To solve this problem, we need to start by visualizing the scenario. We have a rectangular sheet of tin, and we need to cut squares of side length "a" from each corner. The remaining flaps of the sheet will then be folded up to form an open-top box.
Let's assign variables to the different quantities mentioned in the problem:
- Length of the rectangular sheet of tin: L (in inches)
- Width of the rectangular sheet of tin: W (in inches)
- Side length of the squares cut from the corners: a (in inches)
- Height of the box: H (in inches)
The length of the box will be equal to the original width of the rectangular sheet (W) minus two times the side length of the square (2a). So, the length of the box can be written as:
L = W - 2a
The width of the box will be equal to the original length of the rectangular sheet (L) minus two times the side length of the square (2a). So, the width of the box can be written as:
W = L - 2a
The height of the box (H) can be obtained by folding the flaps of the tin, and it will be equal to the side length of the square (a). So, the height of the box is:
H = a
The volume of a box is given by the formula: Volume = Length * Width * Height.
Substituting the expressions for length, width, and height into the volume formula, we have:
420 = (W - 2a) * (L - 2a) * a
Now, we need to solve this equation for the dimensions of the rectangular sheet, given that a = 3.
Substituting a = 3 into the equation, we have:
420 = (W - 2*3) * (L - 2*3) * 3
Simplifying further:
420 = (W - 6) * (L - 6) * 3
Dividing both sides of the equation by 3:
140 = (W - 6) * (L - 6)
Now, we need to find values for W and L that satisfy this equation. We can do this by observing that 140 is the product of two numbers, and one of the factors must be W - 6, while the other factor must be L - 6.
Factoring 140, we find that the possible pairs of factors are:
1 * 140
2 * 70
4 * 35
5 * 28
7 * 20
10 * 14
Since W and L represent lengths, they should be positive. So, we can eliminate pairs like 1 * 140 and the negative factored pairs.
Let's plug in each of the remaining pairs of factors into the equation:
(W - 6) * (L - 6) = 2 * 70
(W - 6) * (L - 6) = 4 * 35
(W - 6) * (L - 6) = 5 * 28
(W - 6) * (L - 6) = 7 * 20
(W - 6) * (L - 6) = 10 * 14
For each equation, we solve for W and L. Take note that when solving, we add 6 to both sides of the equation to isolate (W - 6) and (L - 6):
W - 6 = 2, L - 6 = 70
W - 6 = 4, L - 6 = 35
W - 6 = 5, L - 6 = 28
W - 6 = 7, L - 6 = 20
W - 6 = 10, L - 6 = 14
Solving for W and L, we have:
W = 8, L = 76
W = 10, L = 41
W = 11, L = 34
W = 13, L = 26
W = 16, L = 20
Out of these solutions, we need to choose the dimensions that make sense in our scenario. Since the length of the rectangular sheet is twice its width (mentioned in the problem statement), we can eliminate the solutions where L is not approximately twice the value of W.
From the remaining solutions:
W = 8, L = 76 (Not valid, L is not approximately twice W)
W = 13, L = 26 (Not valid, L is not approximately twice W)
W = 16, L = 20 (Valid, L is approximately twice W)
Therefore, the size of the rectangular sheet that will produce a box with a volume of 420 in^3 when a = 3 is a sheet with dimensions 16 inches by 20 inches.