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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=

A Company Manufacturers And Sells A Electric Drills Per Month The Monthly Cost And Pricedemand Equations AreCx 45000 802p 220 50 Lt 2 Lt 5000A Find The Producti class=

Sagot :

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

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