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Math Help - LaGrange multiplier question

  1. #1
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    LaGrange multiplier question

    I have a function of f (x,y) = x^2 + 2y^2 which has a constraint of x^2 + y^2 is less than or equal to 4

    My LaGrange function is: x^2 + 2y^2 - lambda(x^2 + y^2 - 4)

    dL/dx 2x - lambda 2x = 0
    dL/dy 4y - lambda 2y = 0
    dL/lambda x^2 + y^2 - 4 less than or = 0

    So, (0,0) is one possible critical point as is (0,2) and (0, -2) and I used these and got a maximum of f(x,y) = 8 for (0,2) and (0, -2) and min f(x,y) = 0 at (0,0) but....
    why can we not use (2, 0) and ( -2,0)????? Can someone please explain why?????

    Thanks for your efforts, Frostking
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  2. #2
    MHF Contributor matheagle's Avatar
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    There are two parts to this.

    Case 1 is x^2 + y^2 = 4, i.e., the boundary.
    Here f=x^2 + y^2 +y^2=4+y^2.
    So get the max/min of g(y)=4+y^2 where -2\le y \le 2
    By inspection the max occurs at y=+-2 and min is at y=0 (on the boundary).

    Case 2, the interior of this circle.
    you can get the global max by calc 3, but clearly the function is increasing as you move away from the origin.
    So the max's occur on the boundary, see case 1.
    And the min of a sum of squares is zero at the origin.
    So no calculus is even needed here.
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