# Differentiation - Application

• Dec 25th 2010, 11:52 AM
elder
Differentiation - Application
Hello! I would be incredibly thankful if somebody could show step by step how they solve this question:

A new car hire company has just opened and wants to make the maximum amount of profit. The amount of profit, P, is dependent on two factors; x, the number of tens of cars rented; and y, the distance travelled by each car. The profit is given by the formula: P=xy+40, where y=sinx. Find the smallest number of cars that the comapny needs to rent to give the maximum amount of profit.

This is how i've done it so far:

P = xy+40
P = x*sinx+40
dp/dx=sinx+xcosx (product rule)
sinx+xcosx = 0
x =0

My answer is obviously wrong, so I'd appreciate some help!

Thanks!
• Dec 25th 2010, 12:06 PM
pickslides
x=0 is a solution, what does this mean in terms of the problem? What is the expected profit?

For other solutions

$\displaystyle \sin x+x\cos x = 0$

$\displaystyle x\cos x = -\sin x$

$\displaystyle x = \frac{-\sin x}{\cos x}$

$\displaystyle x = -\tan x$

Solve this numerically
• Dec 25th 2010, 12:37 PM
elder
Thanks for the response, but i still dont understand how x=0 can be used to solve the problem.
• Dec 25th 2010, 12:58 PM
dwsmith
P=xsin(x)+40

P(0)=0*0+40=40

Seems strange but it is better for this business to stay out of business.

Also, there are other solutions which are 2 pi periodic.
• Dec 26th 2010, 01:05 AM
HallsofIvy
Setting the derivative equal to 0 will find values of x that make the expression a maximum or a minimum. x= 0 gives minimum profit.