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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.
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Solution 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.
Solution 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
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