The Savemore Home Improvement Center has an exclusive franchise for distribution
ID: 1206156 • Letter: T
Question
The Savemore Home Improvement Center has an exclusive franchise for distribution of Lawn King lawn mowers in the Kansas City metro area. Lawn King lawn mowers are generally acknowledged to be the best in the industry and there are no close substitutes. A market-research firm engaged by Savemore has estimated the monthly market demand for Lawn King lawn mowers to be: Q = 2000 - 5P. Savemore estimates its total monthly cost for purchasing, warehousing, and marketing the Lawn King line to be: C(Q) = 100 + 4Q + 0.4 Q^2 (a) How many Lawn King lawn mowers should the firm purchase and sell in order to maximize profit? What should the selling price be, and how much profit can be made? (b) If Savemore lowers the price for the Lawn King mower to 300, can a profit be made? If so, how much? (c) If Savemore wants to maximize revenue, what should it do about price? What is the revenue maximizing price? How many lawn-mowers will be sold at this price? Would maximization of revenue still be profitable? (d) Savemore has decided to run a month-long sale of Lawn King lawn mowers as a loss leader to promote sales of other merchandise in its store. They want to sell the mowers "at cost, but not below cost." What should the sale price be?Explanation / Answer
Demand: Q = 2000 - 5P
5P = 2000 - Q
P = 400 - 0.2Q
(a) Profit is maximized by equating marginal revenue (MR) with marginal cost (MC).
Total revenue, TR = P x Q = 400Q - 0.2Q2
MR = dTR / dQ = 400 - 0.4Q
MC = dTC / dQ = 4 + 0.8Q
Equating,
400 - 0.4Q = 4 + 0.8Q
1.2Q = 396
Q = 330
P = 400 - (0.2 x 330) = 400 - 66 = 334
TC = 100 + (4 x 330) + (0.4 x 330 x 330) = 100 + 1,320 + 43,560 = 44,980
TR = P x Q = 334 x 330 = 110,220
Profit = TR - TC = 110,220 - 44,980 = 65,240
(b) When P = 300,
Q = 2000 - (5 x 300) = 2000 - 1500 = 500
TR = 300 x 500 = 150,000
TC = 100 + (4 x 500) + (0.4 x 500 x 500) = 100 + 2,000 + 100,000 = 102,100
Profit = 150,000 - 102,100 = 47,900
(c) Revenue is maximized when MR = 0
400 - 0.4Q = 0
0.4Q = 400
Q = 1,000
P = 400 - (0.2 x 1,000) = 400 - 200 = 200
TR = 200 x 1,000 = 200,000
TC = 100 + (4 x 1,000) + (0.4 x 1,000 x 1,000) = 100 + 4,000 + 400,000 = 404,100
Profit = 200,000 - 404,100 = - 204,100 (Loss)
(d) In this case, TR = TC
400Q - 0.2Q2 = 100 + 4Q + 0.4Q2
0.6Q2 - 396Q + 100 = 0
Solving this quadratic equation using online solver,
Q = 659 or Q = 0.25 (Which is negligible, so dismissed).
When Q = 659, P = 400 - (0.2 x 659) = 400 - 131.8 = 268.2
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