SOLUTION: 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 the flaps are folded upward to form an open box. If t
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Question 1051101: 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 the flaps are folded upward to form an open box. If the volume of the box is 528 in3, what were the original dimensions of the piece of metal?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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 the flaps are folded upward to form an open box.
If the volume of the box is 528 in3, what were the original dimensions of the piece of metal?
:
let x = the width of the rectangular piece of metal
then
(x+10) = the length
:
Removing the 2" squares would make the dimensions of the box:
(x-4) by (x+10-4) or (x-4) by (x+6)
the height of the box = 2"
:
The volume equation L * W * h
(x+6)*(x-4)*2 = 528
divide both sides by 2
(x+6)(x-4) = 264
FOIL
x^2 - 4x + 6x - 24 = 264
x^2 + 2x - 24 - 264 = 0
a quadratic equation
x^2 + 2x - 288 = 0
You can use the quadratic formula a=1; b=2; c=-288, but this will factor to
(x+18)(x-16) = 0
the positive solution
x = 16 in is the width of the rectangular metal
then
26 in = the length
:
:
Check this, the dimensions of the box will be 4' less and the length and width
22 * 12 * 2 = 528 cu in
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