Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimeters, find the dimensions of the can which has the minimum surface area?

Solution:

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

Then, the surface area (S) of the cylinder is given by,

V = πr²h = 100

⇒ h = 100/πr²

Surface area (S) is given by:

S = 2π r² + 2π rh

= 2π r² + 200

Hence,

dS/dr = 4π r - 200/r²

d²S/dr² = 4π r + 400/r³

Now,

dS/dr = 0

⇒ 4πr - 200/r² = 0

4πr = 200/r²

⇒ r³ = 200/4π = 50/π

r = (50/π)^1/3

When r = (50/π)^1/3

Then,

d²S/dr² = 0

By the second derivative test, the surface area is the minimum when the radius of the cylinder is (50/π)^1/3 cm.

When r = (50/π)^1/3

Then,

h = 100/π(50/π)^2/3

= (2 x 50)/(50/π)^2/3(π)^{1 - 2/3}

= 2(50/π)^1/3

Hence, the required dimensions of the can which has the minimum surface area are given by radius (50/π)^1/3 cm and height 2(50/π)^1/3 cm

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 21

Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimeters, find the dimensions of the can which has the minimum surface area?

Summary:

The required dimensions of the can which has the minimum surface area are given by radius (50/π)^1/3 cm and height 2(50/π)^1/3 cm