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Question
The volume of a cylinder is given by the formula V = `pi"r"^2"h"`. Find the greatest and least values of V if r + h = 6
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Solution
Given r + h = 6
⇒ r = 6 – h
Volume V = πr2h
V = π(6 – h)2h
`"dV"/"dh"` = π[(6 – h)2(1) + 2h(6 – h)(– 1)]
= π(6 – h)[6 – 3h]
For maximum or minimum,
`"dV"/"dh"` = 0
⇒ π(6 – h)(6 – 3h) = 0
⇒ h = 6, h = 2
h = 6 is not possible as r + h = 6
∴ h = 2
`("d"^2"V")/("dh"^2)` = π[(6 – h)(– 3) + (6 – 3h)(–1)]
= π[6h – 24]
At h = 2, `("d"^2"V")/("dh"^2) < 0`
∴ Volume of the cylinder is maximum when h = 2 and r = 6 – 2 = 4
Greatest value of V = π(4)2(2) = 32 π
Least value of V = 0
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