A sheet of copper is twice as long as it is wide. From each corner a 3 inch square is cut out, and the end ends are then turned up to form a tray. If the volume of the tray is 324 cubic inches, what were the original dimensions of the sheet if copper.

The original length of the sheet = 24 inches

The original width of the sheet = 12 inches

Step-by-step explanation:

 Let us denote the original width of the sheet of the copper by "x". Since the length of that sheet of copper is twice as long as it is wide (width), then the length of this sheet of copper (rectangle) should be "2x".

  Now, since 3 inch squares were cut out at the four ends of the rectangular sheet of copper, then the new length of the copper should be 2x - 6 inches.

 Reason:- A rectangle has 4 sides. Two of the sides are the length and the remaining two sides are the width or breadth. Now, for one of the sides (length), two 3 - inch squares from the two ends are cut out. This means that each of the length sides will lose (3 inches + 3 inches = 6 inches).

    The new width of the rectangular sheet of copper will then be x - 6 inches (the same reason as the length).

  Later the ends are then turned up to form a tray. This means that the height of the tray will now become 3 inches.

 Since the volume of this tray (cuboid shape) is 324 cubic inches, we can calculate the original dimensions of the sheet:-

  Volume of a cuboid = L*w*h

L = 2x - 6

w = x - 6

h = 3

Then:

   (2x - 6)(x - 6) × 3 = 324

   (2x^2 - 12x - 6x + 36) × 3 = 324

   (2x^2 - 18x + 36) × 3 = 324

   6x^2 - 54x +108 = 324

   6x^2 - 54x + 108 - 324 = 0

   6x^2 - 54x - 216

  Divide through by 6

= x^2 - 9x - 36

 We will now solve for x using any formula or method of solving quadratic equations:

           x = - 3 or 12

 Since x = a positive integer then the width of the rectangular sheet of copper was 12 inches originally.

 The length = 2x

 Therefore the length = 2 × 12 = 24 inches.

L = 24 inches

w = 12 inches.