harsh.champ wrote:A store owner plans to sell two types of MP3 players. One type costs the store owner $200 each; the other type costs $400 each. The $200 models yield a profit of $25 dollars each and the $400 models a profit of $30 each. The store owner estimates that the total demand for the MP3 players will not exceed 300 units. The store owner can invest up to $80,000 on the inventory of MP3 players. The store owner sells some models of each type to maximize the profit. What is the value of the maximum possible profit the store owner can make?


(A)$8000
(B)$7000
(C)$9000
(D)$7500
(E)$6000

MP3 Player: 1st player 2nd player
Cost $200 $400
Profit $25 $30

Total Demand <= 300, Investment <= 80000
--> X+Y<=300 -------------------------------------(i)
--> X(200)+Y(400)<=80000 --> X+2Y<=400 ---------(ii)

From (ii) - (i), we get Y<=100.
Now since Y gives more profit i.e. $30, so we will consider its max value i.e. 100.
so X= 300-100 = 200 for max profit.

Hence max profit = 25X + 30Y = 25*200 + 30*100 = 5000 + 3000 = $8000.

Ans is [spoiler](A)$8000.[/spoiler]

harsh.champ wrote:A store owner plans to sell two types of MP3 players. One type costs the store owner $200 each; the other type costs $400 each. The $200 models yield a profit of $25 dollars each and the $400 models a profit of $30 each. The store owner estimates that the total demand for the MP3 players will not exceed 300 units. The store owner can invest up to $80,000 on the inventory of MP3 players. The store owner sells some models of each type to maximize the profit. What is the value of the maximum possible profit the store owner can make?


(A)$8000
(B)$7000
(C)$9000
(D)$7500
(E)$6000

Say they sell x 200 dollar MP3 players and y 400 dollar models
we want to maximize the profits = 25x + 30y
x+y<= 300
x*200+y*400 <= 80000
x+2y <= 400

Now he is can buy all 200 dollar MP3s he can buy 300 profit will be 300*25 = 7500
or
He can buy all 200 dollar MP3s he can buy 200 profit will be 200*30 =6000

Or he can buy 200, 200 dollar Mp3 players and 100, 400 dollar mp3 [ by solving x+y= 300 and x+2y = 400]
Profit in this case = 200*25 + 100*30 = 8000

A
[spoiler]
This is best solved by something called Linear programming and that way out of scope[/spoiler]