Maximum and bound for a multivariate function

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

Verify that the maximum of

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

occurs for $\displaystyle x=y$, and then that

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

Thanks guys.

Quote:

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Originally Posted by HTale

Verify that the maximum of

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

occurs for $\displaystyle x=y$, and then that

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

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There is an important condition missing from this question. As it stands, the function is unbounded. For example, it goes to +∞ if x=y=½ 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 $\displaystyle f(x,y,w) = \frac{x(1-x)y(1-y)w(1-w)}{(1-(1-xy)w)}$. 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

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

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

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