I have a piece of cardboard that is twice as long as it is wide .I f I cut a 1-inch by 1-inch square from each corner and fold up the resulting flaps ,I get a box with a volume of 40 cubic inches.what are the dimensions of the cardboard?
width = w
length = 2w
(w-2)(2w-2)(1) = 40
(w-1)(w-2) = 20
20 = 4x5, so w = 6
The box is 6 x 12
check: cut off the 1" corners and you get a box 4x10
width --- x
length ---- 2x
after cut-out
width = x-2
length = 2x-2
height = x
volume = x(x-2)(2x-2) = 40
2x^3 - 6x^2 + 4x = 40
x^3 - 3x^2 + 2x - 20 = 0
tried ±1, ±2, ±4, ±5 and could not find a "nice" solution, so I ran it through
Wolfram ,,,,,,
http://www.wolframalpha.com/input/?i=x%5E3+-+3x%5E2+%2B+2x+-+20+%3D+0
and got
x = appr. 3.837
so the cardboard was 3.837 by 7.674
check:
width of box = 3.837-2 = 1.837
length of box =7.674-2 = 5.674
height = 3.837
vol = 1.837x5.674x3.837 =39.99 , not bad
forget my solution,
I was thinking of an entirely different question.
thank you
oddly enough, I did exactly what you did! Then when I saw the messy answer, I reread the problem.
To find the dimensions of the cardboard, we can set up the problem step by step.
Let's call the width of the cardboard 'w'. Since the length is twice as long as the width, the length would be '2w'.
Now, we need to figure out the height of the box. Since we cut out 1-inch by 1-inch squares from each corner, the height would be equal to 1 inch.
The volume of a rectangular box is calculated by multiplying the length, width, and height. In this case, the volume is given as 40 cubic inches. So, we have the equation:
Volume = length * width * height
40 = 2w * w * 1
Simplifying the equation:
40 = 2w^2
Divide both sides by 2:
20 = w^2
Now, take the square root of both sides to solve for 'w':
√20 = w
Therefore, the width of the cardboard is √20 inches.
Since the length is twice the width, the length would be 2 times √20, which is equal to 2√20 inches.
So, the dimensions of the cardboard are approximately:
Width: √20 inches
Length: 2√20 inches