Question

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 $p(x) = 250 - (\frac{x}{20}), 0 \leq x \leq 5000$.​(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.

Answer 1

Answer:

a) The maximum revenue is ​$312500

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 * $(250-\frac{x}{20})$ = $(250x-\frac{x^{2} }{20})$

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

R'(x) = $(250-\frac{2x}{20}) = 0$

$(250-\frac{2x}{20}) = 0$

$250=\frac{2x}{20}$

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) = $(250*2500-\frac{2500^{2} }{20})=312500$

b) Profit P(x)

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

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

P'(x) = $180-\frac{2x }{20}=0$

$180=\frac{2x }{20}$

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)$=-73000+180*1800-\frac{1800^{2} }{20}$ = 89000

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

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) = $(250x-\frac{x^{2} }{20})-(73000+125x) = -73000+125x-\frac{x^{2} }{20}$

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

P'(x) = $125-\frac{2x }{20}=0$

$125=\frac{2x }{20}$

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) =$-73000+125*1250-\frac{1250^{2} }{20}$ = 70825

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