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Optimization - Minimizing the cost of making a cyclindrical can

  • Thread starter roman15
  • Start date
  • #1
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Homework Statement


The can will hold 280 mL of juice. The metal for the side of the can costs $0.75/m^2. The metal for the top and bottom costs $1.4/m^2. The side of the can is one rectangular sheet. The top and bottom are stamped out from another rectangular sheet, the unused metal from this rectangle is donated to a charity. The charity exchanges scrap metal for money. What dimensions for the can will minimize the cost of materials?

I didnt think the part about the charity was important, but I wrote it in just incase.


Homework Equations





The Attempt at a Solution


SA=2pir^2 +2pirh
so the cost would be
C=1.4(2pir^2) + 0.75(2pirh)
=2.8pir^2 + 1.5pirh

I didnt really know where to go from here because I dont know how to get rid of h. I thought that h=2r because I figured out that to minimize the SA of a cyclinder the height:diameter ratio would be 1:1 and if SA is minimized then cost would also be minimized...but that didnt work out because if SA is minimized it doesnt mean that less of the expensive pieces of metal are being used. And Im not sure if the mL was important because I dont know if I can relate mL to m.
 

Answers and Replies

  • #2
1,033
1
See, you can't use that height to diameter minimizing, because you aren't minimizing surface area, you are minimizing the cost of the entire thing. what you are given is a volume. you know the volume of a cylinder is 2*pi*r*h. So you can set that volume = 280mL, which you will have to convert to m^3. once you do that, you can get rid of this h, and then solve.

cheers.
 

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