Find the dimensions of the rectangle of perimeter 36 cm

Solution:

Let $l, b$ and $V$ be the length, breadth and volume of the rectangle, respectively. Then,

$2(l+b)=36$

$\Rightarrow l=18-b$                .....(1)

Volume of the cylinder when revolved about the breadth, $V=\pi l^{2} b$

$\Rightarrow V=\pi(18-b)^{2} b$           $[$ From eq. $(1)]$

$\Rightarrow V=\pi\left(324 b+b^{3}-36 b^{2}\right)$

$\Rightarrow \frac{d V}{d b}=\pi\left(324+3 b^{2}-72 b\right)$

For the maximum or minimum values of $V$, we must have

$\frac{d V}{d b}=0$

$\Rightarrow \pi\left(324+3 b^{2}-72 b\right)=0$

$\Rightarrow 324+3 b^{2}-72 b=0$

$\Rightarrow b^{2}-24 b+108=0$

$\Rightarrow b^{2}-6 b-18 b+108=0$

$\Rightarrow(b-6)(b-18)=0$

$\Rightarrow b=6,18$

Now,

$\frac{d^{2} V}{d b^{2}}=\pi(6 b-72)$

At $b=6:$

$\frac{d^{2} V}{d b^{2}}=\pi(6 \times 6-72)$

$\Rightarrow \frac{d^{2} V}{d b^{2}}=-36 \pi<0$

At $b=18:$

$\frac{d^{2} V}{d b^{2}}=\pi(6 \times 18-72)$

$\Rightarrow \frac{d^{2} V}{d b^{2}}=36 \pi>0$

Substituting the value of $b$ in eq. (1), we get

$l=18-6=12$

So, the volume is maximum when $l=12 \mathrm{~cm}$ and $b=6 \mathrm{~cm}$.