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Math Help - multivariable function maximisation and (?) Lagrange multiplier

  1. #1
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    multivariable function maximisation and (?) Lagrange multiplier

    This is a sample exam question to which I don't have a model answer. I'd appreciate your feedback.


    Question.

    It is thought that a consumer measures the utility u of posessing a quantity x of apples and a quantity y of oranges by the formula:

    u=u(x,y)=x^{\alpha}y^{1-\alpha}.

    It is know that, when a consumer's budget for apples and oranges is $1, he will buy 2 apples and 1 orange when they are equally priced. Find \alpha.

    The price of oranges falls to half that of apples with the price of apples unchanged. How many apples and oranges will the consumer buy for $10?

    [Hint. First solve a utility maximisation problem with \alpha as a parameter.


    Answer.

    First part, finding alpha.

    I will maximise u(x,y)==x^{\alpha}y^{1-\alpha} by
    first - find critical points by setting the first derivative to zero, then
    second - evaluate the second derivative at this critical point(s)

    \nabla{u(x,y)}=({\frac{\partial{u}}{\partial{x}}}{  \frac{\partial{u}}{\partial{y}}})^T=(\alpha{x}^{\a  lpha-1}y^{1-\alpha}, x^{\alpha}(1-\alpha)y^{-\alpha})^T=(0,0)^T

    \alpha{x}^{\alpha-1}y^{1-\alpha}=x^{\alpha}(1-\alpha)y^{-\alpha}

    Then I use the fact that for $1 consumer would buy 2 apples and 1 orange, so I substitute x=2 and y=1:

    \alpha{2}^{\alpha-1}1^{1-\alpha}=2^{\alpha}(1-\alpha)1^{-\alpha}

    \alpha2^{\alpha-1}=2^{\alpha}-2{\alpha}\alpha

    \alpha*2^{\alpha-1}-2^{\alpha}+2^{\alpha}*{\alpha}=0

    2^{\alpha}(\frac{\alpha}{2}-1+\alpha)=0

    Now when I try to evaluate the second derivative, it is Hessian matrix and it's determinant is 0. Therefore, the test is inconclusive.

    What can I do here to show that alpha=2/3 gives the maximum value of the function? (if it does...)

    Given \alpha=2/2 and the utility function is u=u(x,y)=x^{\frac{2}{3}}y^{1-\frac{2}{3}}

    I will post the second part as a separate post, as this one is growing long.
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  2. #2
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    Now that I have the alpha I will find x,y subject to the budgetary constraint of $10. I plan to use Lagrange multiplier method.

    The price of apples is $1/3 as before, but the price of oranges is $1/6 (half of the price of apples).

    Let budget function g(x)=\frac{1}{3}x+\frac{1}{6}y=10

    Then I need to find a common tangent point of u(x,y) and g(x,y) subject to the constraint g(x,y)=10. Let \lambda be the Langrange multiplier:

    \nabla{u}=\lambda\nabla{g}

    (2/3x^{-1/3}y^{1/3}, 1/3x^{2/3}y^{-2/3})^T=\lambda(1/3, 1/6)^T

    2/3x^{-1/3}y^{1/3}=\lambda{1/3}

    1/3x^{2/3}y^{-2/3}=\lambda{1/6}

    After raising into 3rd power, I have y/x=\lambda^3/8 and I am not sure how to proceed next...
    Last edited by Volga; March 14th 2011 at 10:48 PM.
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  3. #3
    MHF Contributor matheagle's Avatar
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    "solve" for lambda and form a relationship between x and y, then insert that into g.

    \lambda ={2y^{1/3}\over x^{1/3}} ={2x^{2/3}\over y^{2/3}}

    so y=x now insert that into g.
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  4. #4
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    Indeed!

    If x=y, 1/3x+1/6x=10 then x=20 and y=20. Solved!
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  5. #5
    MHF Contributor matheagle's Avatar
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    you should have taken my calculus and probability classes
    it would have been fun having you as a student
    I just published another paper in Taiwan, my 18th at Academia Sinica
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  6. #6
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    I wish I could! well i am grateful for the opportunity to learn from you here
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