= Solution
The feasible set is closed and contained in the rectangle
$$
0\leq x\leq p^{1/p},\qquad0\leq y\leq q^{1/q}.
$$
It 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 are
$$
y=\lambda x^{p-1},\qquad x=\lambda y^{q-1},\qquad
\frac{x^p}{p}+\frac{y^q}{q}=1.
$$
The 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 gives
$$
r\left(\frac1p+\frac1q\right)=1,\qquad r=1.
$$
Thus the sole interior candidate is $(1,1)$. Existence of a maximum and the boundary comparison prove the unique global result
$$
\boxed{\max xy=1,\qquad\text{attained exactly at }x=y=1.}
$$
This <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.
\Image[/past-exam-of-the-short-options-course-of-the-university-of-oxford/2015-young-inequality-constraint.png]
{title=The constrained product reaches one where the two level curves are tangent}
{description=Original illustration for p=3 and q=3/2. The constraint arc meets the product level xy=1 only at (1,1), where both gradients point along (1,1). The red dashed line is their common tangent.}
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