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 = πx²y   ...(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 equation (i) we get

V = πx²(18 – x)

⇒ V = π(18x² – x³)

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

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

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

∴ π(36x – 3x²) = 0

⇒ 36x – 3x² = 0

⇒ 3x(12 – x) = 0

⇒ x ≠ 0

∴ 12 – x = 0

⇒ x = 12

From equation (ii)

y = 18 – 12

y = 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)`

= π(36 – 6 × 12)

= π(36 – 72)

= – 36π < 0 maxima

Now volume of the cylinder so formed = πx²y

= π × (12)² × 6

= π × 144 × 6

= 864π cm³

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

Answer 2

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

Given, P = 36

2(x + y) = 36

x + y = 18

y = 18 – x

Let the rectangle revolved around side x and form cylinder with radius r and height (18 – x)

2πr = x

`r = x/(2π)`

Volume, V = πr²h

= `π(x/(2π))^2y`

= `π x^2/(4π^2) (18 - x)`

= `1/(4π) x^2 (18 - x)`

`V = 1/(4π) (18x^2 - x^3)`

Differentiate w.r.t. x

`(dV)/(dx) = 1/(4π) (36x - 3x^2)`

`(dV)/(dx) = 0`

For maxima and minima,

`1/(4π) (36x - 3x^2) = 0`

x ≠ 0, x = 12

`(d^2V)/(dx^2) = 1/(4π) (36 - 6x)`

`(d^2V)/(dx^2)]_(x = 12) = 1/(4π) [36 - 6(12)]`

= `1/(4π) (-36) < 0`

Hence, Volume is maximum when x = 12

y = 18 – x

= 18 – 12

= 6

Required length and breadth of rectangle are 12 and 6 respectively.

Volume `V = π(12/(2π))^2 (6)`

= `36/22 xx 6 xx 7`

= 68.72 unit³