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Math Help - optimization problem

  1. #1
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    optimization problem

    I am totally lost with this problem.

    The length of a cedar chest is twice its width. The cost/dcm^2 of the lid is four times the cost/dcm^2 of the rest of the cedar chest. If the volume of the cedar chest is 1440 dcm^3, find the dimensions so that the cost is a minimum.
    Attached Thumbnails Attached Thumbnails optimization problem-cedar.png  
    Last edited by raymac62; October 3rd 2011 at 08:16 AM.
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  2. #2
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    Re: optimization problem

    Quote Originally Posted by raymac62 View Post
    I am totally lost with this problem.

    The length of a cedar chest is twice its width. The cost/dcm^2 of the lid is four times the cost/dcm^2 of the rest of the cedar chest. If the volume of the cedar chest is 1440 dm^3, find the dimensions so that the cost is a minimum.
    C = 4(2x^2) + 2x^2 + 2(2xh) + 2(xh) = .....

    where 1440 = 2x^2h.

    Therefore C(x) = .......

    and proceed in the usual way.

    If you need more help, please show all your work and say where you get stuck.
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  3. #3
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    Re: optimization problem

    Quote Originally Posted by raymac62 View Post
    I am totally lost with this problem.

    The length of a cedar chest is twice its width. The cost/dcm^2 of the lid is four times the cost/dcm^2 of the rest of the cedar chest. If the volume of the cedar chest is 1440 dm^3, find the dimensions so that the cost is a minimum.
    What are dcm^2 and dm^3?

    CB
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  4. #4
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    Re: optimization problem

    They are typos. I was very tired when I made the original post. I should have wrote "dcm" in both cases. Even on the diagram I have labelled the cedar chest to have a a volume in cubic centimeters where it should be decimeters. 1440dcm cubed
    One cubic decimeter is: 10cm * 10cm * 10cm

    ** Of course I have edited it now.
    Last edited by raymac62; October 3rd 2011 at 08:40 AM.
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  5. #5
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    Re: optimization problem

    Quote Originally Posted by mr fantastic View Post
    C = 4(2x^2) + 2x^2 + 2(2xh) + 2(xh) = .....

    where 1440 = 2x^2h.

    Therefore C(x) = .......

    and proceed in the usual way.

    If you need more help, please show all your work and say where you get stuck.
    C = 11x^2 + 6hx + h^2

    Why is the volume expressed: 2x^2h? Should it not be: 2x^2 * h ?
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  6. #6
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    Re: optimization problem

    Quote Originally Posted by raymac62 View Post
    C = 11x^2 + 6hx + h^2

    Why is the volume expressed: 2x^2h? Should it not be: 2x^2 * h ?
    The implied multiplication ought to be crystal clear to you from the context. What I posted was help, not a pristine solution.
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  7. #7
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    Re: optimization problem

    Quote Originally Posted by mr fantastic View Post
    The implied multiplication ought to be crystal clear to you from the context. What I posted was help, not a pristine solution.
    Let me explain what I know (or at least think I know) so that there is a better understanding of context:

    C = the total cost for producing the cedar chest.

    V = 1440 - In someway I have to take this value and substitute it into the equation so I can determine the value of at least 1 variable.

    Then, I have to find the derivative, and solve for x when it is equal to zero.

    I am not sure how I can go about expressing the total cost with only a single variable.

    ---
    Obviously, what you posted was not a pristine solution. Nonetheless, why have you expressed the volume as 2x^2h? That makes absolutely no sense to me. I do not see how that is algebraically equivalent to expressing V as 2x^2 * h. However, it must mean something to you. If you would care to explain...

    Thanks in advance.

    Sincerely,

    Raymond
    Last edited by raymac62; October 3rd 2011 at 04:14 PM.
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  8. #8
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    Re: optimization problem

    Quote Originally Posted by raymac62 View Post
    Let me explain what I know (or at least think I know) so that there is a better understanding of context:

    C = the total cost for producing the cedar chest.

    V = 1440 - In someway I have to take this value and substitute it into the equation so I can determine the value of at least 1 variable.

