a rectangular piece of metal is 10 in. longer than it is wide. squares with sides 2 in. long are cut from the four corners and t
he flaps are folded upward to form an open box. if the volume of the box is 832 in. cubed, what were the original dimensions of the piece of metal
1 answer:
Volume of the box = length * width * height
length of original peice = x + 10 , width = x
length of box (x + 10) - 4 = x + 6
width of box = x - 4
height of box = 2
so we have
2 * (x -4) * ( x + 6) = 832
2x^2 + 4x - 48 = 832
2x^2 + 4x - 880 = 0
x^2 + 2x - 440 = 0
(x + 22)(x - 20) = 0
x = 20
so width of original piece of metal = 20 ins , length = 30 inches.
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