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Math Help - Find the max/min values of a function when (x,y) is a point in the set...

  1. #1
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    Find the max/min values of a function when (x,y) is a point in the set...

    This normally wouldn't be so hard for me, except for the awkward constraint we've been given.

    Find the maximum and minimum values of the function
    f(x,y)=2x^4-4x^2y^2+5y^4
    when (x,y) is a point in the set
    \{(x,y):x^4+y^4\leq 80\}.

    I've been told that using Lagrange Multipliers is the proper method to solve this, but I'm not so sure. I'm new to using Lagrange Multipliers, and I haven't seen a question for which the constraint is a set that doesn't have a definite value.
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  2. #2
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    you just apply the method for x^4+y^4=80, once got the max. or min. points, keep those which satisfy x^4+y^4\le80.
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  3. #3
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    I'm not sure what Krizalid is saying. If you "apply the method" (Lagrange multipliers) for x^4+ y^4= 80, any points you get will satisfy x^4+ y^4= 80 and so x^4+ y^4\le 80.

    Here's what I would do. Find all x and y such that \nabla f= 0, that is, such that \frac{\partial f}{\partial x}= 0 and \frac{\partial f}{\partial y}= 0. Note those (x, y) that satisfy x^4+ y^4< 80.

    Then use Lagrange multipliers to find those (x, y) that maximize or minimize f with the constraint x^4+ y^4= 80.

    Finally, check the value of f at each of those (x, y) points, both inside and on x^4+ y^4= 80, to determine the maximum and minimum values of f.
    Last edited by HallsofIvy; October 25th 2010 at 04:49 AM.
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    HallsofIvy, either I did your method wrong, or something else happened that I'm not familiar with. I've not seen the method you're using in our course material yet.

    When I went for \frac{\partial f}{\partial x}=0 and \frac{\partial f}{\partial y}=0, I end up with \frac{\partial f}{\partial x}=8x(x^2-y^2) and \frac{\partial f}{\partial y}=4y(5y^2-2x^2), which leads to x=0,y=0 in both cases.

    Did I do this incorrectly somehow, or this some sort of a "trick question" thing?

    EDIT: Okay, using the Lagrange method of u=f(x,y)+\lambda g(x,y) seems to either be malfunctioning on me, or I'm doing it wrong. Again, our source textbook is next-to-worthless.

    Using your constraint x^4+ y^4= 80, I have u=2x^4-4x^2y^2+5y^2+\lambda (x^4+y^4). Taking partial derivatives gets me \frac{\partial u}{\partial x} =8x^3-8xy^2+4\lambda x^3 and \frac{\partial u}{\partial x} =20y^3-8x^2y+4\lambda y^3. Finding \lambda, however, gets me \lambda =\frac{2x^4-4x^2y^2+5y^2}{80}, which I'm guessing is a mistake of some sort.

    Is this method incorrect?
    Last edited by Runty; October 23rd 2010 at 03:50 PM.
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    for fx=fy=0, when x= -1/2 and y=-1/5 is also sasitfy the equation other x=y=0, so will that be 4 critipal points?
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  6. #6
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    f(x,y)=2x^4-4x^2y^2+5y^4 so \nabla f= (8x^3- 8xy^2)\vec{i}+ (-8x^2y+ 20y^3)\vec{j}. At any max or min in the interior of the set we must have 8x^2- 8xy^2= 8x(x^2- y^2)= 8x(x- y)(x+ y)= 0 so one of x= 0, x=y, x=-y must be true. Also -8x^2y+ 20y^3= 4y(5y^2- 2x^2)= 4y(\sqrt{5}y- \sqrt{2}x)(\sqrt{5}y+ \sqrt{2}x)= 0 so one of y= 0, \sqrt{5}y= \sqrt{2}x, or \sqrt{5}y= -\sqrt{2}x must be true.

    If x= 0, only y= 0 satisfies the other equations. If y= x again we have only x= y= 0. I don't see how x= -1/2, y= -1/5 satisfy those equations. I find only x= y= 0.

    On the boundary, g(x, y)= x^4+ y^4= 80, \nabla g= 4x^3\vec{i}+ 4y^3\vec{j} so we must have (8x^3- 8xy^2)\vec{i}+ (-8x^2y+ 20y^3)\vec{j}= \lambda(4x^3\vec{i}+ 4y^3\vec{j} or 8x^3- 8xy^2= \lambda(4x^3) and -8x^2y+ 20y^3= \lambda(4y^3).

    It's never necessary to find \lambda since that is not relevant to the solution. Instead, eliminate \lambda by dividing one equation by the other:
    \frac{8x^3- 8xy^2}{-8x^2y+ 20y^3}= \frac{4x^3}{4y^3}
    \frac{2(x- y)(x+ y)}{(\sqrt{5}y- \sqrt{2}x)(\sqrt{5}y- \sqrt{2}x)}= \frac{x^2}{y^2}.
    That, together with x^4+ y^4= 80 gives two equations to solve for x and y.
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    f(x,y)=2x^4-4x^2y^2+5y^4 so \nabla f= (8x^3- 8xy^2)\vec{i}+ (-8x^2y+ 20y^3)\vec{j}. At any max or min in the interior of the set we must have 8x^2- 8xy^2= 8x(x^2- y^2)= 8x(x- y)(x+ y)= 0 so one of x= 0, x=y, x=-y must be true. Also -8x^2y+ 20y^3= 4y(5y^2- 2x^2)= 4y(\sqrt{5}y- \sqrt{2}x)(\sqrt{5}y+ \sqrt{2}x)= 0 so one of y= 0, \sqrt{5}y= \sqrt{2}x, or \sqrt{5}y= -\sqrt{2}x must be true.

    If x= 0, only y= 0 satisfies the other equations. If y= x again we have only x= y= 0. I don't see how x= -1/2, y= -1/5 satisfy those equations. I find only x= y= 0.

    On the boundary, g(x, y)= x^4+ y^4= 80, \nabla g= 4x^3\vec{i}+ 4y^3\vec{j} so we must have (8x^3- 8xy^2)\vec{i}+ (-8x^2y+ 20y^3)\vec{j}= \lambda(4x^3\vec{i}+ 4y^3\vec{j} or 8x^3- 8xy^2= \lambda(4x^3) and -8x^2y+ 20y^3= \lambda(4y^3).

    It's never necessary to find \lambda since that is not relevant to the solution. Instead, eliminate \lambda by dividing one equation by the other:
    \frac{8x^3- 8xy^2}{-8x^2y+ 20y^3}= \frac{4x^3}{4y^3}
    \frac{2(x- y)(x+ y)}{(\sqrt{5}y- \sqrt{2}x)(\sqrt{5}y- \sqrt{2}x)}= \frac{x^2}{y^2}.
    That, together with x^4+ y^4= 80 gives two equations to solve for x and y.
    Alright, so using your method, I ended up with the following roots:
    x=0,y=0
    x=-2\sqrt{2},y=-2
    x=-2\sqrt{2},y=2
    x=2\sqrt{2},y=-2
    x=-2\sqrt{2},y=2

    According to this result I got off WolframAlpha, however, I have the following max/min values:
    Max: f(x,y)=400 at (x,y)=(0,-2\sqrt[4]{5})
    Min: f(x,y)=0 at (x,y)=(0,0)
    See this link

    WolframAlpha isn't 100% reliable at times, but I calculated x^4+y^4 using the max roots and it worked out to equal 80. However, I'm wondering if the program misinterpreted what I put in.

    Can someone confirm which of my answers would be correct?
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