Discover a world of knowledge and community-driven answers at IDNLearn.com today. Find accurate and detailed answers to your questions from our experienced and dedicated community members.
Sagot :
Sure, let's solve each part of the question step-by-step.
### Part A: Find the Marginal Profit Function
Given:
- Revenue function: [tex]\( R(q) = -q^3 + 370q^2 \)[/tex]
- Cost function: [tex]\( C(q) = 270 + 18q \)[/tex]
First, we need to find the profit function which is defined as:
[tex]\[ P(q) = R(q) - C(q) \][/tex]
Substituting the given functions into the profit function:
[tex]\[ P(q) = (-q^3 + 370q^2) - (270 + 18q) \][/tex]
[tex]\[ P(q) = -q^3 + 370q^2 - 18q - 270 \][/tex]
Next, we find the marginal profit function [tex]\( MP(q) \)[/tex] by calculating the first derivative of the profit function [tex]\( P(q) \)[/tex] with respect to [tex]\( q \)[/tex]:
[tex]\[ MP(q) = P'(q) = \frac{d}{dq} (-q^3 + 370q^2 - 18q - 270) \][/tex]
Taking the derivative term-by-term:
[tex]\[ MP(q) = -3q^2 + 2 \cdot 370q - 18 \][/tex]
[tex]\[ MP(q) = -3q^2 + 740q - 18 \][/tex]
So, the simplified expression for the marginal profit function is:
[tex]\[ MP(q) = -3q^2 + 740q - 18 \][/tex]
### Part B: Number of Items to Maximize Profit
To find the number of items [tex]\( q \)[/tex] that maximize profit, we need to determine the critical points of the marginal profit function [tex]\( MP(q) \)[/tex]. Critical points occur where the derivative (marginal profit) is zero or undefined. Therefore, we solve the equation:
[tex]\[ MP(q) = -3q^2 + 740q - 18 = 0 \][/tex]
The solutions to this quadratic equation can be found using the quadratic formula:
[tex]\[ q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = -3 \)[/tex], [tex]\( b = 740 \)[/tex], and [tex]\( c = -18 \)[/tex].
Plugging in these values:
[tex]\[ q = \frac{-740 \pm \sqrt{740^2 - 4 \cdot (-3) \cdot (-18)}}{2 \cdot (-3)} \][/tex]
[tex]\[ q = \frac{-740 \pm \sqrt{547600 - 216}}{-6} \][/tex]
[tex]\[ q = \frac{-740 \pm \sqrt{547384}}{-6} \][/tex]
[tex]\[ q = \frac{-740 \pm 739.84}{-6} \][/tex]
[tex]\[ q \approx \frac{-740 + 739.84}{-6} \quad \text{or} \quad q \approx \frac{-740 - 739.84}{-6} \][/tex]
[tex]\[ q \approx \frac{-0.16}{-6} \approx 0.027 \quad \text{or} \quad q \approx \frac{-1479.84}{-6} \approx 246.64 \][/tex]
These calculations yield the critical points:
[tex]\[ q = 0.027 \][/tex]
[tex]\[ q = 246.64 \][/tex]
To determine which of these critical points maximizes the profit, we evaluate the second derivative of the profit function [tex]\( P''(q) \)[/tex]:
[tex]\[ P''(q) = MP'(q) = \frac{d}{dq} (-3q^2 + 740q - 18) \][/tex]
[tex]\[ P''(q) = -6q + 740 \][/tex]
Evaluate [tex]\( P''(q) \)[/tex] at each critical point:
For [tex]\( q = 0.027 \)[/tex]:
[tex]\[ P''(0.027) = -6(0.027) + 740 \approx 739.84 \][/tex]
For [tex]\( q = 246.64 \)[/tex]:
[tex]\[ P''(246.64) = -6(246.64) + 740 \approx -740 + 740 = -739.84 \][/tex]
Since the second derivative is negative at [tex]\( q \approx 246.64 \)[/tex], it is a local maximum. The second derivative is positive at [tex]\( q = 0.027 \)[/tex], indicating a local minimum.
### Conclusion
Thus, the number of items that need to be sold to maximize the profit is approximately:
[tex]\[ 246.64 \][/tex]
So the answer is[tex]\(\boxed{ 246.64 } \)[/tex] hundred units must be sold.
