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Math Help - optimization

  1. #1
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    optimization

    a rectangular storage container with an open top is to have a volume of 10 cubic meters. the length of its 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.

    how do you find the length, width and height for volume?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: optimization

    What is the area of the rectangular base in terms of the width?
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    Re: optimization

    the area would be 2w^2 but i need volume with height in it how do i do that?
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: optimization

    Good. Now we know the volume in cubic meters is:

    V=2w^2h=10

    So, we have:

    h=\frac{5}{w^2}

    Now, what is the combined surface area of the 4 sides?
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  5. #5
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    Re: optimization

    20/w^2? is that right? im not really understanding that part
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: optimization

    The total surface area of the sides is:

    2(2wh)+2(wh)=6wh

    Do you see how I found this?

    Now, substitute for h to get the area of the sides as a function of w.
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  7. #7
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    Re: optimization

    but the box is suppose to have an open top so should i do it with 5 sides?
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: optimization

    Yes, there will be the base, whose area you have already found is A_b=2w^2 and the 4 sides. Now, you should have the area of the sides is:

    A_s=\frac{30}{w}

    Both of these are in square meters. So, now you want to construct the cost function. Take the cost of the material per square meter times the area in square meters to get the cost, then sum these together to get the total cost in dollars:

    C(w)=10A_b+6A_s

    Now, minimize this function.
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  9. #9
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    Re: optimization

    i got 3.5569 as the minimum value but where do i plug that back into to get the cost of the materials?
    Quote Originally Posted by MarkFL2 View Post
    Yes, there will be the base, whose area you have already found is A_b=2w^2 and the 4 sides. Now, you should have the area of the sides is:

    A_s=\frac{30}{w}

    Both of these are in square meters. So, now you want to construct the cost function. Take the cost of the material per square meter times the area in square meters to get the cost, then sum these together to get the total cost in dollars:

    C(w)=10A_b+6A_s

    Now, minimize this function.
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  10. #10
    MHF Contributor MarkFL's Avatar
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    Re: optimization

    You should have:

    C(w)=20(w^2+9w^{-1}) and so:

    C'(w)=20(2w-9w^{-2})=\frac{20(2w^3-9)}{w^2}=0

    Since we want 0<w, we take the critical value from the numerator:

    w=\left(\frac{9}{2} \right)^{\frac{1}{3}

    Using the first derivative test, we find this critical number is at a minimum. Hence:

    C_{\text{min}}=C\left(\left(\frac{9}{2} \right)^{\frac{1}{3} \right)

    Now evaluate the cost function at this critical value to find the minimum cost in dollars.

    Please show how you arrived at 3.5569 as your critical number, so we can address the error in the method/computation.
    Last edited by MarkFL; November 1st 2012 at 04:41 PM.
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  11. #11
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    Re: optimization

    i thought the c(w)=10(2w^2)+6(30/w)

    how did you get a different c(w) ?
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  12. #12
    MHF Contributor MarkFL's Avatar
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    Re: optimization

    My apologies, a post made in haste. I have edited it.
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  13. #13
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    Re: optimization

    would the cheapest cost then be $163.54?
    Quote Originally Posted by MarkFL2 View Post
    My apologies, a post made in haste. I have edited it.
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  14. #14
    MHF Contributor MarkFL's Avatar
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    Re: optimization

    Yes, that's what I get too.
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  15. #15
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    Re: optimization

    Yes this is a exactly easy way for solve this problem.
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