# Local and global minimum/maximum and continuity

My task was to determine the global minimum of a function $f(x,y) = x^3 + y^3 - 3xy$ on the square $[0,2] \times [0,2]$.

I first calculated the points where the gradient $(grad f)(x,y) = (3x^2 - 3y, 3y^2 - 3x)$ becomes the zero vector. That's the case for $(0,0)$ and $(1,1)$. I then took a look at the Hessian matrix in these points, which was indefinite in $(0,0)$ and positive definite in $(1,1)$. I concluded that in $(1,1)$ there is a local minimum and the function has no other turning points.

The sample solution then proceeds to scan the boundary points of the square $[0,2]^2$. However, I think that this is not required. I know that the function has no other turning points (especially no local maximum) that could make the function value drop again, hence the local minimum also has to be the global minimum.

Is my conclusion correct for multidimensional functions? Does it matter whether or not the function of interest is continuous?

## 1 Answer The point (0,0) looks like a saddle-point and the point (1,1) a local minimum. How closely did you look at the saddle-point. Are you sure it was not decreasing along the edges? As it happens, it wasn't. But it might have been. In general, it is usually quicker and safer to check the boundary of a region than to make arguments about why the max/min was inside the region.

• You seem to be right. I just saw the following figure in our script. So obviously it seems to be possible for the function value to drop without a local maximum in between. – Christian Schnorr Sep 8 '14 at 18:52