Question 1200523
let L = the length and W equal the width of the rectangular piece of cardboard.
when you cut out 3 inch square from each corner of the cardboard, and fold up 3 inches from the length and the width, you are left with an open box that has the following dimensions.
length = L - 6
width = W - 6
height = 3
the volume of the open top box is equal to length * width * height which is equal to (L - 6) * (W - 6) * 3
since the length of the box is twice as long as it is wide, then you get L = 2W and the dimensions of the open top box are (2W - 6) * (W - 6) * 3
simplify this to get 2W * (W - 6) - 6 * (W - 6) * 3
simplify further to get (2W^2 - 12W - 6W + 36) * 3
simplify further to get 6W^2 - 36W -  18W + 108
combine like terms to get 6W^2 - 54W + 108
that's the volume of the open top box.
since the volume is 5940 cubic inches, you get:
6W^2 - 54W + 108 = 5940
subtract 5940 from both sides of the equation to get:
6W^2 - 54W - 5832 = 0
divide both sides of this equation by 6 to get:
W^2 - 54W - 972 = 0
factor this quadratic eqution to get:
W = -27 or plus 36.
W can't be negative, you are left with W = 36
since length is twice the width, you have:
W = 36
L = 72
the dimensions of the open top box are:
length = L - 6 = 66
width = W - 6 = 30
height = 3
the volume of the open top box is therefore 66 * 30 * 3 = 5940 cubic inches.
the original piece of cardboard should have a length of 36 and a width of 72.