Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.

Let the two numbers be $x$ and $y .$ Then,

$x+y=15$               ......(1)

Now,

$z=x^{2} y^{3}$

$\Rightarrow z=x^{2}(15-x)^{3}$   $[$ From eq. $(1)]$

$\Rightarrow \frac{d z}{d x}=2 x(15-x)^{3}-3 x^{2}(15-x)^{2}$

For maximum or minimum values of $z$, we must have

$\frac{d z}{d x}=0$

$\Rightarrow 2 x(15-x)^{3}-3 x^{2}(15-x)^{2}=0$

$\Rightarrow 2 x(15-x)=3 x^{2}$

$\Rightarrow 30 x-2 x^{2}=3 x^{2}$

$\Rightarrow 30 x=5 x^{2}$

$\Rightarrow x=6$ and $y=9$

$\frac{d^{2} z}{d x^{2}}=2(15-x)^{3}-6 x(15-x)^{2}-6 x(15-x)^{2}+6 x^{2}(15-x)$

At $x=6:$

$\frac{d^{2} z}{d x^{2}}=2(9)^{3}-36(9)^{2}-36(9)^{2}+6(36)(9)$

$\Rightarrow \frac{d^{2} z}{d x^{2}}=-2430<0$

Thus, $z$ is maximum when $x=6$ and $y=9$.

So, the required two parts into which 15 should be divided are 6 and 9 .