Question

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

Answer 1

$$\begin{gathered} \text{Suppose 64 is divided into two parts}\mathit{\text{x}}\text{and 64-}\mathit{\text{x.}}\text{Then,} \\ z=x^3+{\left(64-x\right)}^3 \\ \Rightarrow \frac{dz}{dx}=3x^2+3{\left(64-x\right)}^2 \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}z,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{\mathrm{d}\mathrm{z}}{\mathrm{d}\mathrm{x}}=0 \\ \Rightarrow 3{\mathrm{x}}^2+3{\left(64-\mathrm{x}\right)}^2=0 \\ \Rightarrow 3x^2=3{\left(64-x\right)}^2 \\ \Rightarrow x^2=x^2+4096-128x \\ \Rightarrow x=\frac{4096}{128} \\ \Rightarrow x=32 \\ \mathrm{Now}, \\ \frac{d^{2}z}{d{x}^2}=6x+6\left(64-x\right) \\ \Rightarrow \frac{d^{2}z}{d{x}^2}=384>0 \\ \text{Thus,}\mathit{\text{z}}\text{is minimum when 64 is divided into two equal parts, 32 and 32.} \end{gathered}$$.