Expert: Paul Klarreich - 11/5/2008

Question
Paul,

I checked the archives and found a problem like this that you worked and I feel that I followed it perfectly but I can't seem to get the correct answer. I've done about 20 of these problems over the last several days and haven't yet got one correct. I'm usually off by a couple of feet and I am not rounding anything until I input the answer to the nearest hundredth as required.

Here is the latest problem I've attempted.

"A silo is to be constructed in the form of a cylinder (only 1 of 2 bases) topped by a hemisphere.  The construction cost per unit surface area for the hemisphere is 2.2 times as much as for the cylinder and the volume must be 730,000 ft^3.  If the construction costs are to be minimized, what should the radius be?  Round to the nearest hundredth as necessary.

The answer I got when I tried to mirror your previous example was 36.36 ft.  The answer that it was looking for was 38.52 ft.  I would appreciate any help you could give me.
Thanks,
Scott

Answer
Questioner:   Scott
Category:  Calculus
Private:  No
 
Subject:  Minimizing cost of constructing a silo with a dome top
Question:  Paul,

I checked the archives and found a problem like this that you worked and I feel that I followed it perfectly but I can't seem to get the correct answer. I've done about 20 of these problems over the last several days and haven't yet got one correct.

>> 20 problems involving silos?  What are you studying -- theoretical agriculture?

I'm usually off by a couple of feet and I am not rounding anything until I input the answer to the nearest hundredth as required.

Here is the latest problem I've attempted.

"A silo is to be constructed in the form of a cylinder (only 1 of 2 bases) topped by a hemisphere.  The construction cost per unit surface area for the hemisphere is 2.2 times as much as for the cylinder and the volume must be 730,000 ft^3.  If the construction costs are to be minimized, what should the radius be?  Round to the nearest hundredth as necessary.

The answer I got when I tried to mirror your previous example was 36.36 ft.  The answer that it was looking for was 38.52 ft.  I would appreciate any help you could give me.
Thanks,
Scott
.......................
HI, Scott,

OK, ok -- I'll see what I can do.  I have not looked up the previous problem, and I will NOT assume I did it correctly. (Frequently a bad assumption.)

Let  r = radius of the cylinder (and the hemisphere)
    h = height ...............

Area of base of cylinder = pi r^2
Area of side of cylinder = circumference * height = 2 pi r h
Area of surface of top = 1/2 sphere = 2 pi r^2
Cost of surface of top = 2 pi r^2 * 2.2 = 4.4 pi r^2

C = Total cost = 5.4 pi r^2 + 2 pi r h


Constraint: V = 730K

V = V(cyl) + V(hemi)

V(cyl) = pi r^2 h
V(hemi) =(1/2)* 4/3 pi r^3 = 2/3 pi r^3

V = pi r^2 h + 2/3 pi r^3 = 730000

pi r^2 h  = 730000 - 2/3 pi r^3
    730000 - 2/3 pi r^3
h = --------------------
        pi r^2


Back to cost:
                         730000 - 2/3 pi r^3
C = 5.4 pi r^2 + 2 pi r (--------------------)
                              pi r^2

                   730000 - 2/3 pi r^3
C = 5.4 pi r^2 + 2 (--------------------)
                             r
               
C = 5.4 pi r^2 +  1460000r^-1 - 4/3 pi r^2

OK, differentiate:

dC/dr = 10.8 pi r -  1460000r^-2 - 8/3 pi r

dC/dr = 10.8 pi r -  1460000r^-2 - 2.67 pi r

dC/dr = 8.13 pi r -  1460000r^-2

Set that = 0 and solve:

8.13 pi r -  1460000r^-2 = 0

8.13 pi r^3 -  1460000 = 0

r = cubrt(1460000/8.13 pi)

AND THE CALCULATOR GIVES: (big drum roll here....)

r = 38.521584790155260198722680867687 or so.

Best I can do.