How do I prove the function $f(x,y)=(3−x−y)xy$ has a local maximum?

Question

I have calculated the Hessian matrix for such a function, which turns out to be,

H$f(x)$ = \begin{bmatrix}-2y&3-2x-2y\\3-2x-2y&-2x\end{bmatrix}

How can one prove that such a function reaches a maximum? Is there a way to prove that it is negative definite or negtaive semi-definite at this point?

Thank you.

Answer 1

Hint: there are four critical points to consider. The Hessian is negative definite for one of them.

Answer 2

How about this:

Your function can be regarded as a function of three variables with a constraint!

You try to find out extrama of $F(x,y,z) = xyz \, $ with the constraint $x+y+z = 3$.

And you definetly have a max when $x,y,z \geq 0$ by the Arithmetic-Geometric mean inequality.

Answer 3

$$\nabla f=[3y-2xy-y^2,3x-x^2-2xy]$$ $$=[y(3-2x-y),x(3-x-2y)]$$ Thus the critical points, i.e. the points where $\nabla f=[0,0]$ are (0,0),(0,3),(3,0) and (1,1). At each of the first three critical points, the determinant of the Hessian is negative, so those points are all saddle points. At (1,1), the Hessian matrix is $$\begin{bmatrix}-2&-1\\-1&-2 \end{bmatrix}$$. Among the many equivalent tests for negative-definiteness, we choose, for a 2x2 matrix, top left entry<0, det(Hessian)>0. Both conditions are satisfied, so the Hessian matrix is negative-definite, which means that $f$ has a local maximum at (1,1).

Answer 4

Took me a while, I came up with a rotationally symmetric version, $$ f(x,y) = x^3-3x^2 y - 3 x y^2 + y^3 - 6(x^2 + y^2) + 16 $$ This is zero on three straight lines, is positive inside the little equilateral triangle they bound, has a local maximum at the origin, where the function takes value $16.$ I deliberately displayed the points where the function is slightly smaller, namely $15.$ The function $f$ is also positive in three of the infinite wedges depicted, and negative in the other three infinite regions.