**A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are C(x)=74,000+60x and p(x)=300-(x/20), 0≤x≤6000**

**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.**

**Answer:**

a.) Revenue function is R(x) = x*p(x) = x*( 300 - (x/20) )

R(x) = 300x - xx/20

R’(x) = 300 - (2x)/20 = 300 - x/10

R '(x) = 0 = 300 - x/10

Solve for x.

300 - x/10 = 0

x = 3,000 tv sets.

R(3000) = 300*3000 - 3000*3000/20 = 300*3000 - 150*3000 = (300 - 150)*3000 = 150*3000 = $450,000

Therefore the maximum revenue is $450,000

b.) Profit function is F(x) = x*p(x) - C(x)

F(x) = x(300 - (x/20)) - (74,000 + 60x)

F(x) = 300x - xx/20 - 74,000 - 60x = 240x - xx/20 - 74000

Find max F(x) by solving: F '(x) = 0 or we can simply find the vertex of the parabola.

x = -b/2a is the location of the maximum.

x = -240/(-1/20) = 4,800 television sets.

The maximum revenue is when x = 4,800 tv sets are sold.

F(4,800) = 240*4800 - (4800)^2 / 20 - 74000

F(4800) = 240*4800 - 1226000 = -74,000

The maximum profit is -$74,000