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Math Help - maximization problem

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    20

    maximization problem


    Please help on this one....
    Don't really understand the problem. What they ask me to do.
    It says..

    Solve the following maximization problem:
    max xy^(1\2)
    x,y
    subject to X^(2) +(c*y)^2 <= 1,x,y => 0, where
    c> 0 .
    Let
    V(c) denote the maximized value of the objective function (as a function of c).

    Calculate
    V ' (c).
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  2. #2
    Senior Member
    Joined
    Dec 2008
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    As always, we must check all critical points in the domain. For a differentiable function, the critical points are the boundary points and the points inside at which \nabla z=0. In our case,

    <br />
\begin{aligned}<br />
z&=xy^{\frac{1}{2}}\\<br />
\nabla z&=y^{\frac{1}{2}}\textbf i\,+\frac{1}{2}xy^{-\frac{1}{2}}\textbf j,\\<br />
\end{aligned}<br />

    so the only time \nabla z ever vanishes is when y=0 (and thus z=0) at a boundary point, leaving us with only boundary points for the maximum. Our domain is an elliptical region in the first quadrant bounded by the x and y axes. When x or y equals 0, z=xy^{\frac{1}{2}}=0, leaving us with the boundary points at g(x,y)=x^2+(cy)^2=1. By Lagrange's Method, we know that if (x,y) is a maximum, then

    \begin{aligned}<br />
\nabla z&=\lambda\nabla g\\<br />
{y}^{\frac{1}{2}}\textbf i\,+\frac{1}{2}{x}{y}^{-\frac{1}{2}}\textbf j\,&=\lambda(2x\textbf i\,+2c^2y\textbf j).<br />
\end{aligned}<br />

    Solving this vector equation for x and y will give you a point at which z attains its maximum.
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  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
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    Hey Scott, thx for the help. So if i solve your equation what i get eventually for V(c) is:
    c=
    Code:
    (1\2)^1\2 *x\y
    and finally i receive that the first derivative of c- , which is the maximized value of the objective function is :
    c' =
    Code:
    2x * 1\(2^1\2 * y) +2y * 1\(2^1\2 * x)
    Do you agree ? Am i right.
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  4. #4
    Senior Member
    Joined
    Dec 2008
    Posts
    319
    We can begin with

    \begin{aligned}<br />
y^{\frac{1}{2}}&=2\lambda x\\<br />
\frac{y^{\frac{1}{2}}}{2\lambda}&=x<br />
\end{aligned}<br />

    and substitute for x:

    \begin{aligned}<br />
\frac{1}{2}xy^{-\frac{1}{2}}&=2\lambda c^2y\\<br />
\frac{1}{2}\left(\frac{y^{\frac{1}{2}}}{2\lambda}\  right)y^{-\frac{1}{2}}&=2\lambda c^2y\\<br />
\frac{1}{4\lambda}&=2\lambda c^2y\\<br />
\frac{1}{8\lambda^2c^2}&=y.<br />
\end{aligned}
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