    Then, I have to find the derivative, and solve for x when it is equal to zero.

    I am not sure how I can go about expressing the total cost with only a single variable.

    ---
    Obviously, what you posted was not a pristine solution. Nonetheless, why have you expressed the volume as 2x^2h? That makes absolutely no sense to me. I do not see how that is algebraically equivalent to expressing V as 2x^2 * h. However, it must mean something to you. If you would care to explain...

    Thanks in advance.

    Sincerely,

    Raymond
    V = (2)*(x^2)*(h).

    Better now? (I guess in the age of the CAS calculator, implied multiplication is not as clearly understood is it once was).

    1440 = 2x^2h => h = 720/x^2. Substitute into C etc. etc.
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  9. #9
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    Re: optimization problem

    Quote Originally Posted by mr fantastic View Post
    V = (2)*(x^2)*(h).

    Better now? (I guess in the age of the CAS calculator, implied multiplication is not as clearly understood is it once was).

    1440 = 2x^2h => h = 720/x^2. Substitute into C etc. etc.
    This is what I have and somethings not working out.
    Could you please show me where I went wrong.
    Thanks in advance!

    C=4(2x^2) + 2x^2 + 2(2hx) + 2(hx)=11x^2+6hx+h^2

    h= \frac {1440}{2x^2}

     C(x)=11x^2 + \frac {6(1440x)}{x^2} + [\frac {1440}{x^2}]^2 ...sorry don't know large bracket command

    = 11x^2 + \frac {8640}{x} + \frac {2073600}{x^4}  = 11x^2 + 8640(x)^{-1} + 2073600(x)^{-4}

    C'(x)=22x - \frac {8640}{x^2} - \frac {8294400}{x^5}

    Solve for x when the derivative is equal to zero.

    22x^6 = \frac {8640x^5}{x^2} + 8,294,400

    22x^6 -8640x^3 - 8294400 = 0

    x \approx 9.44

    Dimensions obtained: W= 9.44 L=18.88 h=16.16

    These don't work out to equal 1440, So I know I am going wrong somewhere. Can't seem to pinpoint exactly where.

    Once again, thanks a lot for your help.

    Sincerely,

    Raymond
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  10. #10
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    Re: optimization problem

    Think i have spotted where I went wrong. 2(hx) should be (h^2)(x^2) where I went h^2 +2hx +x^2 in the original Cost equation.
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  11. #11
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    Re: optimization problem

    Quote Originally Posted by raymac62 View Post
    Think i have spotted where I went wrong. 2(hx) should be (h^2)(x^2) where I went h^2 +2hx +x^2 in the original Cost equation.
    The above makes no sense to me. Neither does your expansion and simplification in your previous post.

    I would have thought that after expanding, C = 8x^2 + 2x^2 + 4hx + 2hx = ....

    so I'm not sure how you got your result ....
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  12. #12
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    Re: optimization problem

    Looks like I need to review my algebra. :P I'm over complicating things. I got it now.

    2(xy) is not equal to (x^2)(y^2).... I was drained.
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  13. #13
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    Re: optimization problem

    Quote Originally Posted by mr fantastic View Post
    The above makes no sense to me. Neither does your expansion and simplification in your previous post.

    I would have thought that after expanding, C = 8x^2 + 2x^2 + 4hx + 2hx = ....

    so I'm not sure how you got your result ....
    C=4(2x^2) + 2x^2 + 2(2hx) + 2(hx) = 10x^2 + 6hx

    h= \frac {1440}{2x^2} = \frac {720}{x}

     C(x)=10x^2 + \frac {6(720)x}{x^2} .

    = 10x^2 + \frac {4320}{x}  = 10x^2 + 4320(x)^{-1}

    C'(x)=20x - \frac {4320}{x^2}

    Solve for x when the derivative is equal to zero.

    20x^3 = 4320

    x^3 = 216

    x = 216^{1/3}

    x = 6

    Thus,  W=6, L=12, H=20

    6*12*20= 1440 ...yay.
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