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**garymarkhov** $\displaystyle f(x,y)=(4-x^2)y-2y^2-1$ on the set $\displaystyle \{ x\geq 1, y\geq 1 \}$

Can someone tell me if this is right?

$\displaystyle \frac{\partial f}{\partial x} = -2xy $

$\displaystyle \frac{\partial f}{\partial y} = 4-x^2-4y$

Setting the partials equal to zero, we can find that

$\displaystyle y = 1 - \frac{x^2}{4} $

and

$\displaystyle -2x(1 - \frac{x^2}{4}) = 0 $

So

$\displaystyle x(2x^2-8)=0$

and we can see that x is either 0, 2, or -2. So the inflection points are (0,1), (2,0), (-2,0). So $\displaystyle f(0,1)=1$ is the greatest value we can get.

Checking the boundary conditions, we see that things only get worse if we make x or y larger.

Forming the Hessian at this point, $\displaystyle \left(\begin{array}{cc} -2y & -2x \\ -2x & -4 \end{array}\right) = \left(\begin{array}{cc} -2 & 0 \\ 0 & -4 \end{array}\right)$

This is a symmetric matrix, so the diagonal values are the eigenvalues. Both eigenvalues being negative means that the matrix is negative definite and (0,1) is a max of this function. Put another way, the leading principle minors alternate in sign so the matrix must be negative definite.

Can someone confirm that my calculations are right and make a note if I could have done something more efficiently? Thanks.