SOLUTION: A rectangle has one vertex in quadrant I on the graph of y=10-x^2, another at the origin, one on the positive x-axis, and one on the positive y-axis. a.) Express the area A of t

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A rectangle has one vertex in quadrant I on the graph of y=10-x^2, another at the origin, one on the positive x-axis, and one on the positive y-axis. a.) Express the area A of t      Log On


   



Question 823745: A rectangle has one vertex in quadrant I on the graph of y=10-x^2, another at the origin, one on the positive x-axis, and one on the positive y-axis.
a.) Express the area A of the rectangle as a function of x.
b.) Find the largest area A that can be enclosed by the rectangle?

Found 2 solutions by TimothyLamb, KMST:
Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
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y = 10 - x^2
y = -x^2 + 10
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it is well known that the area of a rectangle is maximized when its sides are of equal length, in other words, when the rectangle is a square
as such:
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y = x = -x^2 + 10
-x^2 - x + 10 = 0
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the above quadratic equation is in standard form, with a=-1, b=-1, and c=10
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
-1 -1 10
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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this quadratic has two real roots (the x-intercepts), which are:
x = -3.70156212
x = 2.70156212
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negative length doesn't make sense for this problem, so use the positive root:
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x = length = 2.70156212 units
y = width = 2.70156212 units
NOTE: the problem statement does not specify units, but these could be any linear unit, such as m, cm, ft, etc.
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answerA:
express the area A of the rectangle as a function of x:
A(x) = x * (-x^2 + 10)
A(x) = -x^3 + 10x
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answerB:
find the largest area A that can be enclosed by the rectangle:
A = maximum area = (2.70156212)*(2.70156212)
A = maximum area = 7.298437888218896 sq.units
NOTE: the problem statement does not specify units, but these could be any square unit, such as m^2, cm^2, ft^2, etc.
---
Solve and graph linear equations:
https://sooeet.com/math/linear-equation-solver.php
---
Solve quadratic equations, quadratic formula:
https://sooeet.com/math/quadratic-formula-solver.php
---
Solve systems of linear equations up to 6-equations 6-variables:
https://sooeet.com/math/system-of-linear-equations-solver.php

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The parabola looks like this
graph%28300%2C450%2C-5%2C5%2C-2%2C13%2C10-x%5E2%29 , with x-intercepts at sqrt%2810%29 and -sqrt%2810%29 .
( sqrt%2810%29 = approx. 3.16 ).
The rectangle in the first quadrant looks like this:
We want a formula to calculate the area of that rectangle, and we want to find the maximum area.

a.) A%28x%29=x%2Ay=x%2810-x%5E2%29=10x-x%5E3 for 0%3Cx%3Csqrt%2810%29 .
We have to put that restriction on the domain, because P%28x%2Cy%29 must be in the first quadrant.
So we can write the function as
highlight%28A%28x%29=-x%5E3%2B10x%29 for highlight%280%3Cx%3Csqrt%2810%29%29 .
It is a polynomial function.
Fully factored, it can be written as
A%28x%29=-x%28x%2Bsqrt%2810%29%29%28x-sqrt%2810%29%29
If the domain were not restricted, we would say that it has zeros at x=-sqrt%2810%29%7D%7D%2C+%7B%7B%7Bx=0 and x=sqrt%2810%29 .
We would figure out that it is positive and decreasing for x%3C-sqrt%2810%29%7D%7D%2C%0D%0Awhere+%7B%7B%7Bx%3C0 , x%2Bsqrt%2810%29%3C0 and x%2Bsqrt%2810%29%3C0 .
It is negative for -sqrt%2810%29%3Cx%3C0 and going through a minimum in that interval.
For 0%3Cx%3Csqrt%2810%29 , , the function is negative and decreasing.
Here's the graph graph%28200%2C300%2C-5%2C5%2C-15%2C15%2C-x%5E3%2B10x%29
andpositive for x=0 and x%3Esqrt%2810%29 .

b.) The only way that I know to find the maximum area of such a rectangle requires using calculus and finding the derivative of A%28x%29
dA%2Fdt=-3x%5E2%2B10
The zeros of that derivative show us the location of the minimum and maximum.
-3x%5E2%2B10=0
10=3x%5E2
10%2F3=x%5E2
The solutions are:
x=-sqrt%2810%2F3%29=-sqrt%2830%29%2F3 or x=sqrt%2810%2F3%29=sqrt%2830%29%2F3
The maximum occurs at x=sqrt%2830%29%2F3 , where x%5E2=10%2F3
Substituting those values into
A%28x%29=x%2810-x%5E2%29 we can easily calculate

The approximate value is highlight%2812.17%29 .