Question

Divide 15 into two parts such that the square of one multiplied with the cube of the other is minimum.

Answer 1

$$\begin{gathered} \text{Let the two numbers be}\mathit{\text{x}}\text{and}\mathit{\text{y.}}\text{Then,} \\ x+y=15...\left(1\right) \\ \mathrm{Now}, \\ z=x^2{y}^3 \\ \Rightarrow z=x^2{\left(15-x\right)}^3\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow \frac{dz}{dx}=2x{\left(15-x\right)}^3-3x^2{\left(15-x\right)}^2 \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}z,\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dz}{dx}=0 \\ \Rightarrow 2x{\left(15-x\right)}^3-3x^2{\left(15-x\right)}^2=0 \\ \Rightarrow 2x\left(15-x\right)=3x^2 \\ \Rightarrow 30x-2x^2=3x^2 \\ \Rightarrow 30x=5x^2 \\ \Rightarrow x=6\mathrm{and}y=9 \\ \frac{d^{2}z}{d{x}^2}=2{\left(15-x\right)}^3-6x{\left(15-x\right)}^2-6x{\left(15-x\right)}^2+6x^2\left(15-x\right) \\ \mathrm{At}x=6: \\ \frac{d^{2}z}{d{x}^2}=2{\left(9\right)}^3-36{\left(9\right)}^2-36{\left(9\right)}^2+6\left(36\right)\left(9\right) \\ \Rightarrow \frac{d^{2}z}{d{x}^2}=-2430<0 \\ \mathrm{Thus},z\mathrm{is}\mathrm{maximum}\mathrm{when}x=6\mathrm{and}y=9. \\ \mathrm{So},\mathrm{the}\mathrm{required}\mathrm{two}\mathrm{parts}\mathrm{into}\mathrm{which}15\mathrm{should}\mathrm{be}\mathrm{divided}\mathrm{are}6\mathrm{and}9. \end{gathered}$$