Find the radius and hight of cylinder with voume 64π and radius r between 1 and 5 that has smallest possible surface area. A cylinder of radius r and hight aah 2π^2+2π r h and π r ^(2) h.

Answer 1

V = r ^ 2 ð h

64 ð = r ^ 2 ð h Divide both sides with ð

64 = r ^ 2 h Divide both sides with r ^ 2

64 / r ^ 2 = h

h = 64 / r ^ 2

A = 2 r ^ 2 ð + 2 r ð h

A = 2 r ^ 2 ð + 2 r ð 64 / r ^ 2

A = 2 r ^ 2 ð + 128 r ð / r ^ 2

A = 2 r ^ 2 ð + 128 ð / r

d A / d r = ( 4 ð ( r ^ 3 - 32 ) ) / r ^ 2

Function has extreme value where first derivative = 0
dA / dr = 0

Solutions:

r = -1.5874 + 2.7495 i

r = -1.5874 2.7495 i

and

r = 2 2 ^( 2 / 3 ) = 3,1748

So :

r = 2 2 ^( 2 / 3 ) = 3,1748

Second derivative :

4 ð ( 64 / r ^ 3 + 1 ) =

4 ð ( 64 / 32 + 1 ) =

4 ð ( 2 + 1 ) =

4 ð 3 = 12 ð > 0

Remark : r ^ 3 = [ 2 2 ^( 2 / 3 ) ] ^ 3 = 32

If second derivative > 0

that is minimum value of function.

So :

r = 2 64 / [ 2 2 ^( 2 / 3 ) ] ^ 2

h = 64 / r ^ 2

h = h = 64 / [ 2 2 ^( 2 / 3 ) ] ^ 2

h = 4 2 ^( 2 / 3 )

Answer 2

ð = pi number