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Of all the closed cylindrical cans, which enclose a given volume has the minimum surface area?

Asked by 7th September 2009, 1:09 PM
Answered by Expert
Answer:

Given volume = V

πr2h = V

h = V/πr2

S = total surface area = 2πr2 + 2πrh

= 2πr2 + 2πr(V/πr2)

 = 2πr2 + 2V/r

To minimize the surface area,

dS/dr = 4πr - 2V/r2 = 0

d2S/dr2 = 4π + 4V/r3 > 0  

2πr = V/r2

r = (V/2π)1/3 

r = (100/2π)1/3 

r = 2.52 cm and h = 5.02 cm.

The cylindrical can with r = 2.52 cm and h = 5.02 cm, will minimize the total surface area.

Regards,

Team,

TopperLearning.

Answered by Expert 7th September 2009, 1:44 PM
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