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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.
Solution
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
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