Problema Solution
a rectangular piece of metal is 30 in longer than it is wide. Squares with sides 6in long are cut from the four corners and the flaps are folded upward to form an open box. If the volume of the box is 3354in^3, what were the original dimensions of the piece of metal?
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let
l = the length of the rectangular piece of metal, l>0
w = the width of the rectangular piece of metal, w>0
a rectangular piece of metal is 30 in longer than it is wide
l = 30 + w
Squares with sides 6in long are cut from the four corners and the flaps are folded upward to form an open box thus the height of the open box will be h = 6 in
the volume of the box is calculated by the formula V = w*l*h and equals 3354in^3
w*l*h = 3354
if weplug l = 30 +w and h = 6 in the above equation we get
w*(30 + w)*6 = 3354
by solving the quadratic equation we find the roots and consider only the positive ones since w>0 and l>0
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w = 13 in
l = 30 + w
l = 30 + 13
l = 43 in
the original dimensions of the piece of metal were 43 in in length and 13 in in width.