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anonymous

  • 5 years ago

An open box is to be constructed so that the length of the base is 4 times larger than the width of the base. If the cost to construct the base is 4 dollars per square foot and the cost to construct the four sides is 3 dollars per square foot, determine the dimensions for a box to have volume = 71 cubic feet which would minimize the cost of construction.

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  1. anonymous
    • 5 years ago
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    Which part of the problem is giving you trouble?

  2. anonymous
    • 5 years ago
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    Well, I thought that I'd figured it out.. but when I put my answer into the computer program, it says I'm wrong. lol What I did was create an equation for the surface area. I used the volume for equation to write height in terms of width, then I substituted 4w in for lenth in the s.a. equation as well as 71/(4w^2) for the height in the surface area equation. After that I found the derivative, set it to zero and solved. I think I made a dumb mistake somewhere, so I asked on here to see if someone would just resolve for me, because I'm not catching it.

  3. anonymous
    • 5 years ago
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    You have that the base length is 4 times larger than the width so you can eliminate one or the other. You have an equation for the volume, and can use that to find height in terms of the variable you kept from the first equation.. Then you can rewrite your cost function in terms of just one of the variables and take the derivative to find where it has a minimum.

  4. anonymous
    • 5 years ago
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    thats what I tried to do.. lol, I'm pretty sure my math is just wrong somewhere.

  5. anonymous
    • 5 years ago
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    \(L=4W\) \(LWH=71 \implies 4W^2H = 71\implies H = \frac{71}{4W^2}\) \(Cost = 4LW + 3(2LH) + 3(2(WH))\) \(= 16W^2 + \frac{6(4W)(71)}{4W^2} + \frac{6W(71)}{4W^2}\) etc.

  6. anonymous
    • 5 years ago
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    Right. That's what I did. Could you just continue the algebra to simplify it? Once its simplified I shoouulld be fine taking the derivative and setting it to zero and all that fun.

  7. anonymous
    • 5 years ago
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    Is that the same cost function you got?

  8. anonymous
    • 5 years ago
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    I had written substituted in the 4w for the length and simplified before I put in the substitution for height.. so I had 2wh +2h(4w) +2w(4w), and then made it 10wh +8w^2 before I put in the 71/(4w^2).

  9. anonymous
    • 5 years ago
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    I don't know if that would have made a difference though.

  10. anonymous
    • 5 years ago
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    That looks like a surface area function though, not the cost function. You forgot to factor in the costs of each of the sides.

  11. anonymous
    • 5 years ago
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    ooohhh. I didn't even realize I had left that out.

  12. anonymous
    • 5 years ago
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    Simplifying we just end up with \(Cost =16w^2 + \frac{1065}{2w}\)

  13. anonymous
    • 5 years ago
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    so our derivative would be.. 32 w- 1065/(2^w2)

  14. anonymous
    • 5 years ago
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    Okay I've got it. Thank you :)

  15. radar
    • 5 years ago
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    Just to check myself did you come out with: width=2.5 Length = 10 height =2.84

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