PLEASE SHOW ALL WORK Suppose you are examining a monopolist firm, with the follo
ID: 1205049 • Letter: P
Question
PLEASE SHOW ALL WORK
Suppose you are examining a monopolist firm, with the following average revenue and average cost functions: AR(s) = -s^3 + 12s^2 - 30s + 1000 and AC(s) = 2s + 1000 - 100/s where s are the average sales per day. If the monopolist chooses to produce nothing (s=0), the total costs become negative. Write an expression for the Profits as a function of s. Determine the most profitable level of output. Show that this output level does indeed lead to a maximum rather than a minimum by checking the second order conditions.Explanation / Answer
(i)
Total revenue, TR = AR x s = - s4 + 12s3 - 30s2 + 1000s
Total cost, TC = AC x s = 2s2 + 1000s - 100
Profit, Z = TR - TC = - s4 + 12s3 - 30s2 + 1000s - (2s2 + 1000s - 100)
= - s4 + 12s3 - 30s2 + 1000s - 2s2 - 1000s + 100
= - s4 + 12s3 - 32s2 + 100
(b) Profit is maximized when dZ / ds = 0
- 4s3 + 36s2 - 64s = 0
Since s is not zero, simplying this we get
- 4s2 + 36s - 64 = 0
s2 - 9s + 16 = 0
Solving the quadratic using online solver, we get
s = 7 or s = 2 (Considering integral values)
When s = 7, Z = - s4 + 12s3 - 32s2 + 100 = - 2401 + 4116 - 1568 + 100 = 247
When s = 2, Z = - s4 + 12s3 - 32s2 + 100 = - 16 + 96 - 128 + 100 = 52
Profit is higher at s = 7. This is profit maximizing output.
Second order condition is:
d2Z / ds2 < 0
- 12s2 + 72s - 64 = 0
When s = 7, d2Z / ds2 = - 12 x 49 + 72 x 7 - 64 = - 588 + 504 - 64 = - 148 < 0
So, s = 7 satisfies the minima condition.
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