Suppose that you can sell as much of a product (in integer units) as you like at
ID: 1225348 • Letter: S
Question
Suppose that you can sell as much of a product (in integer units) as you like at $60 per unit. Your marginal cost (MC) for producing the qth unit is given by: MC=7q This means that each unit costs more to produce than the previous one (e.g., the first unit costs 7*1, the second unit (by itself) costs 7*2, etc.). If fixed costs are $100, what is the profit at the optimal output level? Please specify your answer as an integer. Also, assume that a competitive firm has the total cost function: TC = 1q3 - 40q2 + 840q + 1800 Suppose the price of the firm's output (sold in integer units) is $750 per unit. Using tables (but not calculus) to find a solution, what is the total profit at the optimal output level? Please specify your answer as an integer.
Explanation / Answer
(1) Price = $60 and demand curve is horizontal at this price. In this case profit is maximizing be equating P with MC:
7q = 60
q = 60 / 7 = 9
Total cost, TC = MC x q + Fixed cost = (7q x q) + 100
Profit = Revenue - Total cost = (P x q) - [(7q x q) + 100]
= ($60 x 9) - $[(7 x 9 x 9) + 100] = $540 - $(567 + 100) = $(540 - 667) = - $127 (Loss)
(2) TC = q3 - 40q2 + 840q + 1800
MC = dTC / dq = 3q2 - 80q + 840
In profit-maximizing equilibrium, P = MC
3q2 - 80q + 840 = 750
3q2 - 80q + 90 = 0
Solving this quadratic equation using online solver,
q = 25 or 1 = 1
When q = 25, Profit = Revenue - TC = (P x q) - (q3 - 40q2 + 840q + 1800)
= (750 x 25) - [(25 x 25 x 25) - (40 x 25 x 25) + (840 x 25) + 1800]
= 18,750 - (15,625 - 25,000 + 21,000 + 1800)
= 18,750 - 13,425
= 5,325
When q = 1, Profit = (750 x 1) - [(1 x 1 x 1) - (40 x 1 x 1) + (840 x 1) + 1800]
= 750 - (1 - 40 + 840 + 1800)
= 750 - 2,601
= - 1,851 (Loss)
So, profit is maximized when q = 25 with profit = $5,325
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.