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प्रश्न
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?
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उत्तर
\[\text { Let the two numbers bexandy. Then },\]
\[x, y > - 2 \text { and }x + y = \frac{1}{2} ........ (1)\]
\[\text { Now,} \]
\[z = x + y^3 \]
\[ \Rightarrow z = x + \left( \frac{1}{2} - x \right)^3 .........\left[ \text { From eq } . \left( 1 \right) \right]\]
\[ \Rightarrow \frac{dz}{dx} = 1 + 3 \left( \frac{1}{2} - x \right)^2 \]
\[\text { For maximum or minimum values of z, we must have }\]
\[\frac{dz}{dx} = 0\]
\[ \Rightarrow 1 + 3 \left( \frac{1}{2} - x \right)^2 = 0\]
\[ \Rightarrow \left( \frac{1}{2} - x \right)^2 = \frac{1}{3}\]
\[ \Rightarrow \left( \frac{1}{2} - x \right) = \pm \frac{1}{\sqrt{3}}\]
\[ \Rightarrow x = \frac{1}{2} \pm \frac{1}{\sqrt{3}}\]
\[\frac{d^2 z}{d x^2} = 6\left( \frac{1}{2} - x \right)\]
\[ \Rightarrow \frac{d^2 z}{d x^2} = 3 - 6x\]
\[\text { At } x = \frac{1}{2} \pm \frac{1}{\sqrt{3}}: \]
\[\frac{d^2 z}{d x^2} = 3 - 6\left( \frac{1}{2} + \frac{1}{\sqrt{3}} \right)\]
\[ \Rightarrow \frac{- 6}{\sqrt{3}} < 0\]
\[\text { Thus, z is maximum when x } = \frac{1}{2} + \frac{1}{\sqrt{3}} . \]
\[\text { At x } = \frac{1}{2} - \frac{1}{\sqrt{3}}: \]
\[\frac{d^2 z}{d x^2} = 3 - 6\left( \frac{1}{2} - \frac{1}{\sqrt{3}} \right)\]
\[ \Rightarrow \frac{6}{\sqrt{3}} > 0\]
\[\text { Thus, z is minimum when x
}= \frac{1}{2} - \frac{1}{\sqrt{3}} . \]
\[x + y = \frac{1}{2}\]
\[\text { Substituting the value of x in eq }.\left( 1 \right),\text { we get }\]
\[y = - \frac{1}{2} + \frac{1}{\sqrt{3}} + \frac{1}{2}\]
\[y = \frac{1}{\sqrt{3}}\]
\[\text { So, the required two numbers are } \left( \frac{1}{2} - \frac{1}{\sqrt{3}} \right) \text { and } \frac{1}{\sqrt{3}} .\]
