SOLUTION: A rectangular piece of aluminum is to be used to form a box. A 5cm square is to be used from each corner and the ends are to be folded up to form an open box whose volume will be 1

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Question 169927: A rectangular piece of aluminum is to be used to form a box. A 5cm square is to be used from each corner and the ends are to be folded up to form an open box whose volume will be 10500cm ^3. If the original piece of aluminum is twice as long as it is wide, what were the dimensions of the original rectangle?
Found 2 solutions by nerdybill, rajagopalan:
Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
.
Let w = width of aluminum
then
2w = length of aluminum
.
Box AFTER the square is cut:
width = w-5-5 = w-10
length = 2w-5-5 = 2w-10
.
Volume of box:
10500 = (w-10)(2w-10)5
2100 = (w-10)(2w-10)
2100 = (w-10)2(w-5)
1050 = (w-10)(w-5)
1050 = w^2-5w-10w+50
1050 = w^2-15w+50
0 = w^2-15w-1000
.
Factoring the right:
0 = (w+25)(w-40)
.
w = {40, -25}
.
We can toss out the negative solution leaving us with:
w = 40 cm (width of rectangle)
.
length of rectangle:
2w = 2(40) = 80 cm (length of rectangle)

Answer by rajagopalan(174) (Show Source):
You can put this solution on YOUR website!
Let width of original Al sheet =x
Length of original al sheet =2x
Now from the 4 corners cut away 5x5 square pieces
Then fold the sheet in to an open box
Length of box=2x-10 ( you have cut 2 pieces 5 cm long)
Width of Box=x-10
Height of box=5
Volume of box=LBH=(2x-10)*(x-10)*(5)
(2x^2-30x+100)*5=10500
10x^2-150x+500=10500
10*(x^2-15x+50)=10500
x^2-15x+50=1050
x^2-15x+50-1050=0
x^2-15x-1000=0
x^2-40x+25x-1000=0
x(x-40)+25(x-40)=0
(x-40)(x+25)=0
x=-25 or 40
take positive value 40 as length can not be negative.
Antwort:
Width of sheet=40
Length of sheet=80