Given the cost and demand functions:
[tex]\begin{gathered} C(x)=75000+70x \\ p(x)=160-\frac{x}{30} \\ 0\leq x\leq5000 \end{gathered}[/tex]
we can find the revenue with the following expression:
[tex]R(x)=xp(x)=160x-\frac{x²}{30}[/tex]
then, the profit can be calculated with the difference between the revenue and C(x):
[tex]P(x)=R(x)-C(x)=160x-\frac{x²}{30}-75000-70x=90x-\frac{x²}{30}-75000[/tex]
to find the max profit, we can find the derivative of P(x) and find its root to get the value of x that maximizes the function P(x):
[tex]\begin{gathered} P(x)=90x-\frac{x²}{30}-75000 \\ \Rightarrow P^{\prime}(x)=90-\frac{2}{30}x=90-\frac{1}{15}x \\ P^{\prime}(x)=0\Rightarrow90-\frac{1}{15}x=0 \\ \Rightarrow\frac{1}{15}x=90 \\ \Rightarrow x=90*15=1350 \\ x=1350 \end{gathered}[/tex]
therefore, the production level that results in the maximum profit is x = 1350.
Finally, to find the price, we can evaluate x = 1350 on p(x):
[tex]p(1350)=160-\frac{1350}{30}=160-45=115[/tex]
therefore, the price that the company should charge for each drill to maximize profit is $115
