Question

a cylindrical container can hold a maximum of 3l 80ml milk. a new container is made by decreasing the base diameter of the original container by 20% and increasing its height by 25%. find the maximum amount of milk that the new container can hold.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{original}\mathrm{diameter}\mathrm{of}\mathrm{base}=\mathrm{d} \\ \mathrm{original}\mathrm{height}=\mathrm{h} \\ \mathrm{Now},\mathrm{original}\mathrm{volume}=\mathrm{\pi }{\left(\frac{\mathrm{d}}{2}\right)}^2\mathrm{h}=\frac{{\mathrm{\pi d}}^2\mathrm{h}}{4} \\ \mathrm{Now},\mathrm{new}\mathrm{diameter}=\mathrm{d}\text{'}=\mathrm{d}-20\%\mathrm{of}\mathrm{d}=\frac{4\mathrm{d}}{5} \\ \mathrm{new}\mathrm{height}=\mathrm{h}\text{'}=\mathrm{h}+25\%\mathrm{of}\mathrm{h}=\frac{5\mathrm{h}}{4} \\ \mathrm{Now},\mathrm{new}\mathrm{volume}=\mathrm{\pi }{\left(\frac{\mathrm{d}\text{'}}{2}\right)}^2\mathrm{h}\text{'}=\frac{\mathrm{\pi d}{\text{'}}^2\mathrm{h}\text{'}}{4}=\frac{\mathrm{\pi }}{4}\times \frac{16{\mathrm{d}}^2}{25}\times \frac{5\mathrm{h}}{4}=\frac{1}{5}{\mathrm{\pi d}}^2\mathrm{h}=\frac{4}{5}\mathrm{original}\mathrm{volume} \\ =\frac{4}{5}\times 3080=2464\mathrm{ml} \end{gathered}$$