 Conjugate-power constrained product maximum

For conjugate exponents $p,q>1$, the constraint set is compact, so the continuous product has a maximum. Boundary endpoints have product zero, while $(1,1)$ is feasible with product one. An interior maximum therefore satisfies the Lagrange multiplier equations $y=\lambda x^{p-1}$ and $x=\lambda y^{q-1}$. Multiplying by $x$ and $y$ respectively gives $x^p=y^q$, since the multiplier is nonzero. Substitution in the constraint makes both powers equal one. Thus the maximum is one, attained uniquely at $(1,1)$. The boundary check and compactness are what turn the necessary multiplier equations into a global conclusion.