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 x and y be the length and breadth of a given rectangle ABCD as per question, the rectangle be revolved about side AD which will make a cylinder with radius x and height y.

∴ Volume of the cylinder V = `(pi"r"^2)/"h"`

⇒ V = `pix^2y`  .....(i)

Now perimeter of rectangle P = 2(x + y)

⇒ 36 = 2(x + y)

⇒ x + y = 18

⇒ y = 18 – x  ....(iii)

Putting the value of y in eq. (i) we get

V = `pix^2(18 - x)`

⇒ V = `pi(18x^2 - x^3)`

Differentiating both sides w.r.t. x, we get

`"dV"/"dx" = pi(36x - 3x^2)`  ....(iii)

For local maxima and local minima `"dV"/"dx"` = 0

∴ `pi(36x - 3x^2)` = 0

⇒ 36x – 3x² = 0

⇒ 3x(12 – x) = 0

⇒ x  0

∴ 12 – x = 0

⇒ x = 12

From equation (ii)

y = 18 – 12 = 6

Differentiating equation (iii) w.r.t. x,

We get `("d"^2"V")/("dx"^2) = pi(36 - 6x)`

At x = 12 `("d"^2"V")/("dx"^2)`

= `pi(36 - 6 xx 12)`

= `pi(36 - 72)`

= `- 36pi < 0` maxima

Now volume of the cylinder so formed = `pix^2y`

= `pi xx (12)^2 xx 6`

= `pi xx 144 xx 6`

= 864π cm³

Hence, the required dimensions are 12 cm and 6 cm and the maximum volume is 864π cm³