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 $a-\frac{a}{2}=\frac{a}{2}$.