My source question comes from Vector Calculus, sixth edition by Marsden and Tromba:
Consider all rectangles with fixed perimeter $p$. Use Lagrange multipliers to show that the rectangle with maximal area is a square.
I have shown that this indeed the case. My question is not on applying Lagrange multipliers, but that given the perimeter constraint, there exists a maximum area rectangle. I have shown by the bordered Hessian that the square is indeed a local maxima.
My attempt:
The objective function $h(l, w) = lw$ is bounded by $0$, attained as $l$ or $w$ tends to $p/2$. Therefore $l$ and $w$ are bounded above by $p/2$ and below by $0$. However, this constraint set is not closed, since the degenerate solutions of $(l,w) = (0, p/2)$ is not a rectangle, so these bounds are not in the constraint set, and I cannot cite the existence of a maxima by closedness.