The maximum volume of a right circular cylinder if the sum of its radius and height is $6 \mathrm{~m}$ is

The maximum volume of a right circular cylinder if the sum of its radius and height is $6 \mathrm{~m}$ is
  1. $16 \pi m^{3}$
  2. $32 \pi m^{3}$
  3. $4 \pi m^{3}$
  4. $64 \pi m^{3}$

Solution

Let height of cylinder be $\mathrm{h}$ and radius be $\mathrm{r} \Rightarrow \mathrm{r}+\mathrm{h}=6 \Rightarrow \mathrm{h}=6-\mathrm{r}$ Volume of cylinder $\mathrm{V}=\pi \mathrm{r}^{2} \mathrm{~h}$ $=\pi \cdot r^{2} \cdot(6-r)=\pi\left(6 r^{2}-r^{3}\right)$ For the maximum volume, $\frac{\mathrm{dV}}{\mathrm{dr}}=0$ $\pi\left(12 r-3 r^{2}\right)=0 \Rightarrow 12 r=3 r^{2} \Rightarrow r=4 \Rightarrow h=6-4=2$. $\therefore$ Volume of cylinder $=\pi \times 4^{2} \times 2=32 \pi \cdot \mathrm{m}^{3}$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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