### Part A: Find the Marginal Profit Function
Given:
- Revenue function: [tex]\( R(q) = -q^3 + 370q^2 \)[/tex]
- Cost function: [tex]\( C(q) = 270 + 18q \)[/tex]
First, we need to find the profit function which is defined as:
[tex]\[ P(q) = R(q) - C(q) \][/tex]
Substituting the given functions into the profit function:
[tex]\[ P(q) = (-q^3 + 370q^2) - (270 + 18q) \][/tex]
[tex]\[ P(q) = -q^3 + 370q^2 - 18q - 270 \][/tex]
Next, we find the marginal profit function [tex]\( MP(q) \)[/tex] by calculating the first derivative of the profit function [tex]\( P(q) \)[/tex] with respect to [tex]\( q \)[/tex]:
[tex]\[ MP(q) = P'(q) = \frac{d}{dq} (-q^3 + 370q^2 - 18q - 270) \][/tex]
Taking the derivative term-by-term:
[tex]\[ MP(q) = -3q^2 + 2 \cdot 370q - 18 \][/tex]
[tex]\[ MP(q) = -3q^2 + 740q - 18 \][/tex]
So, the simplified expression for the marginal profit function is:
[tex]\[ MP(q) = -3q^2 + 740q - 18 \][/tex]
### Part B: Number of Items to Maximize Profit
To find the number of items [tex]\( q \)[/tex] that maximize profit, we need to determine the critical points of the marginal profit function [tex]\( MP(q) \)[/tex]. Critical points occur where the derivative (marginal profit) is zero or undefined. Therefore, we solve the equation:
[tex]\[ MP(q) = -3q^2 + 740q - 18 = 0 \][/tex]
The solutions to this quadratic equation can be found using the quadratic formula:
[tex]\[ q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = -3 \)[/tex], [tex]\( b = 740 \)[/tex], and [tex]\( c = -18 \)[/tex].
Plugging in these values:
[tex]\[ q = \frac{-740 \pm \sqrt{740^2 - 4 \cdot (-3) \cdot (-18)}}{2 \cdot (-3)} \][/tex]
[tex]\[ q = \frac{-740 \pm \sqrt{547600 - 216}}{-6} \][/tex]
[tex]\[ q = \frac{-740 \pm \sqrt{547384}}{-6} \][/tex]
[tex]\[ q = \frac{-740 \pm 739.84}{-6} \][/tex]
[tex]\[ q \approx \frac{-740 + 739.84}{-6} \quad \text{or} \quad q \approx \frac{-740 - 739.84}{-6} \][/tex]
[tex]\[ q \approx \frac{-0.16}{-6} \approx 0.027 \quad \text{or} \quad q \approx \frac{-1479.84}{-6} \approx 246.64 \][/tex]
These calculations yield the critical points:
[tex]\[ q = 0.027 \][/tex]
[tex]\[ q = 246.64 \][/tex]
To determine which of these critical points maximizes the profit, we evaluate the second derivative of the profit function [tex]\( P''(q) \)[/tex]:
[tex]\[ P''(q) = MP'(q) = \frac{d}{dq} (-3q^2 + 740q - 18) \][/tex]
[tex]\[ P''(q) = -6q + 740 \][/tex]
Evaluate [tex]\( P''(q) \)[/tex] at each critical point:
For [tex]\( q = 0.027 \)[/tex]:
[tex]\[ P''(0.027) = -6(0.027) + 740 \approx 739.84 \][/tex]
For [tex]\( q = 246.64 \)[/tex]:
[tex]\[ P''(246.64) = -6(246.64) + 740 \approx -740 + 740 = -739.84 \][/tex]
Since the second derivative is negative at [tex]\( q \approx 246.64 \)[/tex], it is a local maximum. The second derivative is positive at [tex]\( q = 0.027 \)[/tex], indicating a local minimum.
### Conclusion
Thus, the number of items that need to be sold to maximize the profit is approximately:
[tex]\[ 246.64 \][/tex]
So the answer is[tex]\(\boxed{ 246.64 } \)[/tex] hundred units must be sold.
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Thank you for choosing IDNLearn.com for your queries. We’re here to provide accurate answers, so visit us again soon.
