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Thread: 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 $\displaystyle u$ of posessing a quantity $\displaystyle x$ of apples and a quantity $\displaystyle y$ of oranges by the formula:

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

    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 $\displaystyle \alpha$.

    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 $\displaystyle \alpha$ as a parameter.


    Answer.

    First part, finding alpha.

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

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

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

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

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

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

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

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

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

    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 $\displaystyle g(x)=\frac{1}{3}x+\frac{1}{6}y=10$

    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 $\displaystyle \lambda$ be the Langrange multiplier:

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

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

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

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

    After raising into 3rd power, I have $\displaystyle y/x=\lambda^3/8$ and I am not sure how to proceed next...
    Last edited by Volga; Mar 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.

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

    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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