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Math Help - Optimization Problem

  1. #1
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    Unhappy Optimization Problem

    You have been asked to design a 1-liter cylindrical can made with sheet
    aluminum. What dimensions of the can (radius and height) will use the least total
    amount of aluminum? State dimensions to the nearest hundredth of a
    centimeter.
    Hint 1: The aluminum must cover both ends of the can as
    well as the circular wall.
    Hint 2: 1 liter = 1000 cm^3

    I believe my primary equation would be: 2TTr^2 + 2TTrh

    I think my secondary equation would be: TTr^2h=1000

    Is that right?

    I came up with a radius of 12.62 and a height of 2.00

    We are supposed to have all of the following...
    • Primary equation
    • Secondary equation, if applicable
    • Statement of domain
    • Critical numbers and FDT to establish relative extrema
    • Concluding statement(s) to answer the question.
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    V = \pi r^2 h

    h = \frac{V}{\pi r^2}

    A = 2 \pi r^2 + 2 \pi r h

    A = 2 \pi r^2 + 2 \pi r \left(\frac{V}{\pi r^2}\right)

    A = 2 \pi r^2 + \frac{2V}{r}<br />

    \frac{dA}{dr} = 4\pi r - \frac{2V}{r^2}

    set \frac{dA}{dr} = 0 ...

    4\pi r = \frac{2V}{r^2}

    r^3 = \frac{V}{2\pi}

    r = \left(\frac{V}{2\pi}\right)^{\frac{1}{3}}

    calculate r ... go back and calculate h

    you should see that h = 2r
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  3. #3
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    Ok, so the minimum surface area would be when the radius is 5.42 and the height is 10.84...right?
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  4. #4
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    Lexington, MA (USA)
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    Hello, tradar!

    Your answers are correct . . .


    You are to design a 1-liter cylindrical can made with sheet aluminum.
    What dimensions of the can (radius and height) will use the least total
    amount of aluminum?
    State dimensions to the nearest hundredth of a centimeter.

    Hint 1: The aluminum must cover both ends of the can as well as the wall.
    Hint 2: 1 liter = 1000 cm³.

    I believe my primary equation would be: . A \:=\: 2\pi r^2 + 2\pi rh

    I think my secondary equation would be: . \pi r^2h\:=\:1000

    Is that right? . . . . Yes!

    I came up with a radius of 12.62 and a height of 2.00 . . . . no

    We have: . \pi r^2h \:=\:1000 \quad\Rightarrow\quad h \:=\:\frac{1000}{\pi r^2} .[1]

    Then: . A\;=\;2\pi r^2 + 2\pi r\left(\frac{1000}{\pi r^2}\right) \;=\;2\pi r^2 + 2000r^{-1}

    Differentiate: . A' \;=\;4\pi r - 2000r^{-2} \:=\:0 \quad\Rightarrow\quad 4\pi r^3 - 2000 \:=\:0

    . . 4\pi r^3 \:=\:2000 \quad\Rightarrow\quad r^3 \:=\:\frac{500}{\pi} \quad\Rightarrow\quad r \:=\:\sqrt[3]{\frac{500}{\pi}} \:=\:5\sqrt[3]{\frac{4}{\pi}}

    Substitute into [1] and we get: . h \:=\:10\sqrt[3]{\frac{4}{\pi}}


    Therefore: .  r \:\approx\:5.42\text{ cm},\;\;h \:\approx\:10.84\text{ cm}

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