and $T$ accounts would occur if $T$ was maximized (close to $45,000$) and $S$ was maximized (close to $90,000$), but constrained by $S + 2T = 90,000$. The yield rate for $S$ (6%) is higher than for $T$ (5%).
4. **Yield Required:** $0.06S + 0.05T = 400$.
The maximum possible value of the yield from $S$ and $T$ given $S+T$ must be less than $90,000$ is significantly higher than N\$400, leading to a mathematically impossible distribution.
Question 1202927: CalJuice Company has decided to introduce three fruit juices made from blending two or more concentrates. These juices will be packaged in 2-qt (64-oz) cartons. One carton of pineapple-orange juice requires 8 oz each of pineapple and orange juice concentrates. One carton of orange-banana juice requires 12 oz of orange juice concentrate and 4 oz of banana pulp concentrate. Finally, one carton of pineapple-orange-banana juice requires 4 oz of pineapple juice concentrate, 8 oz of orange juice concentrate, and 4 oz of banana pulp concentrate. The company has decided to allot 16,000 oz of pineapple juice concentrate, 24,000 oz of orange juice concentrate, and 5000 oz of banana pulp concentrate for the initial production run. The company also stipulated that the production of pineapple-orange-banana juice should not exceed 800 cartons. Its profit on one carton of pineapple-orange juice is $1.00, its profit on one carton of orange-banana juice is $0.80, and its profit on one carton of pineapple-orange-banana juice is $0.90. To realize a maximum profit, how many cartons of each blend should the company produce?
pineapple-orange juice ____ cartons
orange-banana juice ___ cartons
pineapple-orange-banana juice 800 cartons
What is the largest profit it can realize?
$ ______
Are there any concentrates left over? (If so, enter the amount remaining. If not, enter 0.)
pineapple concentrate
0 oz
orange concentrate
0 oz
banana pulp concentrate
____ oz
Answer by ikleyn(53748) (Show Source):
You can put this solution on YOUR website! . 1202927
CalJuice Company has decided to introduce three fruit juices made from blending two or more concentrates.
These juices will be packaged in 2-qt (64-oz) cartons.
One carton of pineapple-orange juice requires 8 oz each of pineapple and orange juice concentrates.
One carton of orange-banana juice requires 12 oz of orange juice concentrate and 4 oz of banana pulp concentrate.
Finally, one carton of pineapple-orange-banana juice requires 4 oz of pineapple juice concentrate,
8 oz of orange juice concentrate, and 4 oz of banana pulp concentrate.
The company has decided to allot 16,000 oz of pineapple juice concentrate, 24,000 oz of orange juice concentrate,
and 5000 oz of banana pulp concentrate for the initial production run.
The company also stipulated that the production of pineapple-orange-banana juice should not exceed 800 cartons.
Its profit on one carton of pineapple-orange juice is $1.00, its profit on one carton of orange-banana juice is $0.80,
and its profit on one carton of pineapple-orange-banana juice is $0.90.
To realize a maximum profit, how many cartons of each blend should the company produce?
pineapple-orange juice ____ cartons
orange-banana juice ___ cartons
pineapple-orange-banana juice 800 cartons
What is the largest profit it can realize?
$ ______
Are there any concentrates left over? (If so, enter the amount remaining. If not, enter 0.)
pineapple concentrate
0 oz
orange concentrate
0 oz
banana pulp concentrate
____ oz
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The solution to this problem in the post by @Theo is incorrect.
Indeed, his solution states
the maximum profit is made when 800 cartons of type A and 1250 cartons of type C mixture are made and sold.
no cartons of type B mixture were made and sold.
Then the maximum profit is 800*1 + 1250*0.9 = 1925 dollars.
I will disprove @Theo' solution right now.
For it, I will present another possible solution, which provides greater profit
and still does satisfy the restrictions.
This solution is 2000 cartoons of pineapple-orange juice and 0 cartons for two other juice.
It requires 2000*8 = 16000 oz of pineapple concentrate (which is under restriction)
and 2000*8 = 16000 oz of orange concentrate (which is also under restriction).
But the profit in this case is 1*2000 = 2000 dollars, which is GREATER than 1925 dollars in the @Theo' solution.
Thus I disproved the @Theo solution - - - it does not provide the optimum strategy.
Now, as I proved that the solution in the post by @Theo is incorrect,
I will present my analysis below.
Let x be the number of cartons of pineapple-orange PO-juice;
y be the number of cartons of orange-banana OB-juice;
z be the number of cartons of pineapple-orange-banana POB-juice.
One carton PO-juice = (8 oz P, 8 oz O ).
One carton OB-juice = ( 12 oz O, 4 oz B).
One carton POB-juice = (4 oz P, 8 oz O, 4 oz B).
We want to maximize the profit function P = x + 0.8y + 0.9z dollars.
The restrictions are
8x + 4z <= 16000 (pineapple juice concentrate restriction, in oz);
8x + 12y + 8z <= 24000 (orange juice concentrate restriction, in oz);
4y + 4z <= 5000 (banana pulp concentrate, in oz).
z <= 800,
x >= 0, y >= 0, z >= 0.
I will solve the problem using the simplex method and the online solver at this web-site
https://www.zweigmedia.com/RealWorld/simplex.html
(the same solver, as @Theo used).
It requires the input information be presented in its special format, which you may find
in the description to the solver.
So, the formatted input information for the solver is this
maximize p = x + 0.8y + 0.9z subject to
8x + 4x <= 16000
8x + 12y + 8z <= 24000
4y + 4z <= 5000
z <= 800,
x >= 0,
y >= 0,
z >= 0.
I copy-paste it into the solver.
After one click, the solver generates this OPTIMAL SOLUTION
p = 2640; x = 1600, y = 400, z = 800.
Having it, you can answer all remaining questions.
Question 1160968: Let L be the line spanned by [-1, 4, 9, 0] in R^4
Find a basis of the orthogonal complement L⊥ of L.
Answer by ikleyn(53748) (Show Source):
You can put this solution on YOUR website! .
Let L be the line spanned by [-1, 4, 9, 0] in R^4.
Find a basis of the orthogonal complement L⊥ of L.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The orthogonal complement to the given line L is the set of all vectors (x1,x2,x3,x4) in R^4
such that
-x1 + 4x2 + 9x3 = 0. (1)
Looking at this equation, we can guess mentally its 3 linearly independent vector solutions
v1 = (4,1,0,0),
v2 = (9,0,1,0),
v3 = (0,-9,4,1).
It is clear and easy to check that these vectors satisfy equation (1), so they all belong
to the orthogonal complement L⊥ of L.
It is also easy to check that vectors v1, v2 and v3 are linearly independent,
so they form a basis in the orthogonal complement.
At this point, the problem is solved completely.
The goal of this problem is to help to a student to develop an intuition,
necessary for solving elementary tasks in linear algebra.
To have this intuition is the same as to keep the multiplication table
in your mind: without it, there is no way in the subject.
Question 1167649: Suppose u, v ∈ R3. Determine if the function
<> = 2u1v1 + u2v2 + 4u3v3
is an inner product on R3. If it is not an inner product, list the axioms which do not hold.
