# Math Help - Optimization money problem

1. ## Optimization money problem

A variety store can sell 500 yo-yos, which cost the store 35 cents each, for $1 each. For each penny the store lowers the price it can sell 20 more yo-yo's. For what price should it sell the yo-yos to make a maximum profit?? 2. Originally Posted by Mathforum A variety store can sell 500 yo-yos, which cost the store 35 cents each, for$1 each. For each penny the store lowers the price it can sell 20 more yo-yo's. For what price should it sell the yo-yos to make a maximum profit??
let x = number of 1 cent price reductions

profit = revenue - cost

profit = (number sold)(price each) - (number sold)(cost each)

$P = (500+20x)(1 - .01x) - (500+20x)(.35)$

maximize $P$ w/r to $x$

3. Originally Posted by Mathforum
A variety store can sell 500 yo-yos, which cost the store 35 cents each, for \$1 each. For each penny the store lowers the price it can sell 20 more yo-yo's. For what price should it sell the yo-yos to make a maximum profit??
We express it as a function...

Profit = Amountx(Sale Price) - Amountx0.35 = Amount((Sale Price) - 0.35)

So we need mathematical representations for Sale Price and amount.

Lets just call Sale Price x.

We know that for x = 1, Amount = 500
For x = 0.99, Amount = 520 etc

So we model the amount sold as...

500 + 2000(1-x)

Why this?

Well we start of with 500 at x=1, for every penny less, we add 20 more. So for x=0.98 we have +40 yoyos
x = 0.97 we have +60 yoyos
etc

Lets look at 1-x to see what happens...

for x=0.98, 1-x = 0.02 we have +40 yoyos
for x = 97 we have 1-x = 0.03 and get +60 yoyos

It can now been seen that 1-x multiplied by 2000 = the extra you sell.

The formula also works when you raise the price.

x=1.01,

We get 500 - (0.01)2000 = 500 - 20 = 480 as anticipated.

So anyway, We now have amount = 500 + 2000(1-x)

Lets put this back into our original formula which we shall call P(x) where x was the sale price.

P(Sale Price) = Amount((Sale Price) - 0.35)

=> P(x) = Amount(x - 0.35)

=> P(x) = (500 + 2000(1-x))(x - 0.35) (now lets expand it out)

= 500x - 175 + 2000x(1-x) - 750(1-x)

=> $P(x) = 3200x-875-2000x^2$

Find the derivative of this, set it equal to 0. Should be your answer.

4. Originally Posted by skeeter
let x = number of 1 cent price reductions

profit = revenue - cost

profit = (number sold)(price each) - (number sold)(cost each)

$P = (500+20x)(1 - .01x) - (500+20x)(.35)$

maximize $P$ w/r to $x$

Lol that's a lot simpler. Ah well...