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Math Help - Maximum and bound for a multivariate function

  1. #1
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    Maximum and bound for a multivariate function

    Hi there guys, I'm having alot of trouble verifying the following:

    Verify that the maximum of

    \displaystyle x(1-x)y(1-y)w(1-w)(1-(1-xy)w)^{-1}

    occurs for x=y, and then that

    \displaystyle \frac{x(1-x)y(1-y)w(1-w)}{(1-(1-xy)w)} \leq (\sqrt 2 - 1)^4

    Thanks guys.
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  2. #2
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    Quote Originally Posted by HTale View Post
    Verify that the maximum of

    \displaystyle x(1-x)y(1-y)w(1-w)(1-(1-xy)w)^{-1}

    occurs for x=y, and then that

    \displaystyle \frac{x(1-x)y(1-y)w(1-w)}{(1-(1-xy)w)} \leq (\sqrt 2 - 1)^4
    There is an important condition missing from this question. As it stands, the function is unbounded. For example, it goes to +∞ if x=y= and w→4/3 from above. I think that the missing condition must be that x, y and w must all lie in the unit interval [0,1].

    Let f(x,y,w) = \frac{x(1-x)y(1-y)w(1-w)}{(1-(1-xy)w)}. Then f is identically zero on the boundary of the unit cube, so its maximum must occur at a critical point inside the cube. So put the three partial derivatives equal to 0. Leaving out some details, this gives the equations

    (1-2x)(1-(1-xy)w) = (1-x)xyw,\qquad\qquad(1)
    (1-2y)(1-(1-xy)w) = (1-y)xyw,\qquad\qquad(2)
    (1-2w)(1-(1-xy)w) = -(1-xy)w(1-w).\qquad(3)

    From (1) and (2) it's easy to see that x=y. From equations (1) and (3) (with y=x) you see that w = \frac x{1-x}. Substitute that into (1) and solve for x, to get x = \sqrt2-1,\ w=1/\sqrt2.

    Finally, check that f(\sqrt2-1,\sqrt2-1,1/\sqrt2) = (\sqrt2-1)^4. So that is the maximum value of the function in the cube.
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