Question 904888: (APPLIED PROBLEM) What is the equation and solution:
A rectangular cardboard is to be made into an open box by cutting 2 cm square piece from each corner and turning up the sides. If the perimeter of the cardboard is 60 cm. and the volume of the resulting box is 224 cm^3, what are the dimensions of the cardboard.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A rectangular cardboard is to be made into an open box by cutting 2 cm square piece from each corner and turning up the sides.
If the perimeter of the cardboard is 60 cm. and the volume of the resulting box is 224 cm^3, what are the dimensions of the cardboard.
:
The perimeter of the base equation
2L + 2W = 60
Simplify, divide by 2
L + W = 30
Use this form for substitution
L = (30-W)
:
We know the height of the box will be 2 inches, divide the vol by 2 and we can just solve for the area of the base which would be 112 sq/in
:
(L-4)*(W-4) = 112
replace L with (30-W)
((30-W)-4)*(W-4) = 112
(-W+26)(W-4) = 112
FOIL
-W^2 + 4W + 26W - 104 = 112
-W^2 + 30W - 104 - 112 = 0
-W^2 + 30W - 216 = 0
Multiply equation by -1, easier to factor
W^2 - 30W + 216 = 0
(W-12)(W-18) = 0
Two valid solutions
W = 12' then L
W = 18
:
The dimensions of the card board 18 by 12
:
:
Check this by finding the vol, subtract 4" from the dimensions:
14*8*2 = 224
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