A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are C(x)=74,000+80x and p(x)=300− x 30, 0≤x≤9000. (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 $5 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) $675000

b) $289000 profit,3300 set, $190 per set

c) 3225 set, $272687.5 profit, $192.5 per set

Step-by-step explanation:

a) Revenue R(x) = xp(x) = x(300 - x/30) = 300x - x²/30

The maximum revenue is at R'(x) =0

R'(x) = 300 - 2x/30 = 300 - x/15

But we need to compute R'(x) = 0:

300 - x/15 = 0

x/15 = 300

x = 4500

Also the second derivative of R(x) is given as:

R"(x) = -1/15 < 0 This means that the maximum revenue is at x = 4500. Hence:

R(4500) = 300 (4500) - (4500)²/30 = $675000  

B) Profit P(x) = R(x) - C(x) = 300x - x²/30 - (74000 + 80x) = -x²/30 + 300x - 80x - 74000

P(x) = -x²/30 + 220x - 74000

The maximum revenue is at P'(x) =0

P'(x) = - 2x/30 + 220= -x/15 + 220

But we need to compute P'(x) = 0:

-x/15 + 220 = 0

x/15 = 220

x = 3300

Also the second derivative of P(x) is given as:

P"(x) = -1/15 < 0 This means that the maximum profit is at x = 3300. Hence:

P(3300) =  -(3300)²/30 + 220(3300) - 74000 = $289000  

The price for each set is:

p(3300) = 300 -3300/30 = $190 per set

c) The new cost is:

C(x) = 74000 + 80x + 5x = 74000 + 85x

Profit P(x) = R(x) - C(x) = 300x - x²/30 - (74000 + 85x) = -x²/30 + 300x - 85x - 74000

P(x) = -x²/30 + 215x - 74000

The maximum revenue is at P'(x) =0

P'(x) = - 2x/30 + 215= -x/15 + 215

But we need to compute P'(x) = 0:

-x/15 + 215 = 0

x/15 = 215

x = 3225

Also the second derivative of P(x) is given as:

P"(x) = -1/15 < 0 This means that the maximum profit is at x = 3225. Hence:

P(3225) =  -(3225)²/30 + 215(3225) - 74000 = $272687.5

The money to be charge for each set is:

p(x) = 300 - 3225/30 = $192.5 per set

When taxed $5, the maximum profit is $272687.5

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6