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Thread: critical point!

  1. #1
    Jul 2011

    critical point!

    which of these is not a critical point of the given function subject to the indicated constraint: $\displaystyle z=x^2y $ constraint $\displaystyle x+2y=2 $

    is it (0,1) or $\displaystyle \frac{4}{3}, \frac{1}{3} $ or (1,0)

    what i don't understand is how to approach this problem, thanks.
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  2. #2
    MHF Contributor
    Oct 2008

    Re: critical point!

    Notice that the third pair doen't even satisfy the constraint. Nonetheless,

    Quote Originally Posted by lawochekel View Post
    how to approach this problem
    ... is by visualising, for sure.

    Online 3-D Function Grapher

    ... and try to see the constraint as selecting just points on that surface (z = ...) that intersect with the plane (y = -1/2 x + 1) which is the plane passing vertically through the line y = -1/2 x + 1.

    Then imagine getting the projection of that intersection against the x-z plane, where it will trace a curve which is z = just the result of replacing y with -1/2 x + 1. Sketch this curve, and locate the other two points to see that they are indeed critical.

    PS: this problem may be for introducing you to Lagrange multipliers. Do the above first, though, and then plugging in the two equations as f and g here, Lagrange multiplier - Wikipedia, the free encyclopedia should make good sense.
    Last edited by tom@ballooncalculus; Apr 17th 2012 at 07:03 AM.
    Thanks from lawochekel
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