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Min Cost Of A Can Using Calculus

  1. Oct 11, 2007 #1
    Hello,
    I'm in an applied calculus class and we just recieved a project that my group is having some small issues with. We have find the minimum cost of a tin can using calculus.

    From the text, " You must design a cylindrical tin can which is to hold 100 cubic inches of material. The tin for the side of the can is to cost A cents per square inch and the tin for the ens is to cost b cents per square inch. The seams for the top and bottom of the can will cost C cents while the seam up the side will cost d cents per inch. find the heigh tand radius of the tin can that will cost the least to make.

    After you have computed the dimensions for the can with your information you must make an accurate drawing of the can in proper dimensions. The drawing may be done to some scale if neccessary."

    This is what we have so far.

    Cost should equal A(LxW)+B(2xPixr^2)+C(4xPixr)+D(H)

    L = Length
    W = Width
    r = Radius
    H = Height

    We came up with this because we drew the can flat with a rectangle attached to two circles. The rectangle being the sides of the can and the circles being the top and bottom.

    Cost Evaluations (cents)
    A= .31
    B= .21
    C= .08
    D=.04

    We know we can figure this out easily without using calculus but we just dont know where to use calculus... such as where to find the derivatives ect ect.

    The professor gave us the start of an equation that we dont understand how to use which was...

    C(h,r)=A(2xpixrxH)+... each part
    C(r)
    C'(r)=0


    Any help with this would be greatly appreciated.
     
  2. jcsd
  3. Oct 11, 2007 #2
    I don't understand the geometry of your can. You drew the can rectangular or the problem drew the can rectangular? It sounds like the problem wants you to do a cylindrical can, such as a can is.

    Basically what you need to do is look at the surface area of can with the constraint that it has to hold the amount of volume indicated, and then minimize the cost. I don't know why there are added terms to the C(h,r) equation either, what is the +... each part?
     
  4. Oct 12, 2007 #3

    HallsofIvy

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    The can is a cylinder, of height h and radius r. If you "cut" off the top and bottom and cut the side part along its seam, you can open it up into a rectangle. The area of top and bottom is r so their total area is [itex]2\pi r^2[/itex]. The circumference of the top and bottom is [itex]2\pi r[/itex] so the rectangle forming the side is [itex]2\pi r[/itex] by h and so has area [itex]2\pi rh[/itex]. The seams around the top and bottom have length equal to the circumference of top and bottom, [itex]2\pi r[/itex] and so for both is [itex]2\pi r[/itex]. The seam along the side of the can has length h. The total cost of the can, then, is [itex]2\pi rha+ 2\pi r^2b+ 2\pi rc+ hd[/itex]. (That's mostly for Mindscrape's benefit. I think it is the same formula BuickBoy gave.)

    The volume of the can is, of course, [itex]\pi r^2h= 100[/itex]. You can use that equation to solve for h as a function of r and replace in the formula for area to get a function of r only. Now differentiate that with respect to r and set the derivative equal to 0.

    A minor quibble: please do not interchange small letters and capital letters (a and A, for example) in the same formula. In many formulas, those will have different meanings.)
     
  5. Oct 12, 2007 #4
    Thank you, we are going to try this. The C parts are some equation that some other kid was using somehow and asked the professor who said he was "on the right track". We are still a bit confused but we are using Derive to do this project to hopefully have less stress and potential error on our part. (Or more if we screw up the input HAHA)

    Ivy, keep me in mind because we may be asking for more help if you dont mind!!??
     
  6. Oct 12, 2007 #5
    I see what they meant now, I thought they were trying to do something other than use the surface area of a cylinder, which is of course (something I thought went without saying, i.e. didn't expect a rectangle trick) [itex]2 \pi r h[/tex]. The cost function I thought was saying that at [itex]c(y=h, r) = 2\pi rha + \textrm{terms}[/itex], which would be wrong, since it would just be [itex]C (y=h,r) = B \pi r^2[/itex], but I see now that they are using h as the parameter. I guess I was confused with what they did but you weren't, my bad.
     
  7. Oct 14, 2007 #6
    we're at a stand still... we're going in tomorrow with hardly anything... i guess another couple of guys got a diameter of 22 inches?? which is huge!!! and also a thickness of .25 inches. I guess if the cost numbers are correct it makes sense... i just would like to see the function used... (we dont have the best professor).

    thanks for the help by the way!
     
  8. Oct 15, 2007 #7

    HallsofIvy

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    As I said before, The total cost of the can is [itex]2\pi rha+ 2\pi r^2b+ 2\pi rc+ hd[/itex].
    The volume of the can is [itex]\pi r^2h= 100[/itex]. So [itex]h= 100/(\pi r^2)[/itex].

    replacing h by that in the cost formula,
    [tex]Cost= \frac{200a}{r}+ 4\pi r^2b+ 2\pi rc+ \frac{100d}{\pi r^2}[/tex]
    [tex]Cost = 200ar^{-1}+ 4\pi br^2+ 2\pi cr+ \frac{100d}{\pi}r^{-2}[/tex]
    might be easier to use. Differentiate, set equal to 0, and solve for r.
     
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