# Practical Optimization

• Mar 10th 2007, 10:30 AM
confusedagain
Practical Optimization
A bookstore can obtain a certain book from the publisher at a cost of \$3 per book. The bookstore has been offering the book at a price of \$15 per copy and, at this price, has been selling 200 copies a month. The store is planning on lowering the price to stimulate sales and estimates that for each \$1 reduction in the price, 20 more books will be sold each month. At what price should the bookstore sell the book to generate the greatest possible profit?

Bare with me
the profit is the # of books sold times profit per book
the number of books sold is (200 + 20)(number of \$1 decreases)

200 + 20(x+15)
20[10-(x+15)]
20[-x-5]

Profit per book is (x-3)
P'(x) = 20[(-x-5)(x-3)]
P'(x) = 20(-x-5)(1) + (x-3)(-1)
P'(x) = 20(-x-5-x+3)
P'(x) = 20(-2x-2)
Setting this equal to 0 gives you x = -1

Subtracting \$1 from the current price of \$15 gives \$14
P(14) = 20(-14-5)(14-3)
P(14) = 20(-19)(11)
P(14) = -4180

That can't be correct, its negative.
• Mar 10th 2007, 10:50 AM
CaptainBlack
Quote:

Originally Posted by confusedagain
A bookstore can obtain a certain book from the publisher at a cost of \$3 per book. The bookstore has been offering the book at a price of \$15 per copy and, at this price, has been selling 200 copies a month. The store is planning on lowering the price to stimulate sales and estimates that for each \$1 reduction in the price, 20 more books will be sold each month. At what price should the bookstore sell the book to generate the greatest possible profit?

Bare with me
the profit is the # of books sold times profit per book
the number of books sold is (200 + 20)(number of \$1 decreases)

200 + 20(x+15)
20[10-(x+15)]
20[-x-5]

Profit per book is (x-3)
P'(x) = 20[(-x-5)(x-3)]
P'(x) = 20(-x-5)(1) + (x-3)(-1)
P'(x) = 20(-x-5-x+3)
P'(x) = 20(-2x-2)
Setting this equal to 0 gives you x = -1

Subtracting \$1 from the current price of \$15 gives \$14
P(14) = 20(-14-5)(14-3)
P(14) = 20(-19)(11)
P(14) = -4180

That can't be correct, its negative.

Sales=200 + 20(15 - Price)

Profit = Sales (Price - 3) = (500 - 20 Price)(Price-3)

........= -20 Price^2 + 560 Price - 1500

so d(Profit)/dt = -40 Price + 560, hence if d(Profit)/dt=0, Price = 14 dollars.

RonL
• Mar 10th 2007, 11:00 AM
confusedagain
Wouldn't that increase the price?
The store is trying to decrease the price of the book from the \$15 it's at now. Wouldn't selling the book at \$20 be an increase?
• Mar 10th 2007, 11:18 AM
ticbol
Quote:

Originally Posted by confusedagain
A bookstore can obtain a certain book from the publisher at a cost of \$3 per book. The bookstore has been offering the book at a price of \$15 per copy and, at this price, has been selling 200 copies a month. The store is planning on lowering the price to stimulate sales and estimates that for each \$1 reduction in the price, 20 more books will be sold each month. At what price should the bookstore sell the book to generate the greatest possible profit?

Bare with me
the profit is the # of books sold times profit per book
the number of books sold is (200 + 20)(number of \$1 decreases)

200 + 20(x+15)
20[10-(x+15)]
20[-x-5]

Profit per book is (x-3)
P'(x) = 20[(-x-5)(x-3)]
P'(x) = 20(-x-5)(1) + (x-3)(-1)
P'(x) = 20(-x-5-x+3)
P'(x) = 20(-2x-2)
Setting this equal to 0 gives you x = -1

Subtracting \$1 from the current price of \$15 gives \$14
P(14) = 20(-14-5)(14-3)
P(14) = 20(-19)(11)
P(14) = -4180

That can't be correct, its negative.

If the profit is negative, then the bookstore loses money. :)

Look at this solution.

Profit = Revenue minus Cost
P = R -C
P(t) = R(t) -C(t) --------all sub-t's, or all are functions of time t.

Initially, or at t=0,
P(0) = 200(15) -200(3) = \$2400 a month.

----------------------
At t = x number of \$1-reducing of the initial \$15 price,

Cost per book = \$3 still
Selling price per book = \$15 -x(\$1) = (15 -x) dollars
Number of books sold = 200 +x(20) = (200 +20x) books
So,
P(x) = R(x) -C(x)
P(x) = (200 +20x)(15-x) -(200 +20x)(3)
P(x) = (200 +20x)[(15 -x) -3]
P(x) = (200 +20x)(12 -x) -------------(1)
Differentiate both sides with respect to x,
P'(x) = (200 +20x)[-1] +[20](12 -x)
P'(x) = -200 -20x +240 -20x
P'(x) = 40 -40x
Set P'(x) to zero,
0 = 40 -40x
x = 1 reducing of the initial selling price.
Meaning, maximum profit would be at one time reducing only.

Therefore, for maximum profit, selling price should be (15 -1 =)...\$14 per book. --------answer.
• Mar 10th 2007, 11:32 AM
Soroban
Hello, confusedagain!

Your setup is incorrect . . .

Quote:

A bookstore can obtain a certain book from the publisher at a cost of \$3 per book.
The bookstore has been offering the book at a price of \$15 per copy and,
at this price, has been selling 200 copies a month.
The store is planning on lowering the price to stimulate sales and estimates that,
for each \$1 reduction in the price, 20 more books will be sold each month.
At what price should the bookstore sell the book to generate the greatest possible profit?

Let x = number of one-dollar reductions.

Then each book will sell for: 15 - x dollars each.
. . The profit per book is: .15 - x - 3 .= .12 - x dollars each.
And 200 + 20x books will be sold.

The profit function is: .P(x) .= .(12 - x)(200 + 20x) .= .2400 + 40x - 20x²

Then: .P'(x) .= .40 - 40x .= .0 . . x = 1

They should reduce the price by \$1 per book to \$14 per book.
. . Their profit will be \$11 per book.
They will sell 200 + 20 = 220 books
. . for a total profit of: .220 x 11 .= .\$2,420 (maximum).

• Mar 10th 2007, 11:42 AM
CaptainBlack
Quote:

Originally Posted by confusedagain
The store is trying to decrease the price of the book from the \$15 it's at now. Wouldn't selling the book at \$20 be an increase?

There was a transcription error which propagated through when I was working the problem now corrected.

RonL