Find the dimensions of the rectangle of perimeter 36cm which will sweep out a volume as large as possible when revolved about one of its sides.

Explanation

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

`therefore` Perimeter of the reactangle = 36 cm
`rArr 2x+2y=36`
`rArr x+y=18`
`rArr y=18-x` …………….(i)
Let the rectangle is being revolved about its length y.
Then, volume (V) of resultant cylinder `=pix^(2).y`
`rArr V=pix^(2).(18-x)` `[therefore V=pir^(2)h]` [using Eq. (i)]
`=18pix^(2)-pix^(2)=pi[18x^(2)-x^(3)]`
On differentiating both sides w.r.t. x, we get
`(dV)/(dx) = pi(36pi-3x^(2))`
Now, `(dV)/(dx)=pi(36pi-3x^(2))`
Now, `(dV)(dx) =0`
`rArr 36x=3x^(2)`
`rArr 3x^(2)-36x=0`
`rArr 3(x^(2)-12x)=0`
`rArr 3x(x-12)=0`
`therefore x=12` `[therefore, x ne0]`
Again, differentiating w.r.t. x, we get
`(d^(2)V)/(dx^(2))=pi(36-6x)`
`rArr (d^(2)V)/(dx^(2))_(x=12) = pi(36-6xx12)=-36pi lt0`
At x=12, volume of the resultant cylinder is the maximum.
So, the dimension of rectangle 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 xx 144 xx6`
`=864 pi cm^(3)`