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Math Help - optimization (multivariable) Done 2 ways, but diff answers

  1. #1
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    optimization (multivariable) Done 2 ways, but diff answers

    "Find the point on the paraboloid :

    z=(x^2)/25 + (y^2)/4

    That is closest to the point (0,5,0)"

    ...my solutions manual solves this using symetry. they say that by symetry we see that x=0. And then the problem turns into a two dimensional problem. and optimize the distance squared... they got (0,2.66,1.76) as the closest point.

    ...NOW, i did this in 3D using the paraboloid as my constraint, and did it by the method of lagrange multipliers. but i get a totally diff answer were some of my points are negative.
    Can someone shed some light please. Thanks in advance
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  2. #2
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    You say you used Langrange multipliers but you don't show how you did that. All I can say is that, using Lagrange multipliers, I get (0, 5, 0).
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    I really hope i did some dumb mistake bc this problem is frustating me.

    Ok, my function to be optimized is the square of the distance formula, between the (0,5,0) and a point x,y,z on the paraboloid: d^2= F = x^2 + (y-5)^2 + z^2 and my constraint will be the parabloid function set equal to zero: (x^2)/25 + (y^2)/4 - z = 0. and then i did the partial integrations, setting them equal to the partials of the constraint times lambda. Is there something wrong in my analysis?
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    Quote Originally Posted by HallsofIvy View Post
    You say you used Langrange multipliers but you don't show how you did that. All I can say is that, using Lagrange multipliers, I get (0, 5, 0).
    And i'm not sure (0,5,0) makes sense bc that point isnt in the paraboloid
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  5. #5
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    Your are right. I don't know where my head was! (0, 5, 0) was the point we are measuring the distance to, not the answer.

    The problem is to minimize f(x,y,z)= x^2+ (y- 5)^2+ z^2 subject to the constraint g(x, y, z)= \frac{x^2}{25}+ \frac{y^2}{4}- z= 0. \nabla f= 2x\vec{i}+ 2(y- 5)\vec{j}+ 2z\vec{k}= \lambda\nabla g= \lambda(\frac{2}{25}x\vec{i}+ \frac{1}{4}y\vec{j}- \vec{k} or 2x= \lambda \frac{2}{25}x, 2(y- 5)= \lambda\frac{1}{4}y, and 2z= -\lambda.

    If you divide the first equation by the last, \frac{x}{z}= \frac{2}{25}x so x= 0 or z= \frac{2}{25}. If you divide the first equation by the second, \frac{x}{y- 5}= \frac{20}{25}\frac{x}{y} so 5xy= 4x(y- 5)= 4xy- 20 so xy= 20.
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    Quote Originally Posted by HallsofIvy View Post
    If you divide the first equation by the last, \frac{x}{z}= \frac{2}{25}x so x= 0 or z= \frac{2}{25}. If you divide the first equation by the second, \frac{x}{y- 5}= \frac{20}{25}\frac{x}{y} so 5xy= 4x(y- 5)= 4xy- 20 so xy= 20.
    I'm sorry for the persistance but when i divide the equations like u said, i got z=-25/2 and y=-40/17...now the problem is that when i plug this into the constraint, i get no solution when solving for x
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