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Math Help - finding max/min with Lagrange Multipliers

  1. #1
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    finding max/min with Lagrange Multipliers

    Find the extreme values of f on the region described by the inequality. f(x,y) = 2x^2 + 3y^2 - 4x -9, x^2 + y^2 ≤ 16

    I am very confused with the steps of how to approach these types of problems. I found..

    -partial deriv of x: 4x-4
    -partial deriv of y: 6y

    Then, I followed an example in my book and found the critical point by setting those equations above to 0, getting x=1, y=0; so the point would be (1,0)

    Then, use Lagrange multipliers like this

    <4x-4, 6y>=λ(2x,2y)...
    4x-4=λ2x
    6y=λ2y
    Here's where I get a little lost because I'm not really sure what to solve for (x,y, or λ?) and where/what I plug that into when I find it... I tried solving 6y=λ2y, getting λ = 3, then plugging it into x=4/4-2λ, but I don't think that's right.

    Thank you so much for your help!
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  2. #2
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    Quote Originally Posted by juicysharpie View Post
    Find the extreme values of f on the region described by the inequality. f(x,y) = 2x^2 + 3y^2 - 4x -9, x^2 + y^2 ≤ 16

    I am very confused with the steps of how to approach these types of problems. I found..

    -partial deriv of x: 4x-4
    -partial deriv of y: 6y

    Then, I followed an example in my book and found the critical point by setting those equations above to 0, getting x=1, y=0; so the point would be (1,0)

    Then, use Lagrange multipliers like this

    <4x-4, 6y>=λ(2x,2y)...
    4x-4=λ2x
    6y=λ2y
    Here's where I get a little lost because I'm not really sure what to solve for (x,y, or λ?) and where/what I plug that into when I find it... I tried solving 6y=λ2y, getting λ = 3, then plugging it into x=4/4-2λ, but I don't think that's right.

    Thank you so much for your help!
    What/where is your Lagrangian?

    CB
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  3. #3
    MHF Contributor matheagle's Avatar
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    1 You should complete the square, it usually helps.
    2 You need to examine the max/min in general and on the boundary.
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  4. #4
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    I see no reason for using Lagrange Multipliers. You can, as matheagle suggests, complete the square to see that the boundary is an ellipse. After having found critical points inside the ellipse, reduce the function to a single variable on the boundary and look for critical points there.

    An ellipse, \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1 or, equivalently, b^2x^2+ a^2y^2= a^2b^2 can be written in parametric equations x= a cos(t), y= b sin(t).
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by matheagle View Post
    2 You need to examine the max/min in general and on the boundary.
    The OP has already found the position of the minima interior to the feasible region.

    CB
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