Found 2 solutions by Resolver123, ikleyn:
Answer by Resolver123(6) (Show Source):
You can put this solution on YOUR website! We are given a function defined on  as:
(u, v) = 
We want to determine whether this function defines an inner product on . Recall the inner product axioms.
A function (.,.) :  ->  is an inner product if it satisfies the following axioms for all u, v, w in and all scalars  in .
1. Symmetry: (u, v) = (v, u)
2. Linearity in the first argument (a.k.a. "bilinearity" for real vector spaces): (c*u + w, v) = c*(u, v) + (w, v)
3. Positive-definiteness: (u, u) ≥ 0 and (u,u)=0 if and only if u = 0.
1.) Check Symmetry
Compute both sides:
(u, v) =  = (v,u).
Therefore, symmetry holds.
2.) Check Linearity in First Argument
Let u, w, v be in and c be in . Let’s compute:
(c*u + w, v ) =  = c*(u,v) + (w,v).
Therefore, linearity in the first argument holds.
3.) Check for Positive-Definiteness
It must be shown that:
(u, u) =  ≥ 0 and = 0 iff u = 0.
Note that each term is squared and multiplied by a positive scalar, so the whole expression is non-negative.
ALso, if we let (u,u) =  , then this statement is true if and only if  , i.e., u = (0,0,0).
Therefore, positive-definiteness holds.
Since all three axioms (symmetry, linearity, positive-definiteness) are satisfied, (u,v) = is an inner product on .
Answer by ikleyn(53748) (Show Source):
You can put this solution on YOUR website! .
Suppose u, v ∈ R3. Determine if the function
⟨u, v⟩ = 2u1v1 + u2v2 + 4u3v3
is an inner product on R3. If it is not an inner product, list the axioms which do not hold.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
In , consider linear transformation
A
(  ,  ,  ) ----> Au = (  , ,  ).
It is clear that
 +  +  = (Au,Av).
So, the given function is an inner product.
QED.
Solved.
Question 1167650: Show that the inner product ⟨u, v⟩ = 5u1v1 − u1v2 − u2v1 + 10u2v2 is the inner product on R2 generated by the matrix A=
| 2 1 |
| -1 3 |
Answer by Resolver123(6) (Show Source):
Question 1167394: How does the rank of the following matrix depend on the value of t?
(1,1,t)
(1,t,1)
(t,1,1)
Answer by Resolver123(6) (Show Source):
You can put this solution on YOUR website! We are given the following 3x3 matrix:
Compute the determinant det(A):
 .
Let  , or  .
Hence, det(A) = 0 if and only if t = 1 or t = -2.
Consider 3 cases:
Case 1: t ≠ 1 and t ≠ -2.
Then det(A) ≠ 0, and so the matrix is of full rank, that is, rank(A) = 3.
Case 2: t = 1
Then the matrix is:
All rows being identical means that there is only 1 linearly independent row. Hence, rank = 1.
Case 3: t = -2
Then we get the matrix:
Using the row operations  and  , we get the row equivalent matrix
Using the row operation  , we finally get
This gives 2 linearly independent rows, and therefore, rank = 2.
Thus, the rank of the matrix depends on  as follows:
* Rank = 3 if t ≠ 1 and t ≠ -2.
* Rank = 2 if t = -2, and
* Rank = 1 if t = 1.
Question 1160838: Find an orthonormal basis of the plane x−4y−z=0.
Answer by ikleyn(53748) (Show Source):
You can put this solution on YOUR website! .
Find an orthonormal basis of the plane x−4y−z=0.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider two vectors V1 = (1,0,1) and V2 = (2,1,-2).
You can manually check that both vectors V1 and V2 satisfy the given equation -
- so, they belong to the plane described by this equation.
Next, it is clear that vectors V1 and V2 are linearly independent - hence, they form
a basis in the plane described by the given equation.
The fact is that vectors V1 and V2 are orthogonal.
You may check it on your own.
To get vectors V1 and V2 orthonormal, we should divide each vector by its length.
Doing so, we get orthonormal vectors (  , ,) and ( , , ).
Thus the problem is solved completely.
Question 1165856: Exercise 9 Determine whether the transformation T : R3 → R2 is a linear transformation.
Matrix2x1 T(x)=( x1, x1 + x2 +x3)
Answer by ikleyn(53748) (Show Source):
You can put this solution on YOUR website! .
Different parts of this assignment are not consistent each with the other.
The post is soup of words and symbols with unclear meaning.
Question 1160376: pi =-2Q^(2)+9Q-4 and TC= 4-Q
What is the value of Fixed Cost?
What is the value of variable cost?
Answer by ikleyn(53748) (Show Source):
Question 1160375: pi =-2Q^(2)+9Q-4 and TC= 4-Q
What is the value of the maximum profit?
Answer by ikleyn(53748) (Show Source):
Question 1160374: \pi =-2Q^(2)+9Q-4 and TC= 4-Q
For what values of Q does the firm break even
Answer by ikleyn(53748) (Show Source):
Question 1165009: The Cream and Custard Bakery makes both coffee cakes and Danish in large pans. The main
ingredients are flour and sugar. There are 25 pounds of flour and 16 pounds of sugar available
and the maximum demand for coffee cakes is 8. Five pounds of flour and 2 pounds of sugar
are required to make one pan of coffee cake, and 5 pounds of flour and 4 pounds of sugar are
required to make one pan of Danish. One pan of coffee cake has a profit of PhP 1, and one pan
of Danish has a profit of PhP 5. Determine the number of pans of cake and Danish that the
bakery must produce each day so that profit will be maximized.
Answer by ikleyn(53748) (Show Source):
You can put this solution on YOUR website! .
The Cream and Custard Bakery makes both coffee cakes and Danish in large pans. The main
ingredients are flour and sugar. There are 25 pounds of flour and 16 pounds of sugar available
and the maximum demand for coffee cakes is 8. Five pounds of flour and 2 pounds of sugar
are required to make one pan of coffee cake, and 5 pounds of flour and 4 pounds of sugar are
required to make one pan of Danish. One pan of coffee cake has a profit of PhP 1, and one pan
of Danish has a profit of PhP 5. Determine the number of pans of cake and Danish that the
bakery must produce each day so that profit will be maximized.
~~~~~~~~~~~~~~~~~~~~~~~~~~~
This problem is simple: it is accessible to a third-grader.
Therefore, I will not write a standard solution using Linear programming method,
but will show a simple arithmetic solution.
From the problem, it is clear that the maximum number of cakes is 5.
Indeed, the bakery can make 5 coffee cakes, using 5*5 = 25 pounds of flour and 5*2 = 10 pounds of sugar.
It can not make more than 5 cakes, either coffee cakes or Danish.
So, our task is to make a schedule (a Table) showing possible cakes that can be prepared, and possible profit.
Obviously, we should spend as much of the ingredients as allowed by restrictions.
Also, from the given data, it is obvious, that for any given number of total cakes, the bakery should
make as many Danish pans as possible, then adding coffee cakes until fitting the restrictions.
