Q. 18

# A cylindrical can

Given,

The can is cylindrical with a circular base

The volume of the cylinder is 1 litre = 1000 cm2.

The surface area of the box is minimum as we need to find the minimum dimensions.

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 = 1000cm3

The Volume of the cylinder= πr2h

1000 = 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.

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:

----- (4)

[Since and ]

Now let us find the value of

As , so the function S is minimum at

Now substituting r in equation (1)

Therefore the radius of base of the cylinder, and height of the cylinder, where the surface area of the cylinder is minimum.

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