The average revenue for product Q it given by AR = 120 - 3Q and the total cost o

Question

The average revenue for product Q it given by AR = 120 - 3Q and the total cost of Q by: STC = 30 + 120Q - 5Q^1 + Q^3 a. What total profit will result from selling the quantity found in (c) at the price found in (d)? Remember, profit is TR - STC. b. At what level of Q is revenue maximized? Remember. let MR = 0 and solve for Q. MR = 0 signals the objective of maximizing revenue. c. At what positive level of Q Is marginal profit maximized? You found the profit function in (e) above. Marginal profit is the first derivative of the profit function (e). Next find the derivative of marginal profit, set it equal to zero, and solve for Q. d. What price per unit should be charged at the quantity found in (g)? Simply plug the Q you got in (g) Into the same price function you found in (a) and also used In (d).

Explanation / Answer

AR = 120 - 3Q

TR = AR * Q = 120Q - 3Q2

STC = 30 + 120Q - 5Q2 + 0.083Q3

Profit = TR - TC

         = 120Q - 3Q2 - 30 - 120Q + 5Q2 - 0.083Q3

Profit = - 0.083Q3 + 2Q2- 30

From (c), we found Q = 16

Thus, Profit = - 0.083Q3 + 2Q2- 30 = - 0.083(16)3 + 2(16)2- 30 = -339.97 + 512 - 30 = $142.03

(b) MR = 120 - 6Q = 0

=> 6Q = 120

=> Q = 120/6 = 20

(c) Marginal Profit = - 0.25Q2 + 4Q = 0

=> Q(-0.25Q + 4) = 0

=> Either Q = 0

or , -0.25Q + 4 = 0

=> 0.25Q = 4

=> Q = 4 / 0.25 = 16

(d) Price = AR = 120 - 3Q = 120 - 3(16) = $72

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