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Optimizing Dimensions to minimize Cost

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data

    A metal can which holds 450 cm^3 is to be manufactured in cylindrical shape. The top and bottom will be cut from two squares and the corner scrap discarded (but paid for). The metal for the sides costs 0.1 cents/cm^2, while the cost for the top and bottom is 0.15cents/cm^2. Allow 0.5cents/cm for the vertical or side seam and 0.6 cents/cm for the seams joining top and bottom to the sides.
    a) Determine the dimensions that would yield a minimum cost. Give dimensions to the nearest thousandth of a centimeter.
    b) Determine the corresponding cost to the nearest hundreth of a cost.

    2. Relevant equations

    Minimizing is to find the values of a unit of function, when the tangent slope is zero.

    therefore,

    If C'(A)=0, then we have to find the dimensions when this happens since this is when cost is minimized.

    V(cylinder)=∏r^2h
    A(cylinder)=2∏r(h) + 2∏r^2

    3. The attempt at a solution

    I notice that part of the problem states that no matter what dimension, V=450cm^3.

    Therefore,

    since this is a can cylinder,

    V=∏r^2h=450cm^3

    and so it is useful to note that:

    (450cm^3)/(∏r^2)=h for all h>0 and r>0
    and
    (√((450cm^3)/(h∏)))=r for all h>0 and r>0

    So, here's how I went along trying to put certain pieces together:

    According to my interpretation of the problem,

    Area(top + bottom)=2(∏r^2)
    So,
    Cost(Area(top + bottom))=[.15cents/cm^2][2∏r^2]

    Area(side)=2∏r(h) since my intuition tells me this represents the area of a cross section of the side of a cylinder where 2∏r is the circumference of the number of circles you multiply by to get the area

    Cost(Area(side))= .15cents/cm^2

    So also since for every radius there is a cost and every height there is a cost (I am very currently skeptical about this intuition!)

    Cost(Area(side + top + bottom))= .3∏r^2[Cost(r)]^2 + .2∏r(h)[Cost(r)][Cost(h)]

    Therefore, by substituting h=(450cm^3)/(∏r^2)

    Cost(Area(side + top + bottom))= .3∏(r)^2[cost(r)] + .2∏r(450cm^3/∏r^2)[cost(r)][cost(h)]

    Cost(h)=(.6cent/cm)h=(.6cent/cm)[(450cm^3)/(∏r^2)]
    cost(r)=(.5cent/cm)r

    Cost(Area(side + top + bottom))=.3∏r^2[(.5cent/cm)r] + .2∏r[450cm^3/∏r^2][(.5cent/cm)r]

    Cost'(Area(side + top + bottom))=.9∏[.5cent/cm]r^3 + 45=0 to minimize cost, but r>0 and h>0 what am I doing wrong?
     
  2. jcsd
  3. Jan 30, 2012 #2

    HallsofIvy

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    Science Advisor

    You said above, "The metal for the sides costs 0.1 cents/cm^2"

    You also need to include the cost of the seams- "Allow 0.5cents/cm for the vertical or side seam and 0.6 cents/cm for the seams joining top and bottom to the sides." The vertical seam has length h so you need to add 0.5h cents and the top and bottom seams have length [itex]2\pi r[/itex], and there are two of them, you need to add [itex](0.5)(4\pi r)= 2\pi r[/itex].

     
  4. Jan 30, 2012 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    The cost of the top + bottom is (0.15)*2*(2r)^2, because you pay 0.15 per cm^2 for the whole 2r by 2r square (you pay for the discarded corners, you said).

    RGV
     
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