# determine the functions for profit

• January 28th 2007, 05:39 PM
helen
determine the functions for profit
a company produces books for a cost of $15 each, and sells the books to bookstores for$40 each.for each one dollar increase in price of books , sales decrease by 100 per week. the firm currently sells 1000 books per week at a price of $40 each. show all calcilations in your work. a) write the total cost function C(x) b) write the total revenue function R(x) c) write the profit function p(X) d) determine the profit maximizing output (x) e) compute the profit maximizing price ( p) • January 29th 2007, 06:30 AM earboth Quote: Originally Posted by helen can you please help me with this one? a company produces books for a cost of$15 each, and sells the books to bookstores for $40 each.for each one dollar increase in price of books , sales decrease by 100 per week. the firm currently sells 1000 books per week at a price of$40 each. show all calcilations in your work.

a) write the total cost function C(x)
b) write the total revenue function R(x)
c) write the profit function p(X)
d) determine the profit maximizing output (x)
e) compute the profit maximizing price ( p)

Hello, Helen,

I'm not quite certain if I understand the words correctly. So to the parts A9 to c) don't trust me:

Let x be the number of books and
let y be the price of one book and
let c be the costs for one book. Then

a) C(x) = c * x
That means here: $C(x)=15 \cdot x$

b) R(x) = y * x
That means here: $R(x)=40 \cdot x$

c) The profit is the difference between revenue and costs:
p(x) = r(x) - C(x)
That means here: $p(x)=40 \cdot x - 15 \cdot x=25x$

I'l do now d) first:

Let a be the amount of the augmentation of the price. Then

$x = 1000 - 100a \Longrightarrow a= 10-\frac{1}{100}x$
$y = 40+a \Longrightarrow y= 40-\frac{1}{100}x+10= 50-\frac{1}{100}x$

The costs per book doesn't change. So the profit is calculated by:

$p(x)=\left(50-\frac{1}{100}x \right) \cdot x - 15 \cdot x=50x-\frac{1}{100}x^2-15x=-\frac{1}{100}x^2+35x$

You get the maximum if the first derivative of this function is zero:

$p'(x)=-\frac{2}{100}x+35$
$0=-\frac{2}{100}x+35$. Solve for x. You'll get 1750 Books.

Now plug in this value to calculate the maximizing price:

$y = 50-\frac{1}{100} \cdot 1750 = 32.50$

EB
• January 29th 2007, 11:13 AM
Soroban
Hello, Helen!

The problem is poorly worded . . .

Quote:

A company produces books for a cost of $15 each, and sells the books to bookstores for$40 each.
For each one dollar increase in price of books , sales decrease by 100 per week.
The firm currently sells 1000 books per week at a price of $40 each. a) Write the total cost function: $C(x)$ What is $x$? The number of books? .The price per book? .The number of$1 increases?

I am familiar with this type of problem.

Let $x$ = number of $1 increases in the price of each book. The price per book is: $40 + x$ dollars. The number of books produced/sold is: $1000 - 100x \:=\:100(10-x)$ $100(10 - x)$ books will cost$15 each.
. . $C(x)\;=\;1500(10-x)$

Quote:

b) Write the total revenue function: $R(x)$
$100(10-x)$ books will bring in $(40 + x)$ dollars each.
. . $R(x) \;=\;100(10-x)(40 + x)$

Quote:

c) Write the profit function: $P(x)$
$\text{Profit }\;=\;\text{Revenue } - \text{ Cost}$

$P(x)\;=\;100(10-x)(40+x) - 1500(10-x)$

which simplifies to: . $P(x)\;=\;100(250-15x-x^2)$

Quote:

d) Determine the profit-maximizing output.
To maximize $P(x)$ solve the equation: $P'(x)= 0$

$P'(x) \;= \;100(-15-2x) \;= \;0\quad\Rightarrow\quad x = -\frac{15}{2}$

Therefore, we should decrease the price by $\7.50$

The output will be: . $100[10 - (-7.5)] \:=\:1750$ books.

Quote:

e) Compute the profit-maximizing price.
The price will be: . $40 + (-7.5) \:=\:\32.50$ per book.