The sum of two positive numbers is given. If the sum of their cubes is minimum, then

Let \(\mathrm{x}\) and \(\mathrm{y}\) be two positive numbers.
\(\therefore\) According to question,
\(\mathrm{x}+\mathrm{y}=\mathrm{a} \quad\) (constant)
Let \(\quad z=x^{3}+y^{3}\)
\(z=x^{3}+(a-x)^{3} \quad \ldots\) (i)
On differentiating Eq. (i) w.r.t. ' \(x\) ', we get
\[ \begin{aligned} \frac{\mathrm{dz}}{\mathrm{dx}} &=3 \mathrm{x}^{2}+3(\mathrm{a}-\mathrm{x})^{2}(-1) \\ &=3\left[\mathrm{x}^{2}-(\mathrm{a}-\mathrm{x})^{2}\right] \\ &=3(\mathrm{x}+\mathrm{a}-\mathrm{x})(\mathrm{x}-\mathrm{a}+\mathrm{x}) \\ &=3 \mathrm{a}(2 \mathrm{x}-\mathrm{a}) \end{aligned} \] For maximum or minimum, \(\frac{\mathrm{dz}}{\mathrm{dx}}=0\) \(\Rightarrow \quad 3 \mathrm{a}(2 \mathrm{x}-\mathrm{a})=0\) \(\Rightarrow \quad 2 \mathrm{x}-\mathrm{a}=0\) \(\Rightarrow \quad \mathrm{x}=\frac{\mathrm{a}}{2}\)
On differentiating Eq. (ii) w.r.t ' \(x\) ', we get
\[
\begin{aligned}
\frac{\mathrm{d}^{2} \mathrm{z}}{\mathrm{dx}}=& 6 \mathrm{a} \\
\therefore \quad & \frac{\mathrm{d}^{2} \mathrm{z}}{\mathrm{dx}^{2}}>0
\end{aligned}
\]
\(\therefore z\) has a minimum value at \(x=\frac{a}{2}\).
Therefore, the numbers are \(\frac{a}{2}\) and \(a-\frac{a}{2}=\frac{a}{2}\).