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A company manufacturers and sells a electric drills per month. The monthly cost and price-demand equations areC(x) = 75000 + 70x,P= 160 - x/30 , 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 75000 70xP 160 X30 0 Lt 2 Lt 5000A Find The Produ class=

Sagot :

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

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