Given,

• The closed is cylindrical can with a circular base and top.

• The volume of the cylinder is 1 litre = 100 cm³.

• The surface area of the box is minimum.

Let us consider,

• The radius base and top of the cylinder be ‘r’ units. (skin coloured in the figure)

• The height of the cylinder be ‘h’units.

• As the Volume of cylinder is given, V = 100cm³

The Volume of the cylinder= πr²h

100 = πr²h

---- (1)

The Surface area cylinder is = area of the circular base + area of the circular top + area of the cylinder

S = πr² + πr² + 2πrh

S = 2 πr² + 2πrh

[substituting (1) in the volume formula]

------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with r and then equating it to zero. This is because if the function f(r) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to r:

[Since and ]

------- (3)

To find the critical point, we need to equate equation (3) to zero.

---- (4)

Now to check if this critical point will determine the minimum surface area of the box, we need to check with second differential which needs to be positive.

Consider differentiating the equation (3) with r:

----- (5)

[Since and ]

Now let us find the value of

As , so the function S is minimum at

As S is minimum from equation (4)

V = 2πr³

Now in equation (1) we have,

h = 2r = diameter

Therefore when the total surface area of a cone is minimum, then height of the cone is equal to twice the radius or equal to its diameter.