Maximizing a product under a linear constraint

Lagrange.
$$\begin{cases} a x^{a-1} y^a = 2\lambda \\ a x^a y^{a-1} = 3\lambda \\ 2x + 3y = 12 \end{cases}$$

Resolution. Dividing the first equation by the second:
$$\frac{y}{x} = \frac{2}{3} \implies 3y = 2x.$$
Substituting into the constraint:
$$2x + 2x = 12 \implies 4x = 12 \implies x = 3, \quad y = 2.$$

Boundary check. On the boundary $x=0$ or $y=0$, $f = 0$. At $(3,2)$, $f = 6^a > 0$. Hence this is the global maximum.