Past exam of the short options course of the University of Oxford / 2015 / 4 / c

The feasible set is closed and contained in the rectangleIt is therefore compact, and the continuous function $xy$ attains a global maximum there by the extreme value theorem. At the two boundary endpoints where a coordinate is zero, the product is zero. Meanwhile $(1,1)$ is feasible because the conjugate exponents satisfy $1/p+1/q=1$, and its product is one. Hence every maximizer has $x,y>0$.

On that positive region the constraint is $C^1$ with nonzero gradient. The Lagrange multiplier equations areThe first two identities imply $xy=\lambda x^p=\lambda y^q$. Since $x,y>0$, the multiplier is nonzero, so $x^p=y^q=r$. The constraint now givesThus the sole interior candidate is $(1,1)$. Existence of a maximum and the boundary comparison prove the unique global resultThis conjugate-power constrained product maximum is valid for all real $p,q>1$ with $1/p+1/q=1$; no integrality of the exponents is needed.