Question

Show that the height of a closed right circular cylinder of given volume and least surface area is equal to its diameter.

Answer 1

Let x be the radius of base, h be the height and S be the surface area of the closed right circular cylinder whose volume is V which is given to be constant.

Then `pir^2h` = V

∴ h = `"V"/(pir^2) = "A"/x^2`,               ...(1)

where A = `"V"/(pir^2)`, which is constant.

Now, S = 2πxh + 2πx²

= `2pix ("A"/x^2) + 2pix^2`         ...[By (1)]

= `(2pi"A")/x + 2pix^2`

∴ `"dS"/dx = d/dx((2pi"A")/x + 2pix^2)`

= `2pi"A"(- 1)x^-2 + 2pi xx 2x`

= `(-2pi"A")/x^2 + 4pix`
and
`(d^2S)/(dx^2) = d/dx((-2pi"A")/x^2 + 4pix)`

= `-2pi"A"(-2)x^-3 + 4pi xx 1`

= `(4pi"A")/x^3 + 4pi`

Now, `"dS"/dx = 0  "gives" (-2pi"A")/x^2 + 4pix`= 0

∴ `-2pi + 4pix^3` = 0

∴ `4pixx^3 = 2pi"A"`

∴ x³ = `"A"/(2)`

∴ x = `("A"/2)^(1/3)` 
and
`((d^2S)/(dx^2))_("at" x  = ("A"/2)^(1/3)`

= `(4pi"A")/(("A"/2)) + 4pi`
= 12π > 0

∴ by the second derivative test, S is minimum when  x = `("A"/2)^(1/3)`

When x = `("A"/2)^(1/3)`, from (1),

h = `"A"/(("A"/2)^(2/3)`

= `2^(2/3)."A"^(1/3)`

= `2.("A"/2)^(1/3)`

∴ h = 2x
Hence, the surface area is least when height of the closed right circular cylinder is equal to its diameter.