Question

5) Maximum profit: You are the manager for a telephone stores. From the previous experience, you know you can sell 12 telephone answering machines per day at a price of $200 each. You estimate that for each $10 price reduction you can sell 2 more per day. The answering machines cost the store $80 each. a) Find the price and the quantity sold per day to maximize the profit, b) What's the maximum profit per day?

Answer 1

SOLUTION :

Let the price be P and quantity sold be Q .

P < 12 .

Additional quantity sold = (200 - P) / 10 * 2 = 40 - 0.2P

So, Q = 12 + (40 - 0.2P) = 52 - 0.2P 

=> P = (52 - Q) / 0.2 = 260 - 5Q 

Revenue, R = P * Q = (260 - 5Q) Q = 260Q - 5Q^2

Cost, C = Q * cost per piece = Q * 80 = 80 Q

So, 

Profit, π 

= R - C

= 260Q - 5Q^2 - 80Q 

= 180Q - 5Q^2

For max. Profit, dπ / dQ = 0

=> dπ / dQ = 180 - 10Q = 0

=> Q = 18 

D2π / dQ2 = - 10 (negative value) . 

Hence ew have maximum value of profit at Q = 18.

=> P = 260 - 5*18 = 170 

So. For maximum profit, price = $170 and quantity = 18 per day (ANSWER).

Maximum profit = 180*18 - 5*18^2 = 1620 ($) 9ANSWER). 

ALTERNATELY :  Maximum profit = 18*170 - 18*80 = 1620 ($)  (ANSWER).