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Question
Divide 64 into two parts such that the sum of the cubes of two parts is minimum.
Solution
\[\text { Suppose 64 is divided into two partsxand 64-x. Then, }\]
\[z = x^3 + \left( 64 - x \right)^3 \]
\[ \Rightarrow \frac{dz}{dx} = 3 x^2 + 3 \left( 64 - x \right)^2 \]
\[\text { For maximum or minimum values of z, we must have }\]
\[\frac{dz}{dx} = 0\]
\[ \Rightarrow 3 x^2 + 3 \left( 64 - x \right)^2 = 0\]
\[ \Rightarrow 3 x^2 = 3 \left( 64 - x \right)^2 \]
\[ \Rightarrow x^2 = x^2 + 4096 - 128x\]
\[ \Rightarrow x = \frac{4096}{128}\]
\[ \Rightarrow x = 32\]
\[\text { Now, }\]
\[\frac{d^2 z}{d x^2} = 6x + 6\left( 64 - x \right) \]
\[ \Rightarrow \frac{d^2 z}{d x^2} = 384 > 0\]
\[\text { Thus, z is minimum when 64 is divided into two equal parts, 32 and 32.}\]
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