Calculus problem, i calculate so many time, but i cannot find the same provided answers, please teach me.?

Question

A cylindrical waste container with no top, a diameter of at least 2 feet, and a volume of 26cubic feet to be constructed. What should its radius be under conditions?
a.65 square feet of the material will be used to construct it ?
b.the smallest possible amount of the material will be used to construct it (how much material is needed?)
the answers are
a. r = 4.09977
b. r = 1.996; 37.566in^2

help me ah, thanks.

I know how to do part a, but i still do not know how to do part b.

what stand for dA/dr?

Answer 1

a. Surface area of cylinder = 2π r h
r = radius
h = height of the cylinder
Total area, A = 65 = 2π r h + πr^2 <--- including bottom
constraints are
diameter = 2r > = 2 feet
Volume = π r^2 h = 26

Therefore from the Area 65 = 2π r h + πr^2, express h in terms of r:
h = (65 - πr^2)/(2π r)

substituting r into the equation for volume
π r^2 h = 26
π r^2 (65 - πr^2)/(2π r) = 26
(0.5)r (65 - πr^2) = 26
32.5 r - 0.5πr^3 = 26
0.5 πr^3 - 32.5 r + 26 = 0
πr^3 - 65r + 52 = 0
(*** Solve for r ***)

b. Area A = 2π r h + πr^2
dA = ∂A/∂r dr + ∂A/∂h dh
dA/dr = ∂A/∂r dr/dr+ ∂A/∂h dh/dr
dA/dr = (2π h + 2πr) + 2πr (dh/dr)
since Volume = π r^2 h = 26 <--- This is a constraint
h = 26/(π r^2)
dh/dr = (26/π) (-2r^-3)
substitute this into dA/dr
dA/dr = (2π h + 2πr) - 2πr (26/π) (2r^-3)
dA/dr = (2π h + 2πr) - 104/r^2
Set dA/dr = 0
(2π h + 2πr) - 104/r^2 = 0
r^2 (2π h + 2πr) = 104
since h = 26/(π r^2), substiture into above eqn
r^2 (2π(26/(π r^2)) + 2πr) = 104
52 + 2πr^3 = 104
2πr^3 = 104 - 52 = 52
r^3 = 52/(2π) = 8.276
r = 2.02 ft. << answer
h = 26/(π r^2) = 2.02
A = 2π r h + πr^2 = 2π(2.02)^2 + π(2.02)^2 = 38.46 << answer

Answer 2

1)pi*r^2*h=26and 2*pi*r*h+pir^2=65 so h = (65-pir^2)/2pi*r
and r*(65-pi*r^2)=52
pi*r^3-65r+52=0which gives r=4.07816
2)pi*r^2*h=26
2pirh+pir^2 =z minimum
h= 26/pir^2 so
52/r+pir^2=z
z´= -52/r^2+2pi*r=0 -26+pir^3=0 so r= (26/pi)^1/3=2.0227feet
h=2.0227 feet and
s= 38.5615 square feet
The calculations are right (I think so)

Answer 3

a