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20 June, 21:30

You are designing a 1000 cm^3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore beA = 8r^2 + 2pi rhWhat is the ratio now of h to r for the most economical can?

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  1. 20 June, 21:55
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    h / r = 2.55

    Step-by-step explanation:

    Area of a can:

    Total area of the can = area of (top + bottom) + lateral area

    lateral area 2πrh without waste

    area of base (considering that you use 2r square) is 4r²

    area of bottom (for same reason) 4r²

    Then Total area = 8r² + 2πrh

    Now can volume is 1000 = πr²h h = 1000/πr²

    And A (r) = 8r² + 2πr (1000) / πr²

    A (r) = 8r² + 2000/r

    Taking derivatives both sides

    A' (r) = 16 r - 2000/r²

    If A' (r) = 0 16 r - 2000/r² = 0

    (16r³ - 2000) / r² = 0 16r³ - 2000 = 0

    r³ = 125

    r = 5 cm and h = (1000) / πr² h = 1000 / 3.14 * 25

    h = 12,74 cm

    ratio h / r = 12.74/5 h / r = 2.55
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