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Math Help - Maximum and minimum of a function of 2 variables (Lagrange multipliers)

  1. #1
    MHF Contributor arbolis's Avatar
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    Maximum and minimum of a function of 2 variables (Lagrange multipliers)

    My question is : Can I solve the problem without passing by Lagrange multipliers? Because I have some difficulties by doing so. If not, then I don't want any tip and I'll work harder.
    I must find the minimum and the maximum of the function f(x,y)=x^2+y^2+\frac{2xy\sqrt{2}}{3} in the ellipse x^2+2y^2 \leq 1.
    So it seems a problem that can be solved via Lagrange multipliers, but I didn't find a way to find \lambda, x and y. (I even had to divide by 0 if \lambda \neq 1 but then I found another condition ( if \lambda \neq 0 then ...) and even with it I couldn't isolate \lambda nor x nor y).
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  2. #2
    Senior Member Spec's Avatar
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    The requirement that the two gradients are parallel can also be expressed as

    \left|\begin{array}{cc}f'_x&f'_y\\g'_x&g'_y\end{ar  ray}<br />
 \right|=0

    This will produce exactly the same result as using the Lagrange multipliers method (once you've eliminated the multiplier).

    Both these methods are a bit impractical here though. Since we're dealing with an ellipse, we can express the main function in terms of one variable.

    f(x,y)=f(\varphi)=\left\{x=\cos\varphi,\ y=\frac{\sin \varphi}{\sqrt{2}} \right\}=\cos^2 \varphi+\frac{\sin^2 \varphi}{2}+\frac{2\cos \varphi\sin \varphi}{3}

    You will of course also need to find the inner points where f'_x=f'_y=0
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Spec View Post
    The requirement that the two gradients are parallel can also be expressed as

    \left|\begin{array}{cc}f'_x&f'_y\\g'_x&g'_y\end{ar  ray}<br />
 \right|=0

    This will produce exactly the same result as using the Lagrange multipliers method (once you've eliminated the multiplier).

    Both these methods are a bit impractical here though. Since we're dealing with an ellipse, we can express the main function in terms of one variable.

    f(x,y)=f(\varphi)=\left\{x=\cos\varphi,\ y=\frac{\sin \varphi}{\sqrt{2}} \right\}=\cos^2 \varphi+\frac{\sin^2 \varphi}{2}+\frac{2\cos \varphi\sin \varphi}{3}

    You will of course also need to find the inner points where f'_x=f'_y=0
    Thank you VERY much. I'll work on that! (I might ask for further help in a few days)
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