SOLUTION: The area of the rectangular piece of cardboard shown on the left is 198 square inches. The cardboard is used to make an open box by cutting a 2 inch square from each corner and tur

Algebra ->  Rectangles -> SOLUTION: The area of the rectangular piece of cardboard shown on the left is 198 square inches. The cardboard is used to make an open box by cutting a 2 inch square from each corner and tur      Log On


   

Question 242218: The area of the rectangular piece of cardboard shown on the left is 198 square inches. The cardboard is used to make an open box by cutting a 2 inch square from each corner and turning up the sides. If the box is to have a volume of 196 cubic inches, find the length and width of the cardboard that must be used.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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The area of the rectangular piece of cardboard shown on the left is 198 square inches.
The cardboard is used to make an open box by cutting a 2 inch square from each corner and turning up the sides.
If the box is to have a volume of 196 cubic inches, find the length and width of the cardboard that must be used.
:
Area of the original piece of cardboard
L * W = 198 sq/in
L = 198%2FW
:
Volume L*W*H = 196 cu/in
(removing 2" squares reduces the length and width by 4", height = 2"
(L-4)*(W-4)*2 = 196
Simplify, divide both sides by 2
(L-4)*(W-4) = 98
FOIL
LW - 4L - 4W + 16 = 98
:
LW - 4L - 4W + 16 - 98 = 0
:
LW - 4L - 4W - 82 = 0
:
From the area equation, replace L with 198/W
198%2FW*W - 4*198%2FW - 4W - 82 = 0
Cancel W, mult by 4
198 - 792%2FW - 4W - 82 = 0
:
Mult by W to get rid of the denominator
198W - 792 - 4W^2 - 82W
:
Arrange as a quadratic equation
-4W^2 + 198W - 82W - 792 = 0
:
-4W^2 + 116W - 792 = 0
Divide by -2, simplify and change the signs
2W^2 - 58W + 396 = 0
This will factor
(2W - 22)(W - 18}) = 0
Two solutions
2W = 22
W = 11
and
W = 18
;
18" by 11" is the cardboard used
:
Check the area: 18 * 11 = 198
:
Check the vol of the box
(18-4)*(11-4)* 2 = 196