Results 1 to 8 of 8
Like Tree2Thanks
  • 1 Post By MarkFL
  • 1 Post By MarkFL

Thread: Marginal Cost derivative

  1. #1
    Member
    Joined
    Jun 2012
    From
    Australia
    Posts
    86

    Marginal Cost derivative

    I have an issue with a problem.

    I have the following function...

    $\displaystyle Cost: C^2 = x^2 + 100(x)^[1/2] + 50$

    I'm asked to find and interpret the marginal cost dC/dx at x = 5.

    Using implicit differentiation, which isn't really too hard because each variable is already on separate sides of the equation, I came up with the answer of

    $\displaystyle x + 25(x)^{-1/2} = C$.

    However, when I plug 5 into the derivative, I get an answer of about 16.18. I should be getting 0.94. Can someone give me a tip on what I've done wrong here?

    Thank you.
    Last edited by skeeter; Sep 27th 2012 at 03:08 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,005
    Thanks
    747

    Re: Marginal Cost derivative

    You neglected to use the chain rule when you differentiated. The right side of your equation after differentiation should be $\displaystyle C\cdot\frac{dC}{dx}$. Now solve for $\displaystyle \frac{dC}{dx}$ by dividing through by $\displaystyle C$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2012
    From
    Australia
    Posts
    86

    Re: Marginal Cost derivative

    Quote Originally Posted by MarkFL2 View Post
    You neglected to use the chain rule when you differentiated. The right side of your equation after differentiation should be $\displaystyle C\cdot\frac{dC}{dx}$. Now solve for $\displaystyle \frac{dC}{dx}$ by dividing through by $\displaystyle C$.
    I don't quite understand what you mean...The way I THOUGHT it was supposed to be worked through as is as follows...

    $\displaystyle c^2 = x^2 + 100(x)^{1/2} + 50$

    $\displaystyle = \frac{d}{dx}C = \frac {d}{dx} x^2 + \frac {d}{dx} 100(x)^{1/2} + \frac {d}{dx} 100$

    $\displaystyle = 2C \frac {dC}{dX} = 2x + \frac {50}{x^{1/2}}$

    $\displaystyle = \frac {dc}{dx} (C) = x + \frac {25}{x^{1/2}}$

    When you say divide through by $\displaystyle C$ what exactly do you mean by that? I thought that once the variables with $\displaystyle \frac {dc}{dx}$ are isolated on one side, that you then factor that out, and then divide the x-valued side, by the C valued side. If I were to do this, wouldn't I end up with this?

    $\displaystyle \frac {dC}{dx} = \frac {X}{C} + \frac {25C}{x^{1/2}}$

    I didn't think this would be correct because I'm supposed to solve for C. Or now do I plug in x=5, THEN isolate C?

    Still a bit lost...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,005
    Thanks
    747

    Re: Marginal Cost derivative

    You are given:

    $\displaystyle C^2=x^2+100x^{\frac{1}{2}}+50$

    Implicitly differentiating with respect to $\displaystyle x$, we find:

    $\displaystyle 2C\cdot\frac{dC}{dx}=2x+50x^{-\frac{1}{2}}$

    $\displaystyle \frac{dC}{dx}=\frac{x+25x^{-\frac{1}{2}}}{C}$

    Now, when $\displaystyle x=5$, we have $\displaystyle C=5\sqrt{4\sqrt{5}+3}$ hence:

    $\displaystyle \frac{dC}{dx}\left|_{x=5}=\frac{5+5 \sqrt{5}}{5\sqrt{4\sqrt{5}+3}}=\frac{1+\sqrt{5}}{ \sqrt{4\sqrt{5}+3}}\approx0.936349095209$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2012
    From
    Australia
    Posts
    86

    Re: Marginal Cost derivative

    Quote Originally Posted by MarkFL2 View Post

    Now, when $\displaystyle x=5$, we have $\displaystyle C=5\sqrt{4\sqrt{5}+3}$ hence:[/TEX]
    Your post makes sense, but I can't see where this value of C is coming from?

    Also, wouldn't $\displaystyle 25x^{-1/2} = 25(5)^{-1/2}$,

    and not $\displaystyle 25x^{-1/2} = 5(5)^{-1/2}$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,005
    Thanks
    747

    Re: Marginal Cost derivative

    To find $\displaystyle C$, use:

    $\displaystyle C=\sqrt{x^2+100x^{\frac{1}{2}}+50}$

    $\displaystyle 25(5)^{-\frac{1}{2}}=\frac{25}{\sqrt{5}}=5\sqrt{5}$
    Thanks from astuart
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jun 2012
    From
    Australia
    Posts
    86

    Re: Marginal Cost derivative

    Quote Originally Posted by MarkFL2 View Post
    To find $\displaystyle C$, use:

    $\displaystyle C=\sqrt{x^2+100x^{\frac{1}{2}}+50}$

    $\displaystyle 25(5)^{-\frac{1}{2}}=\frac{25}{\sqrt{5}}=5\sqrt{5}$
    Thanks so much for all the help (and patience)!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,005
    Thanks
    747

    Re: Marginal Cost derivative

    It takes no patience at all to help someone who shows their work/thoughts and wants to understand the problem!
    Thanks from astuart
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: Oct 5th 2010, 09:11 AM
  2. Marginal cost help
    Posted in the Business Math Forum
    Replies: 1
    Last Post: Aug 12th 2010, 01:38 PM
  3. Marginal Cost/Demand and Cost
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Nov 4th 2009, 06:45 PM
  4. Marginal revenue and the marginal cost
    Posted in the Business Math Forum
    Replies: 1
    Last Post: Nov 28th 2008, 05:22 PM
  5. Marginal Cost
    Posted in the Business Math Forum
    Replies: 4
    Last Post: Nov 11th 2007, 07:11 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum