Question

Divide 64 into two parts such that the sum of the cubes of two parts is minimum.

Answer 1

\[\text { Suppose 64 is divided into two partsxand 64-x. Then, }\]

\[z = x^3 + \left( 64 - x \right)^3 \]

\[ \Rightarrow \frac{dz}{dx} = 3 x^2 + 3 \left( 64 - x \right)^2 \]

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

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

\[ \Rightarrow 3 x^2 + 3 \left( 64 - x \right)^2 = 0\]

\[ \Rightarrow 3 x^2 = 3 \left( 64 - x \right)^2 \]

\[ \Rightarrow x^2 = x^2 + 4096 - 128x\]

\[ \Rightarrow x = \frac{4096}{128}\]

\[ \Rightarrow x = 32\]

\[\text { Now, }\]

\[\frac{d^2 z}{d x^2} = 6x + 6\left( 64 - x \right) \]

\[ \Rightarrow \frac{d^2 z}{d x^2} = 384 > 0\]

\[\text { Thus, z is minimum when 64 is divided into two equal parts, 32 and 32.}\]