Topic: For all you mathematicians
 Message: Posted by: ryay (May 5, 2005 08:17PM)
Mike sells cars. The profit he makes each day can be modeled by the equation:

P(x)=-x^2+240x-700

P stands for the profit and x stands for the price of each car he sells. for what values of x will mike make a profit?
 Message: Posted by: Nir Dahan (May 6, 2005 07:58AM)
This sounds like a puzzle taken from junior high algebra book...
calculate the 2 roots and the solution is everything in between. (the function gets a maximum at x=120)
N.
 Message: Posted by: Nir Dahan (May 6, 2005 08:00AM)
This sounds like a puzzle taken from junior high algebra book...
calculate the 2 roots and the solution is everything in between. (the function gets a maximum at x=120)
N.
 Message: Posted by: leonard (May 6, 2005 08:20AM)
Why would you sell the 121st car, if you are going to lose money? Stop at 120, collect your 13,700 (dollars?), and go home.
 Message: Posted by: Jonathan Townsend (May 6, 2005 02:35PM)
That sounds like a nice way to compute optimal price, though something is off about the demand/supply curves implied there.

IF you sell a car for more then P(120) you start losing money?
 Message: Posted by: GeorgeG (May 6, 2005 03:46PM)
It's easier with simple calculus, get the first differential set to zero, P'(x)=0 so the equation becomes -2x + 240 = 0, and x=120 for a maximum or minimum; P''(x)= -2, so the curve is concave downwards for a maximum. There must be some inherent cost issue with selling more than 120 cars, though I doubt sales and profit has a bell shaped curve. Any business majors in the know?
 Message: Posted by: Heinz Weber (May 17, 2005 10:09AM)
Hi ryay,

I think you want the magic Café to do your homework, right?

;-))
Heinz
 Message: Posted by: idris (May 18, 2005 12:04AM)
The question doesn't say number of cars but dollar value of each car sold.

Therefore he will show a profit with cars which sell for \$3 to \$237. (approximate) with the highest value when he sells a car for \$120 with a profit of \$13700. Boy, I wish I could get a job like that. <G>

Jerry
 Message: Posted by: TomasB (May 18, 2005 03:43AM)
Instead of calculating roots and differentials to find the optimum price you can rewrite the formula by "completing the square". You want to end up with something like b-(x-a)^2 which is the same as b-x^2+2xa-a^2 so you see that the constant a we look for is half of what's in front of x giving us

P(x)=13700-(x-120)^2

By just looking at that you see that the maximum value of P is 13700 and that it happens exactly when the price of a car is 120.