# determining a profit function

• Mar 25th 2011, 03:39 PM
ASUSpro
determining a profit function
I am stuck on this, i thought this is the right method used. Can anyone help me see what i did wrong?

Terry makes and sells necklaces. He has observed over time that when the price is \$ 9 each, he sells an average of 16 per day. If he increases the price, then his average sales fall by 2 per day for each dollar increase. The materials for each necklace cost \$7. Express his profit P as a function of x, the number of necklaces sold.

What i did was:
P(x)=R(x)-C(x)
P(x) = (9+x)(16-2x) - 7(16-2x)
144 - 18x + 16x -2x^2 -112 + 14x

= 32+12x-2x^2
• Mar 25th 2011, 04:25 PM
HallsofIvy
Quote:

Originally Posted by ASUSpro
I am stuck on this, i thought this is the right method used. Can anyone help me see what i did wrong?

Terry makes and sells necklaces. He has observed over time that when the price is \$ 9 each, he sells an average of 16 per day. If he increases the price, then his average sales fall by 2 per day for each dollar increase. The materials for each necklace cost \$7. Express his profit P as a function of x, the number of necklaces sold.

What i did was:
P(x)=R(x)-C(x)
P(x) = (9+x)(16-2x) - 7(16-2x)
144 - 18x + 16x -2x^2 -112 + 14x

= 32+12x-2x^2

The problem is that "x" here is NOT the number of necklaces sold. It is the number of dollars above 9 that the price is: so that the price is 9+ x and the number of necklaces sold is 16- 2x.

You might do this- first, to leave x available, change your x to "y":
P(y)= (9+ y)(16- 2y)- 7(16- 2y). Now let x be the number of necklaces sold: x= 16- 2y so that so that y= (16- x)/2 and 9+ y= (16- x)/2+ 18/2= (34- x)/2. Now you will have P(x)= [(34-x)/2][x]- 7x.
• Mar 25th 2011, 05:01 PM
scounged
Ok, let's see here:
If the price is too high, he won't get any neclaces sold, and thus, no profit. And if the price is low, he will naturally sell a bunch of necklaces, but he won't make a profit. So, if the profit is the function, and the number of necklaces sold will be the x-value, we may be able to determine a thing or two.

\$\displaystyle \mbox{First, let's find the solutions to the equation}~f(x)=0\$
\$\displaystyle \mbox{as the function will be a second degree function,}\$
\$\displaystyle \mbox{we know that the equation above will have two solutions, and these are:}~x=20~and~x=0 \$
\$\displaystyle \mbox{This means that}~k(x-20)(x)=f(x)\$

After this, just solve for k and expand.