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. Also, find the maximum volume.

Answer 1

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

$\because Perimeteroftherectangle=36cm\Rightarrow 2x+2y=36\Rightarrow x+y=18\Rightarrow y=18-x$
The rectangle is being revolved about its length y.
Then, volume (V) of resultant cylinder $=\pi x^2.y\Rightarrow V=\pi x^2.(18-x)=18\pi x^2-\pi x^3=\pi [18x^2-x^3]$
On differentiating both sides w.r.t.x, we get
$\frac{dV}{dx}=\pi \left(36\right(3x^2\left)\right)Now,\frac{dV}{dx}=0\Rightarrow 36x=3x^2\Rightarrow 3x^2-36x=0\Rightarrow 3(x^2-12x)=2\Rightarrow 3x(x-12)=0\Rightarrow x=0,x=12\therefore x=12$
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 rectangles are 12 cm and 6 cm, respectively.

$\therefore$ Maximum volume of resultant cylinder,
$(V{)}_{x=12}=\pi [18.(12{)}^2-(12{)}^3]=\pi [{12}^2(18-12)]=\pi \times 144\times 6=864\pi c{m}^3$