h + 2 pi r = 30
so
h = 30 - 2 pi r
dh/dr = -2 pi
pi r^2 h = v
so yes
dv/dr = pi r^2 dh/dr + h (2 pi r) dr/dr
so
dv/dr = pi r^2 (- 2 pi) + 2 pi r h
that is 0 for maximum of minimum
2 pi^2 r^2 =2 pi r h
pi r =h
r = h/pi
then
h = 30 - 2 pi r = 30 - 2 pi (h/pi)
h = 30 - 2 h
3 h = 30
h = 10
r = 10/pi
Consider all right circular cylinders for which the sum of the height and circumference is 30 centimeters. What is the radius of one with maximum volume.
Work:
V=(pi)r^2(h)
h= V/(pi)(r^2)
2(pi)r + h=30
V=(pi)r^2(h)
V'=2(pi)(r)h+2pi(r^2)=0
3 answers
You have to make V a function of r only abd then set the derivative equal to zero.
V = pi r^2 h
h = 30 - 2 pi r
V = pi r^2 * (30 - 2 pi r)
= 30 pi r^2 - 2 pi^2 r^3
dV/dr = 60 pi r - 6 pi^2 r^2 = 0
divide both sides by 6 pi r
10 - pi r = 0
r = 10/pi
Check my math
V = pi r^2 h
h = 30 - 2 pi r
V = pi r^2 * (30 - 2 pi r)
= 30 pi r^2 - 2 pi^2 r^3
dV/dr = 60 pi r - 6 pi^2 r^2 = 0
divide both sides by 6 pi r
10 - pi r = 0
r = 10/pi
Check my math
Thanks Damon and drwls!