A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are C(x) = 73,000 + 70 x and [tex]p(x) = 250 - (\frac{x}{20}), 0 \leq x \leq 5000[/tex].(A) Find the maximum revenue.
(B) Find the maximum profit, the production level that will realize the maximum profit, and the price the company should charge for each television set.
(C) If the government decides to tax the company $55 for each set it produces, how many sets should the company manufacture each month to maximize its profit? What is the maximum profit? What should the company charge for each set?
(A) The maximum revenue is $_________.
(B) The maximum profit is when sets are manufactured and sold for each.
(C) When each set is taxed at $55, the maximum profit is when sets are manufactured and sold for each.

b) The maximum profit is $89000, the production level that will realize the maximum profit is 1800, and the price the company should charge for each television set is $160.

C) If the government decides to tax the company $55 for each set it produces, the sets should the company manufacture each month to maximize its profit is 1250. the maximum profit is $70825 What should the company charge for each set is $187.5

Step-by-step explanation:

a) Revenue R(x)

R(x) = p(x) * x = x * [tex](250-\frac{x}{20})[/tex] = [tex](250x-\frac{x^{2} }{20})[/tex]

For maximum revenue, the first derivative of R(x) = R'(x) = 0

R'(x) = [tex](250-\frac{2x}{20}) = 0[/tex]

[tex](250-\frac{2x}{20}) = 0\\[/tex]

[tex]250=\frac{2x}{20}[/tex]

x = 2500

the second derivative of R(x)=R''(x)

R''(x) = -1/10 which is less than 0.

Maximum revenue is at x = 2500

R(2500) = [tex](250*2500-\frac{2500^{2} }{20})=312500[/tex]

b) Profit P(x)

P(x) = R(x) - C(x) = [tex](250x-\frac{x^{2} }{20})-(73000+70x) = -73000+180x-\frac{x^{2} }{20}[/tex]

For maximum profit, the first derivative of P(x) = P'(x) = 0

P'(x) = [tex]180-\frac{2x }{20}=0[/tex]

[tex]180=\frac{2x }{20}[/tex]

x = 1800

the second derivative of P(x)=P''(x)

P''(x) = -1/10 which is less than 0.

For maximum profit, x = 1800

Therefore P(1800)[tex]=-73000+180*1800-\frac{1800^{2} }{20}[/tex] = 89000

The price the company should charge for each television set is p(1800) =[tex](250-\frac{1800}{20}) = 160[/tex]

c) f the government decides to tax the company $55 for each set it produces, the new cost C(x) = 73000 + 125x

Profit P(x)

P(x) = R(x) - C(x) = [tex](250x-\frac{x^{2} }{20})-(73000+125x) = -73000+125x-\frac{x^{2} }{20}[/tex]

For maximum profit, the first derivative of P(x) = P'(x) = 0

P'(x) = [tex]125-\frac{2x }{20}=0[/tex]

[tex]125=\frac{2x }{20}[/tex]

x = 1250

the second derivative of P(x)=P''(x)

P''(x) = -1/10 which is less than 0.

For maximum profit, x = 1250 hence 1250 sets should the company manufacture each month to maximize its profit

Therefore P(1800) =[tex]-73000+125*1250-\frac{1250^{2} }{20}[/tex] = 70825

The price the company should charge for each television set is p(1250) =[tex](250-\frac{1250}{20}) = 187.5[/tex]