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Math Help - Utility Function. Getting wrong anwswer.

  1. #1
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    Utility Function. Getting wrong anwswer.

    Greetings.

    Small problem,

    Q) We have U=(x^0.5)*y

    10y+4x^2=400

    and therefore, x=(40-2.5y)^0.5

    Goal is to find the units of y to maximize this the value of output.

    What I did:

    x=40-2.5y^0.5

    du/dy = [(40-2.5y^0.5)]^0.5 * y

    0=1/4(40-2.5y)^-3/4 * y (2.5)
    0=2.5y/(40-2.5y)^3/4

    The working seems incomplete but no matter what I do, I keep ending up with y=0. The answer is 3.20. How would we get this?

    Thank you for any help.
    Last edited by mmh9119; November 23rd 2012 at 05:21 PM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Utility Function. Getting wrong anwswer.

    I would use Lagrange multipliers. We have the objective function:

    U(x,y)=\sqrt{x}y

    subject to the constraint:

    g(x,y)=10y+4x^2-400=0

    Thus, we get the system:

    \frac{y}{2\sqrt{x}}=\lambda 8x

    \sqrt{x}=\lambda 10

    which implies:

    y=\frac{8}{5}x^2

    Substituting into the constraint, we find:

    16x^2+4x^2-400=0

    x^2=20,\,y=32
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    Re: Utility Function. Getting wrong anwswer.

    Thank you for the assist Mark. However, the course I'm taking does not use those methods above so it is not expected from us. Therefore I believe the solution could be found another way. Let me illustrate what I did another similar question:

    Q) V=54x+2y^2
    y=(280-x^3)^0.5
    Find the maximum value of x.

    V=54x+2[(280-x^3)^0.5]^0.5
    V=54x+560-2x^3

    dv/dx = 54-6x^2 =0
    6x^2=54
    x=3

    With the original question, I just have to find y using the equation given in x. I hope this helps.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Utility Function. Getting wrong anwswer.

    Well, we could solve the constraint for y to get:

    y=\frac{2(100-x^2)}{5}

    substitute into the objective function:

    U(x)=\sqrt{x}\left(\frac{2(100-x^2)}{5} \right)=\frac{2}{5}\left(100x^{\frac{1}{2}}-x^{\frac{5}{2}} \right)

    Equating the derivative to zero, we find:

    U'(x)=2\left(\frac{10}{x^{\frac{1}{2}}}-\frac{1}{2}x^{\frac{3}{2}} \right)=0

    This implies:

    \frac{10}{x^{\frac{1}{2}}}-\frac{1}{2}x^{\frac{3}{2}}=0

    20=x^2

    The second derivative test shows this gives a maximum, hence:

    y=\frac{2(100-20)}{5}=32
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    Re: Utility Function. Getting wrong anwswer.

    Thank you. One final thing though. The answer is 3.2, not 32. Any way to have it like that?
    Last edited by mmh9119; November 24th 2012 at 08:05 AM.
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