If each MP3 player costs the store $70, at what price should the store sell the MP3 players to maximize profit in the following problem?
A store sells portable MP3 players for $100 each and, at this price, sells 120 MP3 players every month. The owner of the store wishes to increase this profit, and he estimates that, for every $2 increase in this price of MP3 players, one less MP3 player will be sold each month.
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The store sells 120 MP3 players at $100 each. For every $2 increase in the price, the number of MP3 players sold is reduced by 1. The cost of one MP3 player is $70.
Let the price at which the MP3 players are sold to maximize profit be P. The number of MP3 players sold is 120 - [(P - 100)/2]
The profit per MP3 player is P - 70.
Total profit is (P - 70)(120 - (P - 100)/2)
=> 120P - P(P - 100)/2 - 120*70 + 70(P - 100)/2
=> 120P - P^2/2 + 50P - 8400 + 35P - 3500
=> -P^2/2 + 205P - 11900
To maximise the profit take the derivative of -P^2/2 + 205P - 11900 and solve for P.
-2P/2 + 205 = 0
=>-P + 205 = 0
=> -P = -205
=> P = 205
But only a price increase by multiples of 2 changes the number of MP3 players sold by a whole number. This makes the optimal price either $204 or $206.
The price to maximize profits is either $204 or $206
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