Absolute extrema on a region

The extreme value theorem guarantees that if \(f\) is continuous on a closed, bounded set \(D\), an absolute minimum and maximum exist. The procedure has three steps: find the critical points of \(f\) in the interior of \(D\); find the candidates on the boundary of \(D\); and evaluate \(f\) at all of them. The largest value is \(M\) and the smallest is \(m\).

The boundary is treated as a one-variable problem: parametrize each side or curve (\(y=2-x\), \(x=\cos t,\ y=\sin t\), etc.), substitute into \(f\), and look for extrema of the resulting function on the interval, without forgetting the corners. If the boundary is given by an equation \(g=k\), Lagrange multipliers do the same job.

Guiding example: \(f(x,y)=x^2+y^2-2x-4y+7\) on \([0,3]\times[0,3]\). The critical point \((1,2)\) gives \(f=2\); at the corners \(f\) equals \(7,10,4,7\). Hence \(m=2\) and \(M=10\). Here the minimum is interior and the maximum is on the boundary, which is very common.

Typical mistakes: forgetting the boundary, believing that a critical point is an extremum (a saddle is not), or applying the method to an open or unbounded region, where the maximum or minimum may not exist. There is no need to classify critical points with the Hessian: just compare values.