It is because the first limiting restriction is the amount of floor, and regarding the floor,
the coffee cakes and the Danish pans are in equal position, while Danish provides greater profit.
the number coffee Danish Floor Sugar Profit
of cakes cakes pans (Php)
(total)
--------------------------------------------------------------------------
5 5 0 5*5 = 25 5*2 = 10 5*1 = 5
4 0 4 0*5+4*5 = 20 0*2+4*4 = 16 4*5 = 20 (*)
4 1 3 1*5+3*5 = 20 1+3*5 = 16
4 2 2 2*5+2*5 = 20 2*2+2*4 = 12 2*2+2*4 = 12
There is no sense to continue the table further.
Looking into the table, we mark the optimum solution by (*).
It provides the maximum profit of Php 20 .
ANSWER. 4 Danish pans and 0 (zero) coffee cakes satisfy the restrictions and provide the maximum profit of Php 20.
Solved.
Question 1168301: Fruit & Veg shop sells water in 5-litre bottles.
2.1 On Wednesday Fruit & Veg shop received $2 530 from selling 5-litre bottles of water at $11.50
per bottle. How many litres of water were sold on that day?
2.2 On Thursday, the shop received x $ by selling 5-litre bottles of water at 50 . 11 $ each. In terms
of ,x how many litres of water were sold on that day?
2.3 On Friday the shop received $(x 20) by selling 5-litre bottles of water at $9 each. In terms
of x, how many litres of water were sold on that day?
2.4 If the number of bottles sold on Thursday equal to the number of bottles sold on Friday, how many
bottles of water were sold in each of these two days?
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! Let's break down each part of the problem:
**2.1 On Wednesday:**
* **Price per bottle:** $11.50
* **Total revenue:** $2530
To find the number of bottles sold, we divide the total revenue by the price per bottle:
Number of bottles = Total revenue / Price per bottle
Number of bottles = $2530 / $11.50 = 220 bottles
Since each bottle contains 5 litres of water, the total litres of water sold on Wednesday is:
Total litres = Number of bottles × Litres per bottle
Total litres = 220 bottles × 5 litres/bottle = 1100 litres
**Answer to 2.1: 1100 litres of water were sold on Wednesday.**
**2.2 On Thursday:**
* **Price per bottle:** $11
* **Total revenue:** $x
To find the number of bottles sold, we divide the total revenue by the price per bottle:
Number of bottles = Total revenue / Price per bottle
Number of bottles = $x / $11 = \frac{x}{11}$ bottles
Since each bottle contains 5 litres of water, the total litres of water sold on Thursday is:
Total litres = Number of bottles × Litres per bottle
Total litres = $\frac{x}{11}$ bottles × 5 litres/bottle = $\frac{5x}{11}$ litres
**Answer to 2.2: $\frac{5x}{11}$ litres of water were sold on Thursday.**
**2.3 On Friday:**
* **Price per bottle:** $9
* **Total revenue:** $(x - 20)$
To find the number of bottles sold, we divide the total revenue by the price per bottle:
Number of bottles = Total revenue / Price per bottle
Number of bottles = $\frac{(x - 20)}{9}$ bottles
Since each bottle contains 5 litres of water, the total litres of water sold on Friday is:
Total litres = Number of bottles × Litres per bottle
Total litres = $\frac{(x - 20)}{9}$ bottles × 5 litres/bottle = $\frac{5(x - 20)}{9}$ litres
**Answer to 2.3: $\frac{5(x - 20)}{9}$ litres of water were sold on Friday.**
**2.4 If the number of bottles sold on Thursday equals the number of bottles sold on Friday:**
From our answers to 2.2 and 2.3, the number of bottles sold on Thursday is $\frac{x}{11}$ and the number of bottles sold on Friday is $\frac{x - 20}{9}$. We are given that these are equal:
$\frac{x}{11} = \frac{x - 20}{9}$
To solve for x, we can cross-multiply:
$9x = 11(x - 20)$
$9x = 11x - 220$
Now, isolate x:
$220 = 11x - 9x$
$220 = 2x$
$x = \frac{220}{2}$
$x = 110$
Now that we have the value of x, we can find the number of bottles sold on Thursday and Friday:
Number of bottles on Thursday = $\frac{x}{11} = \frac{110}{11} = 10$ bottles
Number of bottles on Friday = $\frac{x - 20}{9} = \frac{110 - 20}{9} = \frac{90}{9} = 10$ bottles
**Answer to 2.4: 10 bottles of water were sold on each of these two days.**
Question 1168385: A)Consider the vector space P2. Define the inner product,
⟨p, q⟩ = ∫(from0 to1) p(x)q(x) dx
Use the Gram-Schmidt process to transform the standard basis S = {1, x, x^2} into an orthonormal basis.
A.1)Express r(x) = 1 + x + 4x^2 as a linear combination of the vectors in the orthonormal basis for P2 found in the previous exercise.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! Let's solve this problem step by step.
**A) Gram-Schmidt Process**
We are given the standard basis S = {1, x, x²} and the inner product ⟨p, q⟩ = ∫(from 0 to 1) p(x)q(x) dx.
1. **v₁ = u₁ = 1**
2. **v₂ = u₂ - proj_v₁(u₂) = x - ⟨x, v₁⟩ / ⟨v₁, v₁⟩ * v₁**
* ⟨x, v₁⟩ = ∫(from 0 to 1) x * 1 dx = [x²/2] from 0 to 1 = 1/2
* ⟨v₁, v₁⟩ = ∫(from 0 to 1) 1 * 1 dx = [x] from 0 to 1 = 1
* v₂ = x - (1/2) / 1 * 1 = x - 1/2
3. **v₃ = u₃ - proj_v₁(u₃) - proj_v₂(u₃) = x² - ⟨x², v₁⟩ / ⟨v₁, v₁⟩ * v₁ - ⟨x², v₂⟩ / ⟨v₂, v₂⟩ * v₂**
* ⟨x², v₁⟩ = ∫(from 0 to 1) x² * 1 dx = [x³/3] from 0 to 1 = 1/3
* ⟨x², v₂⟩ = ∫(from 0 to 1) x² * (x - 1/2) dx = ∫(from 0 to 1) (x³ - x²/2) dx = [x⁴/4 - x³/6] from 0 to 1 = 1/4 - 1/6 = 1/12
* ⟨v₂, v₂⟩ = ∫(from 0 to 1) (x - 1/2)² dx = ∫(from 0 to 1) (x² - x + 1/4) dx = [x³/3 - x²/2 + x/4] from 0 to 1 = 1/3 - 1/2 + 1/4 = 1/12
* v₃ = x² - (1/3) / 1 * 1 - (1/12) / (1/12) * (x - 1/2) = x² - 1/3 - (x - 1/2) = x² - x + 1/6
Now, normalize the vectors:
1. **e₁ = v₁ / ||v₁||**
* ||v₁|| = √⟨v₁, v₁⟩ = √1 = 1
* e₁ = 1 / 1 = 1
2. **e₂ = v₂ / ||v₂||**
* ||v₂|| = √⟨v₂, v₂⟩ = √(1/12) = 1 / √12 = 1 / (2√3)
* e₂ = (x - 1/2) / (1 / (2√3)) = 2√3(x - 1/2) = 2√3x - √3
3. **e₃ = v₃ / ||v₃||**
* ||v₃|| = √⟨v₃, v₃⟩ = √∫(from 0 to 1) (x² - x + 1/6)² dx = √(1/180) = 1 / √(180) = 1 / (6√5)
* e₃ = (x² - x + 1/6) / (1 / (6√5)) = 6√5(x² - x + 1/6) = 6√5x² - 6√5x + √5
Therefore, the orthonormal basis is {1, 2√3x - √3, 6√5x² - 6√5x + √5}.
