Of all the closed cylindrical cans (right cylinder), of a given volume 100 cm³, find the dimensions of the can which has the minimum surface area.

Let r and h be the radius and height of cylinder respectively.

V and S be the volume and surface area of cylinder respectively.

Given volume = 100 cm³

We know that

Volume of cylinder = πr²h

V = πr²h ⇒ 100 = πr²h

We need to minimize surface area.

Surface area of cylinder = 2πrh + 2πr²

⇒ S = 2πrh + 2πr²

Substituting the value of h,

Differentiating wrt x,

Putting ,

⇒ -200r^-2 + 4πr = 0

⇒ -200 + 4πr³ = 0

⇒ πr³ = 50

Now, finding ,

Substituting the value of r,

Hence, S is minimum at

Now, finding h,

Hence total surface area is least when

Radius of base is and height is cm.