Let $$x, y > 0$$. If $$x^3 y^2 = 2^{15}$$, then the least value of $$3x + 2y$$ is

We are given $$x, y > 0$$ with the constraint $$x^3 y^2 = 2^{15}$$, and we need to find the least value of $$3x + 2y$$.

Since the expression involves five terms, we apply the AM-GM inequality by writing $$ 3x + 2y = x + x + x + y + y $$ and then using $$ \frac{x + x + x + y + y}{5} \ge (x \cdot x \cdot x \cdot y \cdot y)^{1/5}. $$

From this we obtain $$ \frac{3x + 2y}{5} \ge (x^3 y^2)^{1/5}. $$ Substituting $$x^3 y^2 = 2^{15}$$ gives $$ \frac{3x + 2y}{5} \ge (2^{15})^{1/5} = 2^3 = 8, $$ so $$ 3x + 2y \ge 40. $$

Equality holds when all five terms are equal, which requires $$x = y$$. In that case, $$ x^3 \cdot x^2 = x^5 = 2^{15} \implies x = 2^3 = 8, $$ so $$x = y = 8$$ and $$ 3(8) + 2(8) = 24 + 16 = 40, $$ confirming the minimum value.

Therefore, the least value of $$3x + 2y$$ is 40, and the correct option is Option D.