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Homework Help: Calculus Optimization problems

  1. Nov 21, 2005 #1
    Hello Everyone,

    I'm working on optimization problems right now, and let me tell ya, if you thought I was horrible at everything else wait until you see me attempt one of these :grumpy: I have three questions but I'm hoping if I just get help with the one I will be able to get the processes for the rest on my own. Here it goes:

    A rectangular storage container with an open top is to have a volume of 10m^3. The length of the base is twice that of the width. Material for the base costs $10 per square meter and and $6 per square meter for the side. Find the cost of materials for the cheapest such container :yuck:

    OK So first off all I wrote down all my givens and drew myself a little picture which I had hoped would help me visualize the problem. Now I know that the volume of the container is 10 and that V=LWH, but that the length is twice the width allowing me to get rid of one of the variables. Leaving me with 10=2w^2 *h. Ic ould then get rid of the h by saying that h=(5/w^2).

    P=2l +2w +h
    P = 2(2w) +2w +5/w^2

    At this point I end up getting weird numbers, I have a feeling that part of the problem is that I never factored in that the box has an open top but I'm not sure how to go about doing that. I've tried to show what I've done and my reasons for doing it, but unfortunately I know its wrong. Any help you can give I would really appreciate, thanks a lot!
  2. jcsd
  3. Nov 21, 2005 #2


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    Your price equation is a bit odd, would you agree that it should be:

    Total price = (area of base)x$10 + 2x(area of long side)x$6 + 2x(area of short side)x$6.

    Your equation involving the relationship between the side dimensions look good.
  4. Nov 21, 2005 #3
    OK, I hadn't actually gotten to the price part yet, that was actually for perimeter but maybe that wasn't actually needed? I will keep trying to get it just using the price equation you suggested.Thanks
  5. Nov 21, 2005 #4
    OK, no I guess I'm still lost. I can't seem to fit anything in. I'm hopelessly confused.
  6. Nov 21, 2005 #5


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    In an optimization problem, you first need to write the equation for the parameter you want to optimize (obviously), then differentiate it to find its max or min. There must be other constraints involved as well. Here, the constraints are (1) volume must be 10 (2) length must be twice the width. You want to optimize price as a function of 3 basic paramters (L, W, H).

    You have a relationship between two. Try some substitution to see if you can get the price in terms of only 1 variable, then differentiate.
  7. Nov 21, 2005 #6
    OK thanks ill keep working on it!
  8. Nov 21, 2005 #7
    OK so, I went ahead with that price formula and stuck in the numbers as follows:

    (2w^2)(10)+(2)(2wh)(6)+(2)(2h)(6), i then subsituted in my h value of 5/w^2

    Then I simplifed and took the derivative so I could figure out the max and mins. My derivative was (w^-2)(40w^3-180).

    Does this sound right so far? This way I ended up with roots of 0 and cube root 4.5 for my critical values which just doesn't seem right. I'm starting to wonder if the fact that the box has an open top is displayed incorrectly in my work?
  9. Nov 21, 2005 #8


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    From what I get, the price equation simplifies to:

    [tex] P = 44W^2 + \frac{60}{W}[/tex]
  10. Nov 21, 2005 #9
    Oh ok, I guess I'll keep trying to see if I can get that to. Thanks, I'll keep at it
  11. Nov 22, 2005 #10
    Ok I never got the price equation to simplify to what you did, but I did finish the question. Maybe if i post what I did you will be able to see where I went wrong.

    Ok my equation for volume, and my way of isolating h hasn't changed from above. And according to the price equation you listed I end up getting the following for price:

    P = 20w^2+24wh+12wh
    P=20w^2 +36wh

    Then to get rid of the h I substitued in h=5/w^2 which I determined above which gets me:

    P = 20w^2 +36w(5/w^2)
    P = 20w^2 + 180W^-1

    Then I found the derivative of this so I could find the minimum cost.

    P' = 40w-180W^-2
    P' = 20w^-2(2w^3-9)

    Then I found that the critical values of the equation are w=0 and w=cube root (9/2)

    Do I have it now?
  12. Nov 22, 2005 #11


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    My apologies for my calculations earlier. Apparently, I cannot get the same result I did back then ( I kind of did it quickly in my head which failed me miserably ).

    Your answer seems very right! I think you got it ^_^
  13. Nov 22, 2005 #12
    Oh thank you so much!!!! You have no idea how much I appreciate your help!!! Now all I have to do is go back and figure out what the actual cost is going to be. Thanks a lot!!!
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