Expert: Paul Klarreich - 12/5/2008

Question
Paul,

I'm having a bit of trouble applying the Lagrange multiplier method to the following problem:

A barrel in the shape of a closed cylinder must have a volume of 26 ft^3. Using Lagrange multipliers find the minimum surface area.

I appreciate any help you may be able give. \
Thanks,
Scott

Answer
Questioner:   Scott
Category:  Calculus
Private:  No
 
Subject:  Lagrange multipliers - Surface area of a cylinder
Question:  Paul,

I'm having a bit of trouble applying the Lagrange multiplier method to the following problem:

A barrel in the shape of a closed cylinder must have a volume of 26 ft^3. Using Lagrange multipliers find the minimum surface area.

I appreciate any help you may be able give. \
Thanks,
Scott
........................................
Hi, Scott,

Let  h, r = dimensions.

surface area to minimize, =

S(h,r) = 2pi rh + pi r^2
....................
Constraint:  Volume is 26

V(h,r) = pi r^2 h = 26

Now write that in the form 26 - pi r^2 h, and

Set the LaGrange multiplier to L, and write the function to be minimized.

K(h,r,L) = S - L(26 - V)

K(h,r,L) =
2pi rh + pi r^2 - L(26 - pi r^2 h)

Now the gradient is obtained by three partials:

dK/dh = 2 pi r - L(- pi r^2)
dK/dr = 2 pi h + 2 pi r - L(- 2 pi rh)
dK/dL = 26 - pi r^2 h


Now you set that to the zero vector, i.e. set all three equal to zero and solve:

2 pi r - L(- pi r^2) = 0
2 pi h + 2 pi r - L(- 2 pi rh) = 0
26 - pi r^2 h = 0

do some algebra, of course:

2 pi r - L(- pi r^2) = 0  ->
2 + L(r) = 0  ->
L = -2/r

Kr = 2 pi h + 2 pi r - L(- 2 pi rh) = 0

Kr = h + r + L(rh) = 0


   -(h + r)     -2
L = --------- = ----
      rh         r

h = 26/pi r^2

You can probably finish up from here.