calculus
posted by saud .
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.

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 ) 
ð = pi number