Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible, when revolved about one of its sides. Also, find the maximum volume.

Let breadth and length of the rectangle be $x$ and $y$, respectively.

Perimeter of the rectangle $= 36\;{\rm{cm}}$
$\Rightarrow$ $2x + 2y = 36$

$\Rightarrow$ $x + y = 18$
$\Rightarrow$ $y = 18 - x$ ….(i)

Let the rectangle is being revolved about its length $y$.

Then, volume $(V)$ of resultant cylinder $= \pi {x^2} \cdot y$
$\Rightarrow$ $V = \pi {x^2} \cdot (18 - x)$ [using Eq. (i)]

$= 18\pi {x^2} - \pi {x^3} = \pi \left[ {18{x^2} - {x^3}} \right]$

On differentiating both sides w.r.t. $x$, we get
$\frac{{dV}}{{dx}} = \pi \left( {36x - 3{x^2}} \right)$

Now, $\frac{{dV}}{{dx}} = 0$
$\Rightarrow$ $36x = 3{x^2}$

$\Rightarrow$ $3{x^2} - 36x = 0$

$\Rightarrow$ $3\left( {{x^2} - 12x} \right) = 0$
$\Rightarrow$ $3x(x - 12) = 0$

$\Rightarrow$ $x = 0,x = 12$

Therefore,$x = 12$

Again, differentiating w.r.t. $x$, we get
$\frac{{{d^2}V}}{{d{x^2}}} = \pi (36 - 6x)$
$\Rightarrow$ ${\left( {\frac{{{d^2}V}}{{d{x^2}}}} \right)_{x = 12}} = \pi (36 - 6 \times 12) = - 36\pi < 0$

At $x = 12,$ volume of the resultant cylinder is the maximum.

So, the dimensions of rectangle are $12\;{\rm{cm}}$ and $6\;{\rm{cm}}$, respectively. [using Eq. (i)]

Therefore maximum volume of resultant cylinder,

${(V)_{x = 12}} = \pi \left[ {18 \cdot {{(12)}^2} - {{(12)}^3}} \right]$
$= \pi \left[ {{{12}^2}(18 - 12)} \right]$

$= \pi \times 144 \times 6$
$= 864\pi {\rm{c}}{{\rm{m}}^3}$