MHT CET · Maths · Application of Derivatives
The maximum volume of a right circular cylinder if the sum of its radius and height is
\(6 \mathrm{~m}\) is
- A \(16 \pi m^{3}\)
- B \(32 \pi m^{3}\)
- C \(4 \pi m^{3}\)
- D \(64 \pi m^{3}\)
Answer & Solution
Correct Answer
(B) \(32 \pi m^{3}\)
Step-by-step Solution
Detailed explanation
(A)
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 \(=\pi \times 4^{2} \times 2=32 \pi \cdot \mathrm{m}^{3}\)
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 \(=\pi \times 4^{2} \times 2=32 \pi \cdot \mathrm{m}^{3}\)
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