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Math Help - Optimization Problem

  1. #1
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    Optimization Problem

    A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container.

    I understand the calculus for the most part, but creating the initial formula for this is confusing me to no end. So far I've got: Cost = 10(b^2) + 4(6b*h) where b=l*w where l=2w so b = 2w*w = 2w^2.

    Now I assume I mess up somewhere when I go with 10m cubed = (l*w)*h. So 10 = b*h. b = 2w^2, So h = 10/2w^2. I proceed to throw that into the Cost equation so that C = 10(2w^2)^2 + 24(2w^2)(10/2w^2). Which simplifies to 10(2w^2)^2 + 240.

    C' = 20(2w^2) * 4w = 10w^3

    Not quite sure where I messed up, as the problem is kinda long. I would appreciate it if someone could point out my error so I could work on it. I have trouble with these optimization problems as they are mostly word problems that tend to confuse me.
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  2. #2
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    The first term in your cost formula is wrong: if b = lw, then b has units m^2, so b^2 has units m^4, so 10 dollars/m^2 * b^2 has units dollars * m^2...

    If l and w are the length and width of the base, and h is the height of the sides, you must have lwh = 10, l=2w, and your cost function is C := 10 lw + 6(2l+2w)h (if I have apprehended you correctly). Can you go from there?
    Last edited by maddas; April 9th 2010 at 04:14 PM.
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  3. #3
    Junior Member eddie2042's Avatar
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    Quote Originally Posted by abel2 View Post
    A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of materials for the cheapest such container.

    I understand the calculus for the most part, but creating the initial formula for this is confusing me to no end. So far I've got: Cost = 10(b^2) + 4(6b*h) where b=l*w where l=2w so b = 2w*w = 2w^2.

    Now I assume I mess up somewhere when I go with 10m cubed = (l*w)*h. So 10 = b*h. b = 2w^2, So h = 10/2w^2. I proceed to throw that into the Cost equation so that C = 10(2w^2)^2 + 24(2w^2)(10/2w^2). Which simplifies to 10(2w^2)^2 + 240.

    C' = 20(2w^2) * 4w = 10w^3

    Not quite sure where I messed up, as the problem is kinda long. I would appreciate it if someone could point out my error so I could work on it. I have trouble with these optimization problems as they are mostly word problems that tend to confuse me.

    See the attached picture for what I did.

    Optimization Problem-applied-max-min-2.jpg
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