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.
