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Math Help - Differentiation - Application

  1. #1
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    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!
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  2. #2
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    x=0 is a solution, what does this mean in terms of the problem? What is the expected profit?

    For other solutions

    \sin x+x\cos x = 0

    x\cos x = -\sin x

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

    x = -\tan x

    Solve this numerically
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  3. #3
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    Thanks for the response, but i still dont understand how x=0 can be used to solve the problem.
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  4. #4
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    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.
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  5. #5
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    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.
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