Question
Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.
✓
Answer
Let radius of the cylinder = r
Height of the cylinder = h|
$s = 2\pi rh + 2\pi {r^2}$ (1)
$\Rightarrow\frac{{s-2\pi {r^2}}}{{2\pi r}} = h$
Now volume of cylinder is, $v = \pi {r^2}h$
$\Rightarrow$ $v = \pi .{r^2}\left( {\frac{{s - 2\pi {r^2}}}{{2\pi r}}} \right)$
$\Rightarrow v = \frac{1}{2}\left[ {sr - 2\pi {r^3}} \right]$
Now, $\frac{{dv}}{{dr}} = \frac{1}{2}\left[ {s - 6\pi {r^2}} \right]$
$\Rightarrow\frac{{{d^2}v}}{{d{r^2}}} = \frac{1}{2}\left[ {0 - 12\pi r} \right]$
For maximum/minimum
$\frac{{dv}}{{dr}} = 0$
$ \Rightarrow s = 6\pi {r^2}$
From equ (1)
$\Rightarrow 2\pi rh + \ 2π {r^2} = 6\pi {r^2}$
$\Rightarrow r = \frac{h}{2}$
${\left[ {\frac{{{d^2}v}}{{d{r^2}}}} \right]_{r = \frac{h}{2}}} = \frac{1}{2}\left[ {0 - 12\pi \times \frac{h}{2}} \right]$
$ = - 3\pi h < 0$
$\Rightarrow$ s is maximum at $r = \frac{h}{2}$
Hence h = 2r
Height of the cylinder = h|
$s = 2\pi rh + 2\pi {r^2}$ (1)
$\Rightarrow\frac{{s-2\pi {r^2}}}{{2\pi r}} = h$
Now volume of cylinder is, $v = \pi {r^2}h$
$\Rightarrow$ $v = \pi .{r^2}\left( {\frac{{s - 2\pi {r^2}}}{{2\pi r}}} \right)$
$\Rightarrow v = \frac{1}{2}\left[ {sr - 2\pi {r^3}} \right]$
Now, $\frac{{dv}}{{dr}} = \frac{1}{2}\left[ {s - 6\pi {r^2}} \right]$
$\Rightarrow\frac{{{d^2}v}}{{d{r^2}}} = \frac{1}{2}\left[ {0 - 12\pi r} \right]$
For maximum/minimum
$\frac{{dv}}{{dr}} = 0$
$ \Rightarrow s = 6\pi {r^2}$
From equ (1)
$\Rightarrow 2\pi rh + \ 2π {r^2} = 6\pi {r^2}$
$\Rightarrow r = \frac{h}{2}$
${\left[ {\frac{{{d^2}v}}{{d{r^2}}}} \right]_{r = \frac{h}{2}}} = \frac{1}{2}\left[ {0 - 12\pi \times \frac{h}{2}} \right]$
$ = - 3\pi h < 0$
$\Rightarrow$ s is maximum at $r = \frac{h}{2}$
Hence h = 2r
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