**A.1) Express r(x) = 1 + x + 4x² as a linear combination of the vectors in the orthonormal basis.**
Let r(x) = c₁e₁ + c₂e₂ + c₃e₃.
* c₁ = ⟨r(x), e₁⟩ = ∫(from 0 to 1) (1 + x + 4x²) * 1 dx = [x + x²/2 + 4x³/3] from 0 to 1 = 1 + 1/2 + 4/3 = 17/6
* c₂ = ⟨r(x), e₂⟩ = ∫(from 0 to 1) (1 + x + 4x²) * (2√3x - √3) dx = ∫(from 0 to 1) (2√3x - √3 + 2√3x² - √3x + 8√3x³ - 4√3x²) dx = ∫(from 0 to 1) (8√3x³ - 2√3x² + √3x - √3) dx = [2√3x⁴ - 2√3x³/3 + √3x²/2 - √3x] from 0 to 1 = 2√3 - 2√3/3 + √3/2 - √3 = (12√3 - 4√3 + 3√3 - 6√3) / 6 = 5√3 / 6
* c₃ = ⟨r(x), e₃⟩ = ∫(from 0 to 1) (1 + x + 4x²) * (6√5x² - 6√5x + √5) dx = ∫(from 0 to 1) (6√5x² - 6√5x + √5 + 6√5x³ - 6√5x² + √5x + 24√5x⁴ - 24√5x³ + 4√5x²) dx = ∫(from 0 to 1) (24√5x⁴ - 18√5x³ + 4√5x² - 5√5x + √5) dx = [24√5x⁵/5 - 18√5x⁴/4 + 4√5x³/3 - 5√5x²/2 + √5x] from 0 to 1 = 24√5/5 - 9√5/2 + 4√5/3 - 5√5/2 + √5 = (144√5 - 135√5 + 40√5 - 75√5 + 30√5) / 30 = 4√5 / 30 = 2√5 / 15
Therefore, r(x) = (17/6) * 1 + (5√3 / 6) * (2√3x - √3) + (2√5 / 15) * (6√5x² - 6√5x + √5).
Question 1168921: Let vector p=2−x+x^2 and let B be the basis for P2:
{1+x,1+x^2,x+x^2}.
Find [p]B.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! Solution:
Let the basis $B = \{v_1, v_2, v_3\}$, where $v_1 = 1+x$, $v_2 = 1+x^2$, and $v_3 = x+x^2$.
We are given the vector $p = 2 - x + x^2$.
We need to find the coordinates of $p$ with respect to the basis $B$, which means finding scalars $c_1, c_2, c_3$ such that:
$p = c_1 v_1 + c_2 v_2 + c_3 v_3$
$2 - x + x^2 = c_1(1+x) + c_2(1+x^2) + c_3(x+x^2)$
$2 - x + x^2 = c_1 + c_1 x + c_2 + c_2 x^2 + c_3 x + c_3 x^2$
Group the terms by powers of $x$:
$2 - x + x^2 = (c_1 + c_2) + (c_1 + c_3)x + (c_2 + c_3)x^2$
For this equation to hold for all $x$, the coefficients of the corresponding powers of $x$ on both sides must be equal. This gives us a system of linear equations:
1. Coefficient of $x^0$ (constant term): $c_1 + c_2 = 2$
2. Coefficient of $x^1$ (term with $x$): $c_1 + c_3 = -1$
3. Coefficient of $x^2$ (term with $x^2$): $c_2 + c_3 = 1$
We need to solve this system for $c_1, c_2, c_3$.
From equation (1), $c_2 = 2 - c_1$.
From equation (2), $c_3 = -1 - c_1$.
Substitute these expressions for $c_2$ and $c_3$ into equation (3):
$(2 - c_1) + (-1 - c_1) = 1$
$1 - 2c_1 = 1$
$-2c_1 = 0$
$c_1 = 0$
Now, substitute the value of $c_1$ back into the expressions for $c_2$ and $c_3$:
$c_2 = 2 - c_1 = 2 - 0 = 2$
$c_3 = -1 - c_1 = -1 - 0 = -1$
So, the coordinates of $p$ with respect to the basis $B$ are $c_1 = 0$, $c_2 = 2$, and $c_3 = -1$.
The coordinate vector $[p]_B$ is given by:
$[p]_B = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix}$
To verify, we can substitute these values back into the linear combination:
$0(1+x) + 2(1+x^2) + (-1)(x+x^2) = 0 + 2 + 2x^2 - x - x^2 = 2 - x + x^2 = p$.
Final Answer: The final answer is $\boxed{\begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix}}$
Question 1170702: Let M=[{0,-2},{4,6}]. Find formulas for the entries of Mn, where n is a positive integer.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! Let's find the formula for the entries of $M^n$, where $M = \begin{bmatrix} 0 & -2 \\ 4 & 6 \end{bmatrix}$.
**1. Find the Eigenvalues and Eigenvectors of M**
* **Characteristic Equation:**
det(M - λI) = 0
det( $\begin{bmatrix} -\lambda & -2 \\ 4 & 6-\lambda \end{bmatrix}$ ) = 0
(-λ)(6 - λ) - (-2)(4) = 0
-6λ + λ² + 8 = 0
λ² - 6λ + 8 = 0
(λ - 4)(λ - 2) = 0
* **Eigenvalues:**
λ₁ = 4 and λ₂ = 2
* **Eigenvectors:**
* For λ₁ = 4:
(M - 4I)v₁ = 0
$\begin{bmatrix} -4 & -2 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$
-4x - 2y = 0
-2x - y = 0
y = -2x
Let x = 1, then y = -2.
v₁ = $\begin{bmatrix} 1 \\ -2 \end{bmatrix}$
* For λ₂ = 2:
(M - 2I)v₂ = 0
$\begin{bmatrix} -2 & -2 \\ 4 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$
-2x - 2y = 0
x + y = 0
y = -x
Let x = 1, then y = -1.
