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Math Help - optimatization problem

  1. #1
    Junior Member
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    Exclamation word problem

    Hi! can someome help me with this exercise

    An open top box with a rectagular base is to be constructed. The box is to be at least 2 inches wide and twice as long as it is wide and is to have a volume of 150 cubic inches. What should the dimensions of the box be if the surface area is to be
    a. 90 square inches ? b. as small as possible?

    I have already done some of these problems , but I don't know how to set up this one.
    Last edited by skorpiox; February 9th 2009 at 11:21 PM.
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by skorpiox View Post
    Hi! can someome help me with this exercise

    An open top box with a rectagular base is to be constructed. The box is to be at least 2 inches wide and twice as long as it is wide and is to have a volume of 150 cubic inches. What should the dimensions of the box be if the surface are is to be
    a. 90 square inches ? b. as small as possible?

    I have already done some optimatization problems , but I don't know how to set up this one.
    I think that you have your wording wrong in your problem. Are you sure it doesn't say something like, "twice as long as it is tall"?
    Check that, but here is how you need to set this up:

    Volume = length x width x height

    Surface area (box with no top) = 2(length x height) + 2(width x height) + (length x width)

    V = 8h
    A = 2(4h) + 2(2h) + 8
    = 8h + 4h +8
    = 12h + 8

    Double check the question and then I can help you through the rest. IT really isn't an optimization problem the way it's worded.

    Since you are given your volume, you should have two variables in there.
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  3. #3
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    Quote Originally Posted by mollymcf2009 View Post
    I think that you have your wording wrong in your problem. Are you sure it doesn't say something like, "twice as long as it is tall"?
    Check that, but here is how you need to set this up:

    Volume = length x width x height

    Surface area (box with no top) = 2(length x height) + 2(width x height) + (length x width)

    V = 8h
    A = 2(4h) + 2(2h) + 8
    = 8h + 4h +8
    = 12h + 8

    Double check the question and then I can help you through the rest. IT really isn't an optimization problem the way it's worded.

    Since you are given your volume, you should have two variables in there.
    I double checked the problem , and it says
    an open-top box with a rectangular base is to be constructed. the box is to be at least 2 inches wide and twice as long as it is wide and is to have a volume of 150 cubic inches. what should the dimensions of the box be if the surface area is to be
    a. 90 square inches? b.as small as possible?
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  4. #4
    Senior Member mollymcf2009's Avatar
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    Ok, look at it this way:

    Let's call the width (x), so the length is (2x), height we will call (h)

    Volume of the box is:  x \cdot 2x \cdot h = 2x^2h

    Now, your surface area is: 2xh + 4xh + 2x^2
    **That is: 2(width\cdot height) + 2(length \cdot height) + (length \cdot width) 2 short sides + 2 long sides + bottom of box

    So,

    V = 2x^2h
    SA = 6xh + 2x^2

    Using these two equations and the data given to you to figure out the dimensions of your box.

    To find the smallest area, take the area equation we came up with (after you plug in your height you solved for), take the derivative and then set it equal to zero. Solve for x. This is your critical value. Plug it into the original area function and solve for y. This will give you your minimum value (you can do the same thing to find a maximum value too!. Hope that helps you!
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