#### Question

How should we choose two numbers, each greater than or equal to \[-\] 2, whose sum______________ so that the sum of the first and the cube of the second is minimum?

#### Solution

\[\text { Let the two numbers bexandy.Then },\]

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

\[\text { Now,} \]

\[z = x + y^3 \]

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

\[ \Rightarrow \frac{dz}{dx} = 1 + 3 \left( \frac{1}{2} - x \right)^2 \]

\[\text { For maximum or minimum values of z, we must have }\]

\[\frac{dz}{dx} = 0\]

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

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

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

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

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

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

\[\text { At } x = \frac{1}{2} \pm \frac{1}{\sqrt{3}}: \]

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

\[ \Rightarrow \frac{- 6}{\sqrt{3}} < 0\]

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

\[\text { At x } = \frac{1}{2} - \frac{1}{\sqrt{3}}: \]

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

\[ \Rightarrow \frac{6}{\sqrt{3}} > 0\]

\[\text { Thus, z is minimum when x

}= \frac{1}{2} - \frac{1}{\sqrt{3}} . \]

\[x + y = \frac{1}{2}\]

\[\text { Substituting the value of x in eq }.\left( 1 \right),\text { we get }\]

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

\[y = \frac{1}{\sqrt{3}}\]

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