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. 

Answer 1

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.