Answer:
A) MP(q) = -3q² + 440q - 13
B) 146.64 units.
Explanation:
The profit function is given by the revenue minus the cost function:
![P(q) = R(q) - C(q)\\P(q) = -q^3+220q^2-500-13q](https://img.qammunity.org/2021/formulas/mathematics/college/u87gsi79k24py7f4t58u7fjas53whvwacv.png)
A) The Marginal profit function is the derivate of the profit function as a function of the quantity sold:
![P(q) = -q^3+220q^2-500-13q\\MP(q) = (dP(q))/(dq) \\MP(q)=-3q^2+440q-13](https://img.qammunity.org/2021/formulas/mathematics/college/ffzmkktznbjr655yknpfeutg39ct5x6v5x.png)
B) The value of "q" for which the marginal profit function is zero is the number of items (in hundreds) that maximizes profit:
![MP(q)=0=-3q^2+440q-13\\q=(-440\pm √(440^2-(4*(-3)*(-13))) )/(-6)\\q'=146.64\\q'' = - 0.03](https://img.qammunity.org/2021/formulas/mathematics/college/oxxqeahya70nvpusyo0vgmk3yoyqvw456c.png)
Therefore, the only reasonable answer is that 146.64 hundred units must be sold in order to maximize profit.