B. Simple Properties of Cost and Profit Functions Suppose total cost, denoted C,

Question

B. Simple Properties of Cost and Profit Functions Suppose total cost, denoted C, is given as a function of Q to be: 120Q-Q+0.02Q, where Q is output, and price, denoted P, is given by the linear relationship: P = 1 14-0.25Q. 1. Write the equation for cost per unit of output or average cost (AC), denoted C/o 2. Differentiate total cost with respect to Q to compute marginal cost (MC). Differentiate CiQ w.r.t. Q and compute the level of Q that minimizes average cost. show that MC = AC at this level of output. 3. Using the demand relationship, write profit, defined to be PQ-C, as a function of Q. 4. Differentiate this profit function with respect to Q and set your answer equal to zero 5. Solve for the two levels of Q that satisfy the above first-order condition for profit maximization. Hint: Multiply by 100/6 to convert this quadratic for ease of factoring. 6. Determine which of the two levels of Q represents maximum profit by checking the second-order condition for a maximum.

Explanation / Answer

TC= 120Q - Q² + 0.02Q³
AC=TC/Q = 120 - Q + 0.02Q²
AC[MIN] AC/Q = 0
AC/Q = 0.04Q-1 = 0
0.04Q=1
Q= 1/0.04 = 25

B)
Profit MAX if MR=MC
MR=TR/Q
MC=TC/Q
TC/Q = 120-2Q+0.06Q²
TR=P*Q = 114Q-0.25Q²
TR/Q = 114 - 0.5Q
114 - 0.5Q = 120-2Q+0.06Q²
Quadratic equation solutions are:
Q1 = 5 (Second order condition fails here, thus this solution is wrong)
Q2 = 20 this solution maximizes profit

Answ: no, this level (Q=25) does not provides maximum profit, maximum profit is at quantities Q=20.

Q ..... P ............. TR .......... TC ........ Profit
5 ... 112.75 ..... 563.75 ..... 577.50 .... -13.75
20 .. 109.00 ... 2'180.00 .. 2'160.00 ..... 20.00 profit maximizing
25 .. 107.75 ... 2'693.75 .. 2'687.50 ....... 6.25

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