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Thread: deravitive and profit analysis

  1. #1
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    deravitive and profit analysis

    The question is: A fast food restaurant determines the cost and revenue models for its hamburgers to be:

    c = .06X = 7500 ; 0 <= X <= 50000

    r= (1/20000)(65000x - x^2) ; 0 <= X <= 50000

    a. Write the profit function for this situation

    b. determine the intervals the function decreases / increases

    c. determine how many hamburgers the restaurant needs to sell to obtain max profit.

    __
    I have found the profit function to be p= r-c

    so I got : P = (1/20000)(65000x - x^2) - (.6x + 7500)

    and P'= (65000x - x^2) + (1/20000)(65000 - x) - .6


    but this is not the answer in the book. I'm not sure what I am doing wrong.

    Thanks!
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  2. #2
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    Here's a hint: if $\displaystyle p=r-c=\frac{1}{20000}(65000x-x^2)-0.6x-7500$, then

    $\displaystyle
    \begin{aligned}
    p'&=\frac{d}{dx}\left(\frac{1}{20000}(65000x-x^2)-0.6x-7500\right) \\
    &=\frac{1}{20000}(65000-2x)-0.6. \\
    \end{aligned}
    $

    As $\displaystyle p=0$ at the endpoints of the domain $\displaystyle x=0$ and $\displaystyle x=50000$, we know that the maximum profit will be attained at a point where

    $\displaystyle p'=0$.
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  3. #3
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    Quote Originally Posted by craziebbygirl View Post
    The question is: A fast food restaurant determines the cost and revenue models for its hamburgers to be:

    c = .06X = 7500 ; 0 <= X <= 50000

    r= (1/20000)(65000x - x^2) ; 0 <= X <= 50000

    a. Write the profit function for this situation

    b. determine the intervals the function decreases / increases

    c. determine how many hamburgers the restaurant needs to sell to obtain max profit.

    __
    I have found the profit function to be p= r-c

    so I got : P = (1/20000)(65000x - x^2) - (.6x + 7500)

    and P'= (65000x - x^2) + (1/20000)(65000 - x) - .6


    but this is not the answer in the book. I'm not sure what I am doing wrong.

    Thanks!
    first of all, is it ...

    $\displaystyle c = .06x + 7500$ or

    $\displaystyle c = .6x + 7500$ ?


    you set your profit function up correctly, but your derivative is incorrect.

    If $\displaystyle P = (1/20000)(65000x - x^2) - (.6x + 7500)$ , then

    $\displaystyle P' = (1/20000)(65000 - 2x) - .6$
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  4. #4
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    yes it is a .6

    So you do not need to do a product rule for th (1/20000)(65000 - 2X)? I thought that you needed to do the (f' of g) + (g' of f) for that.
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  5. #5
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    Quote Originally Posted by craziebbygirl View Post
    yes it is a .6

    So you do not need to do a product rule for th (1/20000)(65000 - 2X)? I thought that you needed to do the (f' of g) + (g' of f) for that.
    $\displaystyle \frac{1}{20000}$ is just a konstant.

    $\displaystyle \frac{d}{dx}[k \cdot f(x)] = k \cdot f'(x)
    $
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  6. #6
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    When you use the product rule, the result is

    $\displaystyle
    \begin{aligned}
    \frac{d}{dx}\left(\frac{1}{20000}(65000x - x^2)\right) &= \frac{1}{20000}(65000 - 2x)+0\cdot(65000x - x^2) \\
    &= \frac{1}{20000}(65000 - 2x).\\
    \end{aligned}$

    The general rule for derivatives of constant multiples of functions is

    $\displaystyle \frac{d}{dx}(af(x))=af'(x).$
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