Results 1 to 6 of 6

Math Help - deravitive and profit analysis

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    12

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    Here's a hint: if p=r-c=\frac{1}{20000}(65000x-x^2)-0.6x-7500, then

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

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

    p'=0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,623
    Thanks
    428
    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 ...

    c = .06x + 7500 or

    c = .6x + 7500 ?


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

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

    P' = (1/20000)(65000 - 2x) - .6
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2009
    Posts
    12
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,623
    Thanks
    428
    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.
    \frac{1}{20000} is just a konstant.

    \frac{d}{dx}[k \cdot f(x)] = k \cdot f'(x)<br />
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    When you use the product rule, the result is

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

    The general rule for derivatives of constant multiples of functions is

    \frac{d}{dx}(af(x))=af'(x).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How much profit?
    Posted in the Math Puzzles Forum
    Replies: 2
    Last Post: March 15th 2010, 08:22 AM
  2. Profit >=0
    Posted in the Business Math Forum
    Replies: 0
    Last Post: April 21st 2009, 09:48 AM
  3. profit and marginal profit relationship
    Posted in the Business Math Forum
    Replies: 1
    Last Post: January 16th 2009, 08:00 AM
  4. profit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 28th 2008, 10:44 PM
  5. Profit max
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 10th 2007, 12:24 PM

Search Tags


/mathhelpforum @mathhelpforum