Step 1

Perimeter of the rectangle is

$p=36$

Let the length of the rectangle be = $x$

Let the breadth of the rectangle be = $y$

$p=2(x+y)=36$

$ \Rightarrow x+y=18 \Rightarrow y=18-x$

As the rectangle revolves, the

length = $x$

width = $18-x$

height = $x$

$ \therefore$ volume of the cubid is

$ v = x \times (18-x) \times x$

$ = x^2(18-x)$

$ = 18x^2-x^3$

Step 2

Differentiating w.r.t $x$ we get

$ \large\frac{dv}{dx}=36x-3x^2$

differentiating w.r.t $x$ we get,

$ \large\frac{d^2v}{dx^2}=36-6x$

Step 3

For maximum or minimum we must have

$ \large\frac{dv}{dx}=0$

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

$ \Rightarrow 3x=36$

$ x=12$

$ \therefore y=18-12$

$=6$

$ \therefore \bigg( \large\frac{d^2v}{dx^2}\bigg)_{(12, 6)} = 36-6 \times 12$

$=36-72$

$-36<0$

Hence $v$ is maximum when $x=12$

Step 4

$ \therefore $ The dimensions of the rectangle is

$x=12\: and \: y=6$

The maximum volume of the box is given by

$ v=12 \times 12 \times 6$

$ = 864$ cubic units