Show that the height of a closed cylinder of given volume and the leas

1.	Show that the height of a closed cylinder of given volume and the least surface area is equal to its diameter.
Answer» Let r be the radius of the base and h the height of a cylinder. The surface area is given by, S = 2 π r²+ 2 π rh h = S - 2πr²/2πr ………(1) Let V be the volume of the cylinder. Therefore, V = πr²h V = πr² \((\frac{S-2\pi r^3}{2\pi r})\) …….Using equation 1 V = \(\frac{Sr-2\pi r^3}{2}\) Differentiating both sides w.r.t r, we get, \(\frac{dV}{dr}=\frac{S}{2}\) - 3πr² ……….(2) For maximum or minimum, we have, \(\frac{dV}{dr}\) = 0 ⇒ \(\frac{S}{2}\) - 3πr² = 0 ⇒ S = 6πr² 2πr² + 2πrh = 6πr² h = 2r Differentiating equation 2, with respect to r to check for maxima and minima, we get, \(\frac{d^2V}{dr^2}\) = - 6πr < 0 Hence, V is maximum when h = 2r or h = diameter

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

Answer»

Let r be the radius of the base and h the height of a cylinder.

The surface area is given by,

S = 2 π r²+ 2 π rh

h = S - 2πr²/2πr ………(1)

Let V be the volume of the cylinder.

Therefore, V = πr²h

V = πr² \((\frac{S-2\pi r^3}{2\pi r})\) …….Using equation 1

V = \(\frac{Sr-2\pi r^3}{2}\)

Differentiating both sides w.r.t r, we get,

\(\frac{dV}{dr}=\frac{S}{2}\) - 3πr² ……….(2)

For maximum or minimum, we have,

\(\frac{dV}{dr}\) = 0

⇒ \(\frac{S}{2}\) - 3πr² = 0

⇒ S = 6πr²

2πr² + 2πrh = 6πr²

h = 2r

Differentiating equation 2, with respect to r to check for maxima and minima, we get,

\(\frac{d^2V}{dr^2}\) = - 6πr < 0

Hence, V is maximum when h = 2r or h = diameter