SOLUTION: A rectangular piece of matal is 10 in. longer than it is woide. Squares with sides 2 in. lon are cut from the four corners, and the flaps are folded upward to form an open box. if
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Question 343296: A rectangular piece of matal is 10 in. longer than it is woide. Squares with sides 2 in. lon are cut from the four corners, and the flaps are folded upward to form an open box. if the volume of the box is 832 in3(to the third power) whaqt 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 832 in3(to the third power) what were the original dimensions of the piece of metal?
:
We know the height of the box = 2", therefore
= 416 sq/in is the area of the base
:
Let x = the width of the metal
then
(x+10) = the length of the metal
Therefore
dimensions of the box:
(x-4) = the width
and
(x+10-4) = (x+6) the length of the box
:
Use the area of the base to find x:
(x-4)*(x+6) = 416
FOIL
x^2 + 6x - 4x - 24 = 416
:
x^2 + 2x - 24 - 416 = 0
a quadratic equation
x^2 + 2x - 440 = 0
Factor
(x+22)(x-20) = 0
:
the positive solution is what we want here
x = 20 in is the width of the metal
and
20 + 10 = 30 in is the length of the metal
:
Check solution by finding the volume using these dimensions
(20-4)*(30-4) * 2 = 832 cu/in, confirms our solution
:
20" by 30" is the dimensions of the metal
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