Question

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.

• A. $x=4,y=14$
• B. $x=6,y=12$
• C. $x=8,y=10$
• D. $x=9,y=9$

Answer 1

The correct option is B $x=6,y=12$

Let $x$ and $y$ be the dimensions of the rectangle.
Now, perimeter $P=2(x+y)$
$2(x+y)=36$
$\Rightarrow x+y=18.$
When revolved about side length $y$, it will generate a cylinder of height $y$ and radius $x$ so that
$V=\pi r^{2}h=\pi x^{2}y$
$V=\pi x^2(18-x)=\pi (18x^2-x^3)$
$\therefore \frac{dV}{dx}=\pi (36x-3x^2)$
For maximum or minimum,
$\frac{dV}{dx}=0$
$\Rightarrow x=12$
$\frac{d^{2}V}{d{x}^2}=\pi (36-6x)=-ive$
when $x=12\therefore y=6.$