A company manufactures and sells x television sets per month. The monthly cost a

Question

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 0sx S 6000 (A) Find the maximum revenue. (B Find the maximum profit, the production level that will realize the mainunprofit and the price he compan, should e age or eanteen on (C) If the government decides to tax the company $6 for each set it produces, how many sets should the company manufacture each month to maximize its profit? at is the maximum profit? What should the company charge for each set?

Explanation / Answer

a)

Revenue = x * p(x) = x * (300 - x/20) = 300x - x^2/20

To maximize the revenue

dRevenue/dx must be equated to zero

dRevenue/dx = 300 - x/10 = 0

Hence we get the value of x equal to 3000

Checking using second derivative

R''(x) = -1/10 which is less than zero, hence the maxima is obtained at this point

Maximum Revenue = 300(3000) - (3000)^2/20 = 450000 dollars

b)

Profit = Revenue - Cost

Profit = (300x-x^2/20) - (74000+80x)

Profit = 220x - x^2/20 - 74000

dProfit/dx = 220 - x/10= 0

which means x=2200

P''(x) = -1/10 [which means the point is a maxima]

P(2200) = 220(2200) - (2200)^2/20 - 74000 = 168000

Price charged for each set = 300 - 2200/20 = 190 dollars

c)

New cost = 74000 + 80x + 6x = 74000 + 86x

Profit = (300x-x^2/20) - (74000+86x)

Profit = 214x - x^2/20 - 74000

dProfit/dx = 214 - x/10= 0

which means x=2140

P''(x) = -1/10 [which means the point is a maxima]

P(2140) = 214(2140) - (2140)^2/20 - 74000 = 154980

Price charged for each set = 300 - 2140/20 = 193 dollars

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