#### Question

Sum

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 = πR^{2} H

S = πR^{2} + 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.

Is there an error in this question or solution?

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