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23 Maths -- Derivatives and its Application

A closed cylindrical can is to be made so hat its volume is 52 cm3 . Find its dimensions if the surface is to be a minimum.

A closed cylindrical can is to be made so hat its volume is 52 cm3 . Find its dimensions if the surface is to be a minimum.

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Let r be the radius, h be the height and V be the volume of the cylinder.

V=πr2 h

52=πr2 h

h=52/πr2

Let  S be the surface area of cylinder

S=2πr(r+h)

S=2πr2 + 2πrh

S=2πr2 + 2πr * 52/πr2

S=2πr2 + 104/r

dS/dr = 4πr-104/r2 

d2S/dr2 = 4π+208/r3 >0 (Minimum)

Since

dS/dr = 0

4πr-104/r2 =0

4πr3 -104=0

4πr3=104

r3=26/π

r=(26/π)1/3 

And, 

h=52/πr2 

h= 52/{π × (26/πr)}. 

h=2r = 2×(26/π)1/3

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Let x and y be the length and breadth of the rectangle respectively.

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y=60-3x/2

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