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Cylinder Optimization for a Minimum?

  • Thread starter dandaman
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Question: Cost of producing cylindrical can determined by materials used for wall and end pieces. If end pieces are 3 times as expensive per cm2 as the wall, find dimensions (to nearest millimeter) to make a can at minimal cost with volume of 600 cm3.



Relevant equations: a) V=600cm3=[pi]r2h

b) surface area = 2[pi]r2 + [pi]rh




The miraculous attempt:

a) solve h in terms of [pi]r2: 600=[pi]r2h ; 600/[pi]r2=h

b) find surface area equation using h: SA=2[pi]r2 + 2[pi]r(600/[pi]r2 ; SA=2[pi]r2 + 2(600/r) ; SA=2[pi]r2 + 1200/r

c) first derivative: SA'= 4[pi]r - (600/r2) = 0


I am regrettably hitting a brick wall on what should happen next. Additional information is that I think the cost looks something like: 2[pi]r2 = (2[pi]rh)(3), with the multiplication by 3 being the fact that the round end pieces are three times as expensive.

Any and all further help would be greatly appreciated.

Dan
 
Last edited:
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The constraint equation is as you have it
V = pi*r^2*h = 600

The cost function is the surface area, weighted by the production cost factors:
C = 2*pi*r*h + 3*2*pi*r^2
where the first term is the cost of producing the side wall and the second term is the cost of producing the two ends, with the factor 3 included to reflect the higher manufacturing cost for the ends.

Your previous approach was to solve the constraint for h, so let us do this again:
h = V/(pi*r^2)
and then substitute into the cost function
C = 2*pi*r*V/(pi*r^2) + 3*2*pi*r^2
= 2*V/r + 6*pi*r^2

Now differentiate wrt r ...
dC/dr = -2*V/r^2+12*pi*r
set the derivative equal to zero...
0 = -2*V/r^2 + 12*pi*r
=V - 6*pi*r^3
r^3 = V/(6*pi)
and so forth...
(all assuming that I did not make any mistakes!)
 
Thanks a million! The math seems to check out on paper as well, with a radius of 7.3 cm and a height of 3.6 cm. (Volume would end up being 602.7 cm3.)

Dan
 

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