Question

A rectangle of perimeter 36 cm is revolved around one of its sides to sweep out a cylinder of maximum volume.

Find the dimensions of the rectangle.

Answer 1

Let l, w and p be the length and width and perimeter of the rectangle, respectively. and V be the volume of the cylinder.

Now, P = 2(l + w) = 36

⇒ l + w = 18

Volume of the cylinder = V = πr²h   ...{h = w; r = l}

V = πr²w

⇒ V = πr² (18 – l)

⇒ V = 18πr² – πr³ 

For maximum volume differential volume w.r.t ‘l’ both sides,

`(dv)/(dl)=d/(dl)(18pir^2-pir^3)`

`(dv)/(dl)=36pil-3pil^2`

= 3πl (12 – l)

For maximum volume; `(dv)/(dl)=0`

⇒ 3πl (12 – l) = 0

l = 0, 12

again differential both sides w.r.t ‘l’ 

`(d^2v)/(dl^2)=36pi-6pil`

At l = 12,

`(d^2v)/(dl^2)` = 36π – 72π

= –36π < 0

Hence, volume of cylinder is max. at l = 12 cm and width w = 18 – 12 = 6 cm