Question

The demand for product Q is given by Q = 136 -.4P and the total cost of Q by:

STC = 3000 + 40Q - 5Q^2 + 1/3Q^3

A. Find the price function and then the TR function. See Assignment 3 or 4 for an example.

B. Write the MR and MC functions below. Remember: MR = dTR/dQ and MC = dSTC/dQ. See Assignment 5 for a review of derivatives.

C.What positive value of Q will maximize total profit?  Remember, letting MR = MC signals the objective of total profit maximization. Solve MR = MC for Q.   The value of Q you get should not be zero or negative.

D. Use the price function found in (a) to determine the price per unit that will need to be charged at the Q found in (c). This will be the price you should ask for the total profit maximizing quantity.

E. What total profit will result from selling the quantity found in (c) at the price found in (d)? Remember, profit is TR - STC.

F. At what level of Q is revenue maximized? Remember, let MR = 0 and solve for Q. MR = 0 signals the objective of maximizing revenue.

G. 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.

H. 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).

Answer 1

A) Q = 136 - 0.4P

P = 340 - 2.5Q

TR = P x Q = 340Q - 2.5Q²

B) using the formulas given in the question:

MC = 40 - 10Q + Q²

MR = 340 - 5Q

C)

MR = MC:

340 - 5Q = 40 - 10Q + Q²

Q² - 5Q -300 = 0

Q = 20 (total profit would be maximized at this quantity)

D) P = 340 - 2.5 x 20 = 290

E) Total profit = TR - TC

Total profit = 290 x 20 - (3000 + 40 x 20 -5 x 20² + 20³/3)

Total profit = 1333.33

F) MR = 0

340 - 5Q = 0

Q = 68 (revenue is maximized at this quantity)

G) Total profit = 340Q - 2.5Q² - ((3000 + 40Q -5Q² + Q³/3)

MP = d(TP)/dQ = 340 - 5Q - 40 + 10Q - Q²

d(MP)/dQ = 0 = 5 - 2Q = 0

Q = 2.5

F) P = 340 - 2.5 x 2.5 = 333.75