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Math Help - least amount of material

  1. #1
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    least amount of material

    what is the least amount of material that can be used to make a can with a circular base and closed top with capacity of 6 cubic inches?

    can someone help me set this problem up?
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  2. #2
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    Quote Originally Posted by dat1611 View Post
    what is the least amount of material that can be used to make a can with a circular base and closed top with capacity of 6 cubic inches?

    can someone help me set this problem up?
    you'll need to minimize the surface area of the can ...

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

    you'll also need to use the volume formula to get the surface area in terms of a single variable ...

    6 = \pi r^2 h


    let's see your attempt at setting it up.
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  3. #3
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    i have no idea
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  4. #4
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    Quote Originally Posted by dat1611 View Post
    i have no idea
    How is the maximum or minimum of a quadratic equation calculated using calculus?
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    set the derivative = to 0
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  6. #6
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    Quote Originally Posted by dat1611 View Post
    set the derivative = to 0
    Yes,
    but there is a difficulty with differentiating the surface area equation,
    because it contains "h".

    In order for the volume to remain fixed at 6 cubic units,
    as the radius changes, the height also changes.
    Hence, there are 2 variables "r" and "h".

    Hence, skeeter's clue needs to be utilised.
    This is because "h" can be written in terms of "r" using the volume formula.
    Then you can differentiate with respect to "r".
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  7. #7
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    6 = \pi r^2 h

    solving for h ...

    \displaystyle h = \frac{6}{\pi r^2}


    surface area equation ...

    A = 2\pi r^2 + 2\pi r h<br />

    substitute in the expression for h ...

    \displaystyle A = 2\pi r^2 + 2\pi r \cdot \frac{6}{\pi r^2}

    simplify ...

    \displaystyle A = 2\pi r^2 + \frac{12}{r}

    now, find \displaystyle \frac{dA}{dr} , set it equal to 0 and minimize.
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    so dA/dr is 4(pi)r-12/r^2

    4(pi)r-12/r^2 = 0

    i get .98
    .98
    is that it?
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  9. #9
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    Quote Originally Posted by dat1611 View Post
    so dA/dr is 4(pi)r-12/r^2

    4(pi)r-12/r^2 = 0

    i get .98
    .98
    is that it?
    That radius will allow you to calculate the corresponding height "h".
    Then you can use those two parameter values to calculate the minimum amount of material,
    by placing them in the surface area equation.

    The differentiation found the value of "r" for which surface area is a minimum.
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  10. #10
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    Quote Originally Posted by dat1611 View Post
    so dA/dr is 4(pi)r-12/r^2

    4(pi)r-12/r^2 = 0

    i get .98
    .98
    is that it?
    I get \displaystyle r = \sqrt[3]{\frac{3}{\pi}} as the value of the radius that will minimize the surface area.

    now ... what did the original question ask for?
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  11. #11
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    ok i plug it all into the area formula i get 18.28
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  12. #12
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    is that the right answer?
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  13. #13
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    Quote Originally Posted by dat1611 View Post
    is that the right answer?
    Yes,
    but the main thing is to understand the technique.
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