v₂ = $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$
**2. Form the Matrix P and Diagonal Matrix D**
* P = $\begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix}$ (matrix of eigenvectors)
* D = $\begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}$ (diagonal matrix of eigenvalues)
**3. Find P⁻¹**
* det(P) = (1)(-1) - (1)(-2) = -1 + 2 = 1
* P⁻¹ = (1/det(P)) $\begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}$ = $\begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}$
**4. Use the Formula Mⁿ = PDⁿP⁻¹**
* Dⁿ = $\begin{bmatrix} 4ⁿ & 0 \\ 0 & 2ⁿ \end{bmatrix}$
* Mⁿ = $\begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix} \begin{bmatrix} 4ⁿ & 0 \\ 0 & 2ⁿ \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}$
* Mⁿ = $\begin{bmatrix} 4ⁿ & 2ⁿ \\ -2(4ⁿ) & -2ⁿ \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}$
* Mⁿ = $\begin{bmatrix} -4ⁿ + 2(2ⁿ) & -4ⁿ + 2ⁿ \\ 2(4ⁿ) - 2(2ⁿ) & 2(4ⁿ) - 2ⁿ \end{bmatrix}$
* Mⁿ = $\begin{bmatrix} -4ⁿ + 2^(n+1) & -4ⁿ + 2ⁿ \\ 2(4ⁿ) - 2^(n+1) & 2(4ⁿ) - 2ⁿ \end{bmatrix}$
**Formulas for the Entries:**
* Mⁿ₁₁ = -4ⁿ + 2^(n+1)
* Mⁿ₁₂ = -4ⁿ + 2ⁿ
* Mⁿ₂₁ = 2(4ⁿ) - 2^(n+1)
* Mⁿ₂₂ = 2(4ⁿ) - 2ⁿ
Question 1174344: Q−4: [6+4 marks] Let S={v_1,v_2,v_3} be a linearly independent set of vectors in〖 R〗^n and T={u_1,u_2,u_3}, where u_1=v_1+v_2+v_3, u_2=v_2+v_3 and u_3=v_3.
a-Determine whether each of v_1,v_2 and v_3 is a linearly combination of vectors in T.
b-Show that T is linearly independent set.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! You've asked this question before, and I provided a detailed solution. Let's recap the key points:
**a) Determining if v1, v2, and v3 are Linear Combinations of T**
* **v3:**
* Directly, we have u3 = v3. Therefore, v3 = 1 * u3, so v3 is a linear combination of T.
* **v2:**
* We have u2 = v2 + v3.
* We also know v3 = u3.
* Substituting, u2 = v2 + u3.
* Rearranging, v2 = u2 - u3.
* Therefore, v2 is a linear combination of T.
* **v1:**
* We have u1 = v1 + v2 + v3.
* We also know v2 = u2 - u3 and v3 = u3.
* Substituting, u1 = v1 + (u2 - u3) + u3.
* Simplifying, u1 = v1 + u2.
* Rearranging, v1 = u1 - u2.
* Therefore, v1 is a linear combination of T.
**Conclusion:** v1, v2, and v3 are all linear combinations of the vectors in T.
**b) Showing that T is a Linearly Independent Set**
To show that T is linearly independent, we need to prove that the only solution to the equation:
c1 * u1 + c2 * u2 + c3 * u3 = 0
is c1 = c2 = c3 = 0.
* **Substitute u1, u2, and u3:**
* c1 * (v1 + v2 + v3) + c2 * (v2 + v3) + c3 * (v3) = 0
* **Distribute and group:**
* c1 * v1 + (c1 + c2) * v2 + (c1 + c2 + c3) * v3 = 0
* **Use Linear Independence of S:**
* Since S = {v1, v2, v3} is linearly independent, the coefficients must all be zero:
* c1 = 0
* c1 + c2 = 0
* c1 + c2 + c3 = 0
* **Solve for c1, c2, and c3:**
* From c1 = 0, we know c1 is 0.
* Substituting c1 = 0 into c1 + c2 = 0, we get 0 + c2 = 0, so c2 = 0.
* Substituting c1 = 0 and c2 = 0 into c1 + c2 + c3 = 0, we get 0 + 0 + c3 = 0, so c3 = 0.
* **Conclusion:**
* The only solution is c1 = 0, c2 = 0, and c3 = 0. Therefore, T = {u1, u2, u3} is a linearly independent set.
If you have any further questions or would like me to elaborate on any aspect of the solution, feel free to ask.
Question 1174495: Q−4: [6+4 marks] Let S={v_1,v_2,v_3} be a linearly independent set of vectors in〖 R〗^n and T={u_1,u_2,u_3}, where u_1=v_1+v_2+v_3, u_2=v_2+v_3 and u_3=v_3.
a- Determine whether each of v_1,v_2 and v_3 is a linearly combination of vectors in T.
b- Show that T is linearly independent set.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! You've asked this question before, and I provided a detailed solution. Let's recap the key points:
**a) Determining if v1, v2, and v3 are Linear Combinations of T**
* **v3:**
* Directly, we have u3 = v3. Therefore, v3 = 1 * u3, so v3 is a linear combination of T.
* **v2:**
* We have u2 = v2 + v3.
* We also know v3 = u3.
* Substituting, u2 = v2 + u3.
* Rearranging, v2 = u2 - u3.
* Therefore, v2 is a linear combination of T.
* **v1:**
* We have u1 = v1 + v2 + v3.
* We also know v2 = u2 - u3 and v3 = u3.
* Substituting, u1 = v1 + (u2 - u3) + u3.
* Simplifying, u1 = v1 + u2.
* Rearranging, v1 = u1 - u2.
* Therefore, v1 is a linear combination of T.
**Conclusion:** v1, v2, and v3 are all linear combinations of the vectors in T.
**b) Showing that T is a Linearly Independent Set**
To show that T is linearly independent, we need to prove that the only solution to the equation:
c1 * u1 + c2 * u2 + c3 * u3 = 0
is c1 = c2 = c3 = 0.
* **Substitute u1, u2, and u3:**
* c1 * (v1 + v2 + v3) + c2 * (v2 + v3) + c3 * (v3) = 0
* **Distribute and group:**
* c1 * v1 + (c1 + c2) * v2 + (c1 + c2 + c3) * v3 = 0
* **Use Linear Independence of S:**
* Since S = {v1, v2, v3} is linearly independent, the coefficients must all be zero:
* c1 = 0
* c1 + c2 = 0
* c1 + c2 + c3 = 0
* **Solve for c1, c2, and c3:**
* From c1 = 0, we know c1 is 0.
* Substituting c1 = 0 into c1 + c2 = 0, we get 0 + c2 = 0, so c2 = 0.
* Substituting c1 = 0 and c2 = 0 into c1 + c2 + c3 = 0, we get 0 + 0 + c3 = 0, so c3 = 0.
* **Conclusion:**
* The only solution is c1 = 0, c2 = 0, and c3 = 0. Therefore, T = {u1, u2, u3} is a linearly independent set.
If you have any further questions or would like me to elaborate on any aspect of the solution, feel free to ask.
Question 1175538: Q−4: [6+4 marks] Let S={v_1,v_2,v_3} be a linearly independent set of vectors in〖 R〗^n and T={u_1,u_2,u_3}, where u_1=v_1+v_2+v_3, u_2=v_2+v_3 and u_3=v_3.
A- Determine whether each of v_1,v_2 and v_3 is a linearly combination of vectors in T.
B- Show that T is linearly independent set.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! Absolutely, let's break down this problem step-by-step.
**A. Determining if v1, v2, and v3 are Linear Combinations of T**
We need to see if we can express v1, v2, and v3 using linear combinations of u1, u2, and u3.
