SOLUTION: A rectangular piece of metal is 15 inches longer than it is wide. Square with sides 3 inches long are cut from the four corners and the flaps are folded upward and taped to form an

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Question 946856: A rectangular piece of metal is 15 inches longer than it is wide. Square with sides 3 inches long are cut from the four corners and the flaps are folded upward and taped to form an open-topped box. If the volume of the box is 648 cubic inches, what was the original width (shorter side) of the piece of metal?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Just to begin,
w, the width
L, the length
L=w+15
h, the side length of the squares cut out
h=3
v, the volume of the box when the flaps folded
v=648
The only true unknown here is the width, w.

The bottom area will be %28w-2h%29%28L-2h%29=wL-2Lh-2wh%2B4h%5E2;
The volume will be %28wL-2Lh-2wh%2B4h%5E2%29%2Ah=v;
Substitute for L, and simplify. This should be a quadratic equation in the symbolic form with w being the only unknown variable.

You would solve the resulting quadratic equation in w, and plugin the known values any time you are ready.