Question

A company manufacturers and sells a electric drills per month. The monthly cost and price-demand equations areC(x) = 45000 + 802,p = 220 - 5,0 < 2 < 5000.(A) Find the production level that results in the maximum profit.Production Level =(B) Find the price that the company should charge for each drill in order to maximize profit.Price=

Answer 1

Given:

`\begin{gathered} C(x)=45000+80x \\ p=220-(x)/(30),\text{ }0\leq x\leq5000 \end{gathered}`

The revenue is given by;

`\begin{gathered} R(x)=p.x \\ R(x)=(220-(x)/(30))x \\ R(x)=220x-(x^2)/(30) \end{gathered}`

Question A:

The profit is the difference between the revenue and the cost.

Hence,

`\begin{gathered} profit=R(x)-C(x) \\ profit=(220x-(x^2)/(30))-(45000+80x) \\ profit=-(x^2)/(30)+220x-80x-45000 \\ profit=-(x^2)/(30)+140x-45000 \\ P(x)=-(x^2)/(30)+140x-45000 \end{gathered}`

To get the maximum profit, we differentiate the profit function.

`\begin{gathered} P(x)=-(x^2)/(30)+140x-45000 \\ P^(\prime)(x)=-(x)/(15)+140 \\ when\text{ }P^(\prime)(x)=0, \\ 0=-(x)/(15)+140 \\ 0-140=-(x)/(15) \\ -140=-(x)/(15) \\ -140*-15=x \\ x=2100 \end{gathered}`

Therefore, the production level that results in maximum profit is x = 2100

Question B:

The price for each drill to maximize profit is;

`\begin{gathered} p=220-(x)/(30) \\ p=220-(2100)/(30) \\ p=220-70 \\ p=150 \end{gathered}`

Therefore, the price the company should charge for each drill is 150