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Math Help - Optimization Problem

  1. #1
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    Optimization Problem

    Detergent will be packaged into a square-bottomed box with volume of 450 cubic inches. What dimensions should the box have to minimize the surface area?

    I am a bit lost with this. Can someone please show me the steps on how to correctly solve this?

    Thank you for any help.
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  2. #2
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    Quote Originally Posted by ilsj6 View Post
    Detergent will be packaged into a square-bottomed box with volume of 450 cubic inches. What dimensions should the box have to minimize the surface area?

    I am a bit lost with this. Can someone please show me the steps on how to correctly solve this?

    Thank you for any help.
    let x = side length of square bottom

    h = box height

    V = (x^2)h = 450

    A = 2x^2 + 4xh

    using the volume equation, solve for h in terms of x, then substitute for h in the surface area formula to get it in terms of a single variable.

    find dA/dx and minimize
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  3. #3
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    I understand what you wrote and that is how I solved it, initially. However, my answer just didn't seem right because that is what you would get if you solved it without Calculus. My work:

    V = L^2(h)
    450 = L^2(h)
    h = 450/(L^2)

    SA = (2L^2) + (1800L/(L^2))

    I took the derivative:

    SA' = 4L - (1800/(L^2))

    Set it equal to zero and solved for "L:"

    0 = 4L - (1800/(L^2))

    -4L = - (1800/(L^2))

    -4L^3 = -1800

    L^3 = 450

    L = 7.66

    From here, I plugged it back into the volume formula to solve for "h:"

    450 = 58.6756h

    h = 7.669 = 7.67

    Basically, my height, width, and length would approximately be 7.66 inches to minimize the Surface Area.

    Is this correct?
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  4. #4
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    Quote Originally Posted by ilsj6 View Post
    I understand what you wrote and that is how I solved it, initially. However, my answer just didn't seem right because that is what you would get if you solved it without Calculus. My work:

    V = L^2(h)
    450 = L^2(h)
    h = 450/(L^2)

    SA = (2L^2) + (1800L/(L^2))

    I took the derivative:

    SA' = 4L - (1800/(L^2))

    Set it equal to zero and solved for "L:"

    0 = 4L - (1800/(L^2))

    -4L = - (1800/(L^2))

    -4L^3 = -1800

    L^3 = 450

    L = 7.66

    From here, I plugged it back into the volume formula to solve for "h:"

    450 = 58.6756h

    h = 7.669 = 7.67

    Basically, my height, width, and length would approximately be 7.66 inches to minimize the Surface Area.

    Is this correct?
    remember your work on this ... the minimum surface area for a rectangular prism of fixed volume is a cube.
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  5. #5
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    Quote Originally Posted by skeeter View Post
    remember your work on this ... the minimum surface area for a rectangular prism of fixed volume is a cube.
    Cool, so my answer is correct. I figured this would be correct...it just seemed way too easy.
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