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Math Help - Max/Min Problems-$ and Diagonals

  1. #1
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    Max/Min Problems-$ and Diagonals

    These two max min problems involve shapes with surface area, volume, etc.
    The first one is asking for a ratio, not just a simple value. The second one says quadrilateral but does that mean it is a cube or something?

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  2. #2
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    These two max min problems involve shapes with surface area, volume, etc.
    The first one is asking for a ratio, not just a simple value. The second one says quadrilateral but does that mean it is a cube or something?

    Hello,

    to #13:

    let x and y be the lengthes of the diagonals. Then the area of the quadrilateral is calculated by:

    a = * x * y

    From the given conditions you know: x + y = 6 ==> y = 6 - x

    Plug in the term for y into the first equation:

    a(x) = *x*(6 - x) = -x + 3x

    The graph of this function is a parabola opening downwards. The maximum value is at it's vertex: V(3, 4.5)

    You get: x = 3, y = 3 the maximum area is 4.5 cm

    The quadrilateral is a square with diagonals which are 3 cm long.
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  3. #3
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    These two max min problems involve shapes with surface area, volume, etc.
    The first one is asking for a ratio, not just a simple value. The second one says quadrilateral but does that mean it is a cube or something?

    Hello,

    to #12:

    volume of the can: V = k = π*r*h ==> h = k/(π*r) [1]

    complete surface area: a = 2*πr + 2*π*r*h

    costs: c = 2*πr*0.03 + 2*π*r*h*0.06 [2] Plug in the term for h from [1] into [2]:

    c(r) = 2*πr*0.03 + 2*π*r*k/(π*r)*0.06 = 0.06*πr + 0.12k/r

    You'll get an extremum if the first derivation equals zero:

    c'(r) = 0.12πr - 0.12k/r With c'(r) = 0 you get r = √(k)

    (It is necessary to prove if you've got a minimum. I leave this to you )

    Now plug in this value into [1] and calculate h/r = 1/π
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