Find dimensions of a rectangle of perimeter 12 cm which will generate maximum volume when swept along a circular rotation keeping the shorter side…

Let Length and Breadth of the rectangle be ' $\displaystyle x$ ' and ' $\displaystyle y$ ' respectively. Also ' $\displaystyle r$ ' be the radius of the cylinder then, \[2(x+y)=12 \Rightarrow x+y=6,2 \pi r=x \] $\displaystyle V($ Volume of cylinder $\displaystyle )=\pi r^{2} y \Rightarrow V=\pi\left(\frac{x}{2 \pi}\right)^{2}(6-x)=\frac{1}{4 \pi}\left(6 x^{2}-x^{3}\right)$ \[\begin{gathered} V^{\prime}(x)=0 \Rightarrow \frac{1}{4 \pi}\left(12 x-3 x^{2}\right)=0 \Rightarrow x=4,(\because x \neq 0) \\ V^{\prime \prime}(x)=\frac{1}{4 \pi}(12-6 x) \Rightarrow V^{\prime \prime}(4)=-\frac{3}{\pi}<0 \end{gathered} \] The volume of the cylinder obtained by the rotation will be maximum if the dimensions of the rectangle are $\displaystyle x=4 \mathrm{~cm}, y=2 \mathrm{~cm}$.