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Math Help - Optimization Problem

  1. #1
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    Optimization Problem

    I need help with two optimization problems. The first one involves the volume of a box and I have no idea what the second problem is asking.

    1) If 1200 cm^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box?

    2) (a) Show that of all the rectangles with a given area, the one with smallest perimeter is a square.

    (b) Show that of all rectangles with a given perimeter, that one with greatest area is a square.
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  2. #2
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    Quote Originally Posted by zachb View Post
    I need help with two optimization problems. The first one involves the volume of a box and I have no idea what the second problem is asking.

    1) If 1200 cm^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box?
    See diagram below:

    Let S be the surface area
    Let V be the volume of the box

    We have S = 1200
    => 1200 = x^2 + 4xy .........see diagram
    => y = 300/x - x/4

    Now V = x^2 * y
    => V = x^2 (300/x - x/4)
    => V = 300x - (x^3)/4
    => V' = 300 -(3/4)x^2

    For max V, set V' = 0
    => 300 -(3/4)x^2 = 0
    => (3/4)x^2 = 300
    => x^2 = 400
    => x = +/- 20
    => x = 20, since x cannot be negative

    but y = 300/x - x/4
    => y = 300/20 - 20/4
    => y = 10

    so for max volume, the dimensions are x = 20, y = 10

    since V = x^2 * y
    max V = (20)^2 * 10
    => maxV = 4000 cm^3
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by zachb View Post
    2) (a) Show that of all the rectangles with a given area, the one with smallest perimeter is a square.

    (b) Show that of all rectangles with a given perimeter, that one with greatest area is a square.
    (a) Let the dimensions of the rectangle be length = x and width = y.

    then the given area, A = xy
    => y = A/x

    now the perimeter P of the rectangle will be given by
    P = 2x + 2y
    => P = 2x + 2(A/x) = 2x + 2A*x^-1
    => P' = 2 - 2A*x^-2
    for smallest P, set P' = 0
    => 2 - 2A*x^-2 = 0
    => 2A*x^-2 = 2
    => x^-2 = 1/A ............divided both sides by 2A
    => x^2 = A ................took the inverse of both sides
    => x = +/- sqrt(A)

    but y = A/x
    => y = +/- A/sqrt(A)
    => y = +/- sqrt(A)

    so we see that y = x = +/- sqrt(A), so our desired rectangle is actually a square


    b) This is done similar to part (a), try it and post your solution and i'll guide you if you go wrong
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  4. #4
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    P = 2x + 2y
    => y = P - 2x/2 or P/2 - x

    A = xy
    => A = x(P - 2x/2)
    => A = (px - 2x^2)/2
    => A' = -2x (I'm not sure that A' is correct)

    I'm not sure what to do after this point. I set A' = 0, but I don't think that's what I'm supposed to do here. I worked it all the way through but it was not a square.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by zachb View Post
    P = 2x + 2y
    => y = P - 2x/2 or P/2 - x

    A = xy
    => A = x(P - 2x/2)
    => A = (px - 2x^2)/2
    => A' = -2x (I'm not sure that A' is correct)

    I'm not sure what to do after this point. I set A' = 0, but I don't think that's what I'm supposed to do here. I worked it all the way through but it was not a square.
    ok, let's see what happened.

    we are given P = 2x + 2y
    solve for y we obtain, y = (P - 2x)/2 = P/2 - x

    Now the area will be given by
    A = xy
    => A = x(P/2 - x)
    => A = (P/2)x - x^2

    => A' = P/2 - 2x .........ah, so your A' was not correct, that's the problem. anyway, let's continue

    for max A, set A' = 0
    => P/2 - 2x = 0
    => 2x = P/2
    => x = P/4

    but y = P/2 - x
    => y = P/2 - P/4
    => y = P/4

    so we see x = y = P/4, so our desired rectangle is a square
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  6. #6
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    Okay, that make sense and I know why I got A' wrong. Thanks.
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