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