Divide 64 into two parts such that the sum of the cubes of two parts is minimum.

Solution:

Suppose 64 is divided into two parts x and 64-x. Then,

$z=x^{3}+(64-x)^{3}$

 

$\Rightarrow \frac{d z}{d x}=3 x^{2}+3(64-x)^{2}$

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

$\frac{\mathrm{dz}}{\mathrm{dx}}=0$

$\Rightarrow 3 \mathrm{x}^{2}+3(64-\mathrm{x})^{2}=0$

$\Rightarrow 3 x^{2}=3(64-x)^{2}$

$\Rightarrow x^{2}=x^{2}+4096-128 x$

$\Rightarrow x=\frac{4096}{128}$

 

$\Rightarrow x=32$

Now,

$\frac{d^{2} z}{d x^{2}}=6 x+6(64-x)$

 

$\Rightarrow \frac{d^{2} z}{d x^{2}}=384>0$

Thus, $z$ is minimum when 64 is divided into two equal parts, 32 and 32 .