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