Solution:
Given:
[tex]\begin{gathered} C(x)=45000+80x \\ p=220-\frac{x}{30},\text{ }0\leq x\leq5000 \end{gathered}[/tex]The revenue is given by;
[tex]\begin{gathered} R(x)=p.x \\ R(x)=(220-\frac{x}{30})x \\ R(x)=220x-\frac{x^2}{30} \end{gathered}[/tex]Question A:
The profit is the difference between the revenue and the cost.
Hence,
[tex]\begin{gathered} profit=R(x)-C(x) \\ profit=(220x-\frac{x^2}{30})-(45000+80x) \\ profit=-\frac{x^2}{30}+220x-80x-45000 \\ profit=-\frac{x^2}{30}+140x-45000 \\ P(x)=-\frac{x^2}{30}+140x-45000 \end{gathered}[/tex]To get the maximum profit, we differentiate the profit function.
[tex]\begin{gathered} P(x)=-\frac{x^2}{30}+140x-45000 \\ P^{\prime}(x)=-\frac{x}{15}+140 \\ when\text{ }P^{\prime}(x)=0, \\ 0=-\frac{x}{15}+140 \\ 0-140=-\frac{x}{15} \\ -140=-\frac{x}{15} \\ -140\times-15=x \\ x=2100 \end{gathered}[/tex]Therefore, the production level that results in maximum profit is x = 2100
Question B:
The price for each drill to maximize profit is;
[tex]\begin{gathered} p=220-\frac{x}{30} \\ p=220-\frac{2100}{30} \\ p=220-70 \\ p=150 \end{gathered}[/tex]Therefore, the price the company should charge for each drill is 150