How should we choose two numbers,

Solution:

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

$x, y>-2$ and $x+y=\frac{1}{2}$         .....(1)

Now,

$z=x+y^{3}$

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

$\Rightarrow \frac{d z}{d x}=1+3\left(\frac{1}{2}-x\right)^{2}$

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

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

$\Rightarrow 1+3\left(\frac{1}{2}-x\right)^{2}=0$

$\Rightarrow\left(\frac{1}{2}-x\right)^{2}=\frac{1}{3}$

$\Rightarrow\left(\frac{1}{2}-x\right)=\pm \frac{1}{\sqrt{3}}$

$\Rightarrow x=\frac{1}{2} \pm \frac{1}{\sqrt{3}}$

$\frac{d^{2} z}{d x^{2}}=6\left(\frac{1}{2}-x\right)$

$\Rightarrow \frac{d^{2} z}{d x^{2}}=3-6 x$

At $x=\frac{1}{2} \pm \frac{1}{\sqrt{3}}:$

$\frac{d^{2} z}{d x^{2}}=3-6\left(\frac{1}{2}+\frac{1}{\sqrt{3}}\right)$

$\Rightarrow \frac{-6}{\sqrt{3}}<0$

Thus, $z$ is maximum when $x=\frac{1}{2}+\frac{1}{\sqrt{3}}$.

At $x=\frac{1}{2}-\frac{1}{\sqrt{3}}:$

$\frac{d^{2} z}{d x^{2}}=3-6\left(\frac{1}{2}-\frac{1}{\sqrt{3}}\right)$

$\Rightarrow \frac{6}{\sqrt{3}}>0$

Thus, $z$ is minimum when $x=\frac{1}{2}-\frac{1}{\sqrt{3}}$.

$x+y=\frac{1}{2}$

Substituting the value of $x$ in eq. (1), we get

$y=-\frac{1}{2}+\frac{1}{\sqrt{3}}+\frac{1}{2}$

$y=\frac{1}{\sqrt{3}}$

So, the required two numbers are $\left(\frac{1}{2}-\frac{1}{\sqrt{3}}\right)$ and $\frac{1}{\sqrt{3}}$.