Question 1171702: The profit P(x) that company earn for selling x number of toy cars can be modelled by
P(x) =25x^2 +1000 -3000.
1. Write an inequality that models the problem if the profit P has to be at least php 5000
2. what could be the range of possible values of x
3. how many toy cars must be sold for a profit of at least php 5000
4. how many toy cars must be sold to obtain the maximum profit
5. how much is the maximum profit
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step-by-step.
**Given Profit Function:**
* P(x) = 25x² + 1000x - 3000
**1. Inequality for Profit at Least PHP 5000:**
* P(x) ≥ 5000
* 25x² + 1000x - 3000 ≥ 5000
* 25x² + 1000x - 8000 ≥ 0
**2. Range of Possible Values of x:**
* To find the range, we need to solve the inequality 25x² + 1000x - 8000 ≥ 0.
* First, simplify by dividing by 25: x² + 40x - 320 ≥ 0
* Find the roots of the quadratic equation x² + 40x - 320 = 0 using the quadratic formula:
* x = [-b ± √(b² - 4ac)] / 2a
* x = [-40 ± √(40² - 4(1)(-320))] / 2(1)
* x = [-40 ± √(1600 + 1280)] / 2
* x = [-40 ± √2880] / 2
* x = [-40 ± 24√5] / 2
* x = -20 ± 12√5
* Approximate the roots:
* x₁ = -20 - 12√5 ≈ -46.83
* x₂ = -20 + 12√5 ≈ 6.83
* Since x represents the number of toy cars sold, it must be non-negative. Therefore, we only consider the positive root.
* The inequality x² + 40x - 320 ≥ 0 is satisfied when x ≤ -46.83 or x ≥ 6.83.
* Since x must be non-negative, the range of possible values for x is x ≥ 6.83.
* However, since we are dealing with toy cars, x must be a whole number.
* Therefore the range of x is x ≥ 7.
**3. Number of Toy Cars for Profit of at Least PHP 5000:**
* Since x ≥ 6.83, the company must sell at least 7 toy cars to achieve a profit of at least PHP 5000.
**4. Number of Toy Cars for Maximum Profit:**
* To find the maximum profit, we need to find the vertex of the parabola P(x) = 25x² + 1000x - 3000.
* The x-coordinate of the vertex is given by x = -b / 2a, where a = 25 and b = 1000.
* x = -1000 / (2 * 25) = -1000 / 50 = -20
* Since the coefficient of x² is positive (25), the parabola opens upwards, meaning it has a minimum, not a maximum. There must be an error with the problem. I will assume the coefficent of x² is negative.
* Let us assume the profit function is P(x) = -25x² + 1000x - 3000.
* x = -1000 / (2 * -25) = -1000 / -50 = 20
* Therefore, 20 toy cars must be sold to obtain the maximum profit.
**5. Maximum Profit:**
* Using the corrected profit function P(x) = -25x² + 1000x - 3000, substitute x = 20:
* P(20) = -25(20)² + 1000(20) - 3000
* P(20) = -25(400) + 20000 - 3000
* P(20) = -10000 + 20000 - 3000
* P(20) = 7000
* The maximum profit is PHP 7000.
**Summary:**
1. **Inequality:** 25x² + 1000x - 8000 ≥ 0
2. **Range of x:** x ≥ 7
3. **Toy Cars for Profit ≥ PHP 5000:** 7
4. **Toy Cars for Maximum Profit (corrected function):** 20
5. **Maximum Profit (corrected function):** PHP 7000
Answer by ikleyn(52777)  (Show Source):
You can put this solution on YOUR website! .
Hello, in your post I see at least two FATAL ERRORS.
First, as written, this quadratic function represents a parabola opened up.
So, this quadratic function has no a maximum, at all, but has a minimum.
Second notice is that the linear term with x in degree 1 is missed,
and instead, two constant terms are written, making the entire problem UNNATURAL.
So, as presented in your post, the problem is FATALLY DEFECTIVE.
|
|
|
