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Question
Show that the height of a cylinder, which is open at the top, having a given surface area and greatest volume, is equal to the radius of its base.
Solution
Let R be the radius
H be the height
V be the volume
S be the total surface area
V = πR2 H
S = πR2 + 2π RH
⇒ H = `("S" - π"R"^2)/(2π"R")`
Substituting value of H in V
`"V" = 1/2 ("SR" = π"R"^3)`
`(d"V")/(d"R") = 1/2 ("S" -3π"R"^2)`
`(d"V")/(d"R") = 0`
⇒ `1/2 ("S" -3π"R"^2) = 0`
`"R" = sqrt("S"/(3π)`
`(d^2"V")/(d"R"^2) = 1/2 (0 - 6π"R")`
= -3πR
`(d^2"V")/(d"R"^2) = -3πsqrt("S"/(3π)`
= -`sqrt3πS < 0`
V is greatest when R = `sqrt("S"/(3π)`
H = `("S" - π xx ("S")/(3π))/(2πsqrt("S"/(3π))`
H = `((2S)/3)/(2sqrt((piS)/3))`
H = `sqrt("S"/(3π)`
Hence, proved that radius is equal to the height of the cylinder.
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