* **v3:**
* Directly, we have u3 = v3. Therefore, v3 = 1 * u3, so v3 is a linear combination of T.
* **v2:**
* We have u2 = v2 + v3.
* We also know v3 = u3.
* Substituting, u2 = v2 + u3.
* Rearranging, v2 = u2 - u3.
* Therefore, v2 is a linear combination of T.
* **v1:**
* We have u1 = v1 + v2 + v3.
* We also know v2 = u2 - u3 and v3 = u3.
* Substituting, u1 = v1 + (u2 - u3) + u3.
* Simplifying, u1 = v1 + u2.
* Rearranging, v1 = u1 - u2.
* Therefore, v1 is a linear combination of T.
**Conclusion:** v1, v2, and v3 are all linear combinations of the vectors in T.
**B. Showing that T is a Linearly Independent Set**
To show that T is linearly independent, we need to prove that the only solution to the equation:
c1 * u1 + c2 * u2 + c3 * u3 = 0
is c1 = c2 = c3 = 0.
* **Substitute u1, u2, and u3:**
* c1 * (v1 + v2 + v3) + c2 * (v2 + v3) + c3 * (v3) = 0
* **Distribute and group:**
* c1 * v1 + (c1 + c2) * v2 + (c1 + c2 + c3) * v3 = 0
* **Use Linear Independence of S:**
* Since S = {v1, v2, v3} is linearly independent, the coefficients must all be zero:
* c1 = 0
* c1 + c2 = 0
* c1 + c2 + c3 = 0
* **Solve for c1, c2, and c3:**
* From c1 = 0, we know c1 is 0.
* Substituting c1 = 0 into c1 + c2 = 0, we get 0 + c2 = 0, so c2 = 0.
* Substituting c1 = 0 and c2 = 0 into c1 + c2 + c3 = 0, we get 0 + 0 + c3 = 0, so c3 = 0.
* **Conclusion:**
* The only solution is c1 = 0, c2 = 0, and c3 = 0. Therefore, T = {u1, u2, u3} is a linearly independent set.
Question 1178699: this time, our immune system is the best defense . With this, a Melegail wishes to mix two types of foods in such a way that vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food A contains 2 units /kg of Vitamin A and 1 unit of /kg of vitamin C. Food B contains 1 unit/kg of vitamin A and 2 units/kg of Vitamin C. It costs 50.00 per kg to purchase food A and 70.00 per kg to purchase Food B. Formulate this problem as a linear programming problem to minimize the cost of such a mixture.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! Absolutely! Let's formulate this as a linear programming problem.
**1. Define the Decision Variables:**
* Let `x` represent the number of kilograms of Food A.
* Let `y` represent the number of kilograms of Food B.
**2. Define the Objective Function:**
* The objective is to minimize the cost.
* Cost = 50x + 70y
* Minimize Z = 50x + 70y
**3. Define the Constraints:**
* **Vitamin A Constraint:** The mixture must contain at least 8 units of Vitamin A.
* 2x + y ≥ 8
* **Vitamin C Constraint:** The mixture must contain at least 10 units of Vitamin C.
* x + 2y ≥ 10
* **Non-negativity Constraints:** The amount of food cannot be negative.
* x ≥ 0
* y ≥ 0
**Complete Linear Programming Problem:**
**Minimize:** Z = 50x + 70y
**Subject to:**
* 2x + y ≥ 8
* x + 2y ≥ 10
* x ≥ 0
* y ≥ 0
Question 1183383: Let A ∈ Mn×n(F), and let T : Mn×n(F) → Mn×n(F) be the linear transformation X → AXA, for X ∈
Mn×n(F). What is the rank of T? Show that there exists a matrix B such that ABA = A.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! Let's analyze the rank of the linear transformation T and the existence of matrix B.
**1. Rank of T:**
The rank of T depends crucially on the matrix A. There isn't a single, universally applicable rank for T without knowing A. Here's why:
* **Case 1: A is invertible:** If A is invertible (i.e., det(A) ≠ 0), then the rank of T is n². In this case, for any matrix Y in Mₙₓₙ(F), there exists a unique X such that AXA = Y. Specifically, X = A⁻¹YA⁻¹. Thus, T is surjective (onto), and since the domain and codomain have the same dimension (n²), T is also injective (one-to-one). Therefore, T is an isomorphism, and its rank is n².
* **Case 2: A is not invertible:** If A is not invertible, then its rank is less than n. The rank of T will be less than or equal to the rank of A. Let r be the rank of A (r < n). Then the rank of T will be at most r². It might be strictly less than r² because the transformation involves A on both the left and right.
* Example: If A is the zero matrix, then T(X) = AXA = 0 for all X. The rank of T is 0.
* Another example: If A is a projection matrix (A²=A) of rank r, the dimension of the image of T is at most r.
In summary, the rank of T can be anywhere from 0 to n², and it depends directly on the rank of A.
**2. Existence of Matrix B such that ABA = A:**
The statement that there exists a matrix B such that ABA = A is **true**, and it's a consequence of the properties of the Moore-Penrose pseudoinverse, denoted by A⁺. However, we can provide a more elementary argument for this specific case.
* **If A is invertible:** Then we can simply choose B = A⁻¹. Then ABA = AA⁻¹A = IA = A.
* **If A is not invertible:** This is where the argument gets a bit more subtle. We can use the rank factorization of A. Any matrix A of rank r can be factored as A = CR, where C is an n × r matrix of rank r, and R is an r × n matrix of rank r.
We want to find B such that ABA = A, or (CR)B(CR) = CR. If we can find a matrix B' such that RB'C = I (the r × r identity matrix), then we have (CR)B(CR) = C(RB'C)R = CIR = CR = A.
We can always find such a matrix B'. Since C has rank r, its columns are linearly independent, so C has a left inverse L such that LC = I. Since R has rank r, its rows are linearly independent, so R has a right inverse M such that RM = I. Then we can choose B' = MRCL. Then RB'C = RMRCLC = I*I = I.
Therefore, B = B' = MRCL is a matrix such that ABA = A.
In conclusion, such a matrix B always exists, regardless of whether A is invertible. When A is invertible, B is simply A⁻¹. When A is not invertible, a suitable B can be constructed by using the rank factorization of A.
Question 1185460: Table: Rate of the Cricket Chirps
Temperature in ℉
40
60
80
100
120
Rate
(Number of Chirps per Minute)
0
86
172
258
516
a.) Find a formula for g if g(t) represents the number of chirps per minute a cricket makes at temperature t degrees Fahrenheit.
b.) If f(c) represents the Fahrenheit reading that corresponds to a Celsius reading of c, which between the two functions g(f(t)) or f(g(t)) represents the number of chirps per minute a cricket makes when the temperature is c degrees Celsius?
c.) For the function in (b), write a formula for this and name it function h.
d.)Find the rate at which a cricket chirps if the temperature is __℉? __℃?
e.)Find the slope of the function g(t), h(c), and f(c). What does the slope of g(t) mean within the context of the problem?
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! Here's how to solve this cricket chirp problem:
**(a) Formula for g(t):**
The data points suggest a linear relationship. Let's find the slope (m) and y-intercept (b) of the line.
