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a box with an open top is to be made from a rectangular piece of tin by cutting equal squares from the corners and turning up the sides. The piece of tin measures 1mx2m. Find the size of the squares that yields a maximum capacity for the box.
So far i have

  • Calc -

    so, figure dV/dx

    dV/dx = 2(6x^2-6x+1)
    dV/dx = 0 when x = 1/6 (3±√3) = .211 or .789

    Now .789 is impossible, since the width is only 1.

    so, the cuts are .211m

  • Calc -

    how did you go from 0= 1/6 (3+-ã3)

  • Calc -

    dV/dx = 2(6x^2-6x+1
    so, dV/dx = 0 when 6x^2-6x+1

    solve the quadratic to get X = 1/6 (3±√3)

    this is calculus; algebra I should be no problem...

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