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Math Help - Lagrange Multipliers

  1. #1
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    Lagrange Multipliers

    Use the method of lagrange multipliers to determine the maximum value of f(x,y)= x^a y^b (a and b are two fixed positive constants) subject to constraint x + y= 1. (assume x,y bigger than zero)



    Help would be appreciated! I get ax^a-1 = bY^b-1 but dont know how to do the rest. I cant isolate x based on only a and b.
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  2. #2
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    Re: Lagrange Multipliers

    Don't forget your constraint: x+ y= 1 so that y= 1- x. Putting that into ax^{a-1} = by^{b-1} gives ax^{a-1}= b(1- x)^{b-1}. Of course, that is a polynomial of degree a- 1 or b-1, whichever is larger, and there is no simple way of solving that.
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  3. #3
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    Re: Lagrange Multipliers

    Quote Originally Posted by apatite View Post
    Use the method of lagrange multipliers to determine the maximum value of f(x,y)= x^a y^b (a and b are two fixed positive constants) subject to constraint x + y= 1. (assume x,y bigger than zero)



    Help would be appreciated! I get ax^a-1 = bY^b-1 but dont know how to do the rest. I cant isolate x based on only a and b.
    Let g(x,y)=x+y

    \nabla f=-\lambda\nabla g or:
    \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)=-\lambda\left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right)

    \frac{\partial f}{\partial x} = ax^{a-1}y^b
    \frac{\partial f}{\partial y} = bx^ay^{b-1}
    \frac{\partial g}{\partial x} = \frac{\partial g}{\partial y} = 1

    So: \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)=-\lambda\left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right) \implies

    (ax^{a-1}y^b,bx^ay^{b-1})=-\lambda(1,1)=(-\lambda,-\lambda)
    Using the fact that g(x,y)=1 or y= 1 - x then
    ax^{a-1}(1-x)^b=-\lambda
    bx^a(1-x)^{b-1}=-\lambda

    \frac{ax^{a-1}(1-x)^b}{bx^a(1-x)^{b-1}}=1

    \frac{a(1-x)}{bx}=1

    Or: x= \frac{a}{a+b} so that y= \frac{b}{a+b} i.e. f(x,y) is a maximum when x and y take on these values to get

    f_{max} = \left(\frac{a}{a+b}\right)^a \left( \frac{b}{a+b} \right)^b

    As a final step, you must check concavity here to verify that f is a maximum at these values.
    Last edited by majamin; March 12th 2013 at 06:00 AM.
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