1. **Choose two points:** Let's use (40, 0) and (60, 86).
2. **Calculate the slope (m):**
m = (change in y) / (change in x) = (86 - 0) / (60 - 40) = 86 / 20 = 4.3
3. **Find the y-intercept (b):** Use the point-slope form of a line (y - y₁ = m(x - x₁)) and one of the points, or plug one of the points into y=mx+b. Using (40,0):
0 = 4.3 * 40 + b
b = -172
4. **Write the equation:**
g(t) = 4.3t - 172
**(b) Which composite function represents chirps at Celsius temperature?**
* f(c) converts Celsius to Fahrenheit.
* g(t) converts Fahrenheit to chirps per minute.
Therefore, *g(f(c))* is the correct composite function. It first converts Celsius to Fahrenheit using f(c), and then converts the Fahrenheit temperature to chirps per minute using g(t).
**(c) Formula for h(c) = g(f(c)):**
1. **Formula for f(c):** The formula to convert Celsius to Fahrenheit is:
f(c) = (9/5)c + 32
2. **Substitute f(c) into g(t):**
h(c) = g(f(c)) = 4.3 * f(c) - 172
h(c) = 4.3 * ((9/5)c + 32) - 172
h(c) = (38.7/5)c + 137.6 - 172
h(c) = 7.74c - 34.4
**(d) Chirp rate at specific temperatures:**
You need to provide the specific temperatures (one in Fahrenheit and one in Celsius) to calculate the chirp rates. Plug the Fahrenheit temperature into g(t) and the Celsius temperature into h(c).
**(e) Slopes and their meaning:**
* **Slope of g(t):** The slope of g(t) is 4.3. This means that for every 1-degree Fahrenheit increase in temperature, the cricket chirp rate increases by 4.3 chirps per minute.
* **Slope of h(c):** The slope of h(c) is 7.74. This means that for every 1-degree Celsius increase in temperature, the cricket chirp rate increases by 7.74 chirps per minute.
* **Slope of f(c):** The slope of f(c) is 9/5 or 1.8. This means that for every 1-degree Celsius increase in temperature, the Fahrenheit temperature increases by 1.8 degrees.
Question 1186205: The Intellectual Company produces a chemical solution used for cleaning carpets. This chemical is made from a mixture of two other chemicals which contain cleaning agent X and cleaning agent Y. Their product must contain 175 units of agent X and 150 units of agent Y and weigh at least 100 pounds. Chemical A costs ₱ 8 per pound, while chemical B costs ₱ 6 per pound. Chemical A contains one unit of agent X and three units of agent Y. Chemical B contains seven units of agent X and one unit of agent Y.
a. Set up the following:
i. Variables
ii. Constraints
iii. Objective Function
b. Find the minimum cost
c. Determine the best combination of the ingredients to minimize the cost.
Found 2 solutions by ikleyn, CPhill:
Answer by ikleyn(53748) (Show Source):
You can put this solution on YOUR website! .
The Intellectual Company produces a chemical solution used for cleaning carpets. This chemical is made from a mixture of two other chemicals which contain cleaning agent X and cleaning agent Y. Their product must contain 175 units of agent X and 150 units of agent Y and weigh at least 100 pounds. Chemical A costs ₱ 8 per pound, while chemical B costs ₱ 6 per pound. Chemical A contains one unit of agent X and three units of agent Y. Chemical B contains seven units of agent X and one unit of agent Y.
a. Set up the following:
i. Variables
ii. Constraints
iii. Objective Function
b. Find the minimum cost
c. Determine the best combination of the ingredients to minimize the cost.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The solution in the post by @CPhill, giving the answer
x = 43 pounds of chemical A and y = 19 pounds of chemical B
is INCORRECT. It can be easily disproved, since x + y = 43 + 19 = 62, which is less than 100.
Thus the restriction x+y >= 100 of the problem is not satisfied: it is FAILED, instead.
The cause is that @CPhill incorrectly determined the feasibility domain and used WRONG vertices for estimations.
Below is my solution, proper and correct.
**a. Set up:**
**i. Variables:**
* x = pounds of chemical A
* y = pounds of chemical B
**ii. Constraints:**
* Agent X: x + 7y ≥ 175
* Agent Y: 3x + y ≥ 150
* Weight: x + y ≥ 100
* Non-negativity: x ≥ 0, y ≥ 0
**iii. Objective Function:**
Minimize Cost (C) = 8x + 6y
**b. Find the minimum cost:**
1. **Graph the constraints:** Treat each inequality as an equation and plot the lines. Shade the appropriate region based on the inequality.
For example, for x + 7y ≥ 175, plot the line x + 7y = 175, and shade the region *above* and to the *right* of the lines.
Do this for all constraints.
2. **Identify the feasible region:** The feasible region is the area where all the shaded regions overlap.
3. **Find the vertices:** The vertices of the feasible region are the points where the constraint lines intersect.
Solve systems of equations to find these intersection points. The relevant vertices are:
* Intersection of x + 7y = 175 and y = 0: x = 175, y = 0
* Intersection of x + 7y = 175 and x + y = 100: Solving these gives x = 87.5, y = 12.5
* Intersection of 3x + y = 150 and x + y = 100: Solving these gives x = 25, y = 75
* Intersection of 3x + y = 175 and x = 0: Solving these gives x = 0, y = 175
4. **Evaluate the objective function at each vertex:**
* C(175, 0) = 8*175 + 6*0 = 1400
* C( 87.5, 12.5) = 8*81.5 + 6*12.5 = 727
* C(25, 75) = 8*25 + 6*75 = 650
* C(0, 175) = 8*0 + 6*175 = 1050
5. **Determine the minimum cost:** The minimum cost is the smallest value of the objective function.
The minimum cost is ₱650.
**c. Determine the best combination of ingredients:**
The best combination of ingredients is the (x, y) values that correspond to the minimum cost.
The minimum cost of ₱650 occurs when x = 25 and y = 75.
Therefore, the best combination is **25 pounds of chemical A and 75 pounds of chemical B**. <<<---=== ANSWER
Solved.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! Here's how to set up and solve this linear programming problem:
**a. Set up:**
**i. Variables:**
* x = pounds of chemical A
* y = pounds of chemical B
**ii. Constraints:**
* Agent X: x + 7y ≥ 175
* Agent Y: 3x + y ≥ 150
* Weight: x + y ≥ 100
* Non-negativity: x ≥ 0, y ≥ 0
**iii. Objective Function:**
Minimize Cost (C) = 8x + 6y
**b. Find the minimum cost:**
1. **Graph the constraints:** Treat each inequality as an equation and plot the lines. Shade the appropriate region based on the inequality. For example, for x + 7y ≥ 175, plot the line x + 7y = 175, and shade the region *above* and to the *right* of the line. Do this for all constraints.
2. **Identify the feasible region:** The feasible region is the area where all the shaded regions overlap.
