Solution:
Given:
[tex]\begin{gathered} C(x)=55000+80x \\ p=220-\frac{x}{30} \end{gathered}[/tex]To get the production level that will lead to maximum profit, we need to get the revenue.
Revenue (R) is price multiplied by the number of electric drills.
Hence,
[tex]\begin{gathered} R=(220-\frac{x}{30})x \\ R=220x-\frac{x^2}{30} \end{gathered}[/tex]Part A:
Profit is the difference between revenue and cost;
[tex]\begin{gathered} Profit=R-C(x) \\ Profit=220x-\frac{x^2}{30}-(55000+80x) \\ Profit=220x-80x-\frac{x^2}{30}-55000 \\ Profit=140x-\frac{x^2}{30}-55000 \end{gathered}[/tex]Therefore,
[tex]Profit=140x-\frac{x^{2}}{30}-55,000[/tex]To get the production level, we find the maximum value from the profit.
[tex]\begin{gathered} P(x)=140x-\frac{x^2}{30}-55,000 \\ Differentiating\text{ the equation,} \\ P^{\prime}(x)=140-\frac{x}{15} \\ \\ The\text{ maximum occurs when }P^{\prime}(x)=0 \\ 0=140-\frac{x}{15} \\ \frac{x}{15}=140 \\ x=15\times140 \\ x=2100 \end{gathered}[/tex]Therefore, the production level that results in maximum profit is 2100.
Part B:
The price 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]