Question

A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, in cutting off squares from each corners and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?

Answer 1

$$\begin{gathered} \mathrm{Suppose}\mathrm{square\; of\; side}\mathrm{measuring}x\mathrm{cm\; is\; cut\; off}. \\ \mathrm{Then},\mathrm{the}\mathrm{length},\mathrm{breadth}\mathrm{and}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{box}\mathrm{will}\mathrm{be}\left(45-x\right),\left(24-2x\right)\mathrm{and}x,\mathrm{respectively}. \\ \Rightarrow \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{box},V=\left(45-2x\right)\left(24-2x\right)\mathrm{x} \\ \Rightarrow \frac{dV}{dx}=\left(45-2x\right)\left(24-2x\right)-2x\left(45-2x\right)-2x\left(24-2x\right) \\ \mathrm{For}\mathrm{maximum}\mathrm{or}\mathrm{minimum}\mathrm{values}\mathrm{of}\mathrm{V},\mathrm{we}\mathrm{must}\mathrm{have} \\ \frac{dV}{dx}=0 \\ \Rightarrow \left(45-2x\right)\left(24-2x\right)-2x\left(45-2x\right)-2x\left(24-2x\right)=0 \\ \Rightarrow 4x^2+1080-138x-48x+4x^2+4x^2-90x=0 \\ \Rightarrow 12x^2-276x+1080=0 \\ \Rightarrow x^2-23x+90=0 \\ \Rightarrow x^2-18x-5x+90=0 \\ \Rightarrow x\left(x-18\right)-5\left(x-18\right)=0 \\ \Rightarrow x-18=0\mathrm{or}x-5=0 \\ \Rightarrow x=18\mathrm{or}x=5 \\ \mathrm{Now}, \\ \frac{d^{2}V}{d{x}^2}=24x-276 \\ {\overline{)\frac{d^{2}V}{d{x}^2}}}_{x=5}=120-276=-156<0 \\ {\overline{)\frac{d^{2}V}{d{x}^2}}}_{x=18}=432-276=156>0 \\ \mathrm{Thus},\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{box}\mathrm{is}\mathrm{maximum}\mathrm{when}x=5\mathrm{cm}. \\ \mathrm{Hence},\mathrm{the}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{to}\mathrm{be}\mathrm{cut}\mathrm{off}\mathrm{measures}5\mathrm{cm}. \end{gathered}$$