3. **Find the vertices:** The vertices of the feasible region are the points where the constraint lines intersect. Solve systems of equations to find these intersection points. The relevant vertices are:
* Intersection of x + 7y = 175 and 3x + y = 150: Solving these gives x = 43, y = 19
* Intersection of 3x + y = 150 and x + y = 100: Solving these gives x = 25, y = 75
* Intersection of x + 7y = 175 and x + y = 100: Solving these gives x = 12.5, y = 87.5
4. **Evaluate the objective function at each vertex:**
* C(43, 19) = 8(43) + 6(19) = 344 + 114 = ₱458
* C(25, 75) = 8(25) + 6(75) = 200 + 450 = ₱650
* C(12.5, 87.5) = 8(12.5) + 6(87.5) = 100 + 525 = ₱625
5. **Determine the minimum cost:** The minimum cost is the smallest value of the objective function.
The minimum cost is ₱458.
**c. Determine the best combination of ingredients:**
The best combination of ingredients is the (x, y) values that correspond to the minimum cost.
The minimum cost of ₱458 occurs when x = 43 and y = 19.
Therefore, the best combination is **43 pounds of chemical A and 19 pounds of chemical B**.
Question 1186248: The Intellectual Company produces a chemical solution used for cleaning carpets. This chemical is made from a mixture of two other chemicals which contain cleaning agent X and cleaning agent Y. Their product must contain 175 units of agent X and 150 units of agent Y and weigh at least 100 pounds. Chemical A costs ₱ 8 per pound, while chemical B costs ₱ 6 per pound. Chemical A contains one unit of agent X and three units of agent Y. Chemical B contains seven units of agent X and one unit of agent Y.
a. Set up the following:
i. Variables
ii. Constraints
iii. Objective Function
b. Find the minimum cost
c. Determine the best combination of the ingredients to minimize the cost.
Found 2 solutions by ikleyn, CPhill:
Answer by ikleyn(53748) (Show Source):
You can put this solution on YOUR website! .
The Intellectual Company produces a chemical solution used for cleaning carpets.
This chemical is made from a mixture of two other chemicals which contain cleaning agent X and cleaning agent Y.
Their product must contain 175 units of agent X and 150 units of agent Y and weigh at least 100 pounds.
Chemical A costs ₱ 8 per pound, while chemical B costs ₱ 6 per pound.
Chemical A contains one unit of agent X and three units of agent Y.
Chemical B contains seven units of agent X and one unit of agent Y.
a. Set up the following:
i. Variables
ii. Constraints
iii. Objective Function
b. Find the minimum cost
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The solution in the post by @CPhill, giving the answer
x = 43 pounds of chemical A and y = 19 pounds of chemical B
is INCORRECT. It can be easily disproved, since x + y = 43 + 19 = 62, which is less than 100.
Thus the restriction x+y >= 100 of the problem is not satisfied: it is FAILED, instead.
The cause is that @CPhill incorrectly determined the feasibility domain and used WRONG vertices for estimations.
Below is my solution, proper and correct.
**a. Set up:**
**i. Variables:**
* x = pounds of chemical A
* y = pounds of chemical B
**ii. Constraints:**
* Agent X: x + 7y ≥ 175
* Agent Y: 3x + y ≥ 150
* Weight: x + y ≥ 100
* Non-negativity: x ≥ 0, y ≥ 0
**iii. Objective Function:**
Minimize Cost (C) = 8x + 6y
**b. Find the minimum cost:**
1. **Graph the constraints:** Treat each inequality as an equation and plot the lines. Shade the appropriate region based on the inequality.
For example, for x + 7y ≥ 175, plot the line x + 7y = 175, and shade the region *above* and to the *right* of the lines.
Do this for all constraints.
2. **Identify the feasible region:** The feasible region is the area where all the shaded regions overlap.
3. **Find the vertices:** The vertices of the feasible region are the points where the constraint lines intersect.
Solve systems of equations to find these intersection points. The relevant vertices are:
* Intersection of x + 7y = 175 and y = 0: x = 175, y = 0
* Intersection of x + 7y = 175 and x + y = 100: Solving these gives x = 87.5, y = 12.5
* Intersection of 3x + y = 150 and x + y = 100: Solving these gives x = 25, y = 75
* Intersection of 3x + y = 175 and x = 0: Solving these gives x = 0, y = 175
4. **Evaluate the objective function at each vertex:**
* C(175, 0) = 8*175 + 6*0 = 1400
* C( 87.5, 12.5) = 8*81.5 + 6*12.5 = 727
* C(25, 75) = 8*25 + 6*75 = 650
* C(0, 175) = 8*0 + 6*175 = 1050
5. **Determine the minimum cost:** The minimum cost is the smallest value of the objective function.
The minimum cost is ₱650.
**c. Determine the best combination of ingredients:**
The best combination of ingredients is the (x, y) values that correspond to the minimum cost.
The minimum cost of ₱650 occurs when x = 25 and y = 75.
Therefore, the best combination is **25 pounds of chemical A and 75 pounds of chemical B**. <<<---=== ANSWER
Solved.
Answer by CPhill(2189) (Show Source):
You can put this solution on YOUR website! Here's how to set up and solve this linear programming problem:
**a. Set up:**
**i. Variables:**
* x = pounds of chemical A
* y = pounds of chemical B
**ii. Constraints:**
* Agent X: 1x + 7y >= 175
* Agent Y: 3x + 1y >= 150
* Weight: x + y >= 100
* Non-negativity: x >= 0, y >= 0 (We can't have negative amounts of chemicals.)
**iii. Objective Function:**
We want to minimize the cost, which is given by:
Cost (C) = 8x + 6y
**b. Find the minimum cost:**
To find the minimum cost, we need to graph the constraints and find the feasible region. The optimal solution will occur at one of the vertices (corners) of the feasible region.
1. **Graph the constraints:** Treat each inequality as an equation and plot the lines. Then shade the appropriate region based on the inequality.
* x + 7y = 175
* 3x + y = 150
* x + y = 100
2. **Identify the feasible region:** The feasible region is the area where all the shaded regions overlap.
3. **Find the vertices:** The vertices of the feasible region are the points where the constraint lines intersect. You'll need to solve systems of equations to find these intersection points. The relevant vertices are usually where two constraints intersect.
* Intersection of x + 7y = 175 and 3x + y = 150: Solving these equations gives x = 43, y = 19
* Intersection of 3x + y = 150 and x + y = 100: Solving these equations gives x = 25, y = 75
* Intersection of x + 7y = 175 and x + y = 100: Solving these equations gives x = 75/6 = 12.5, y = 87.5
4. **Evaluate the objective function at each vertex:**
* C(43, 19) = 8(43) + 6(19) = 344 + 114 = 458
* C(25, 75) = 8(25) + 6(75) = 200 + 450 = 650
* C(12.5, 87.5) = 8(12.5) + 6(87.5) = 100 + 525 = 625
5. **Determine the minimum cost:** The minimum cost is the smallest value of the objective function at the vertices.
The minimum cost is ₱458.
**c. Determine the best combination of ingredients:**
The best combination of ingredients is the (x, y) values that correspond to the minimum cost.
The minimum cost of ₱458 occurs when x = 43 and y = 19.
Therefore, the best combination is 43 pounds of chemical A and 19 pounds of chemical B.
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