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# 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 = πr^{2}h = 100

⇒ h = 100/πr^{2}

Surface area (S) is given by:

S = 2π r^{2} + 2π rh

= 2π r^{2} + 200

Hence,

dS/dr = 4π r - 200/r^{2}

d^{2}S/dr^{2} = 4π r + 400/r^{3}

Now,

dS/dr = 0

⇒ 4πr - 200/r^{2} = 0

4πr = 200/r^{2}

⇒ r^{3} = 200/4π = 50/π

r = (50/π)^{1/3}

When r = (50/π)^{1/3}

Then,

d^{2}S/dr^{2} = 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

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

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