turned up to make a box with a volume 273 in^3. find the length and width of the original piece of sheet metal.
The hardest part of this problem is taking the information given to you in english and turning it into a math equation. So let's start with naming some variables. We have a width, which we can call W. There is also a length that we can call L. We are also given a volume, which we can call V. The other dimension of the box is the height, which we will figure out later, but we will name that variable H. We are told that a rectangular piece of sheet metal is to be used to make the box, so we will assign the variables L and W as the length and width of the sheel metal, respectively.
We are told that the sheet metal is 6 inches longer than it is wide, or that the length (L) is 6 inches more than the width (W). In algebriac terms, we can write W + 6 = L. This gives us the value of L in terms of W, so we can write the length and width in terms of only one varable, the width. So we have a piece of sheet metal with a width of W and a length of W + 6.
Next, a square section that measures 3 inches on each side is cut from each corner, and the sides are turned up to make the box. So each side of the sheet metal will have 3 + 3 = 6 inches removed from it. We will have to then subract 6 inches from the length and width of the sheet metal to give us the length and width of the box. Thus, our width goes from W to W - 6 and our length goes from W + 6 to W + 6 - 6, or just W. Now the tricky part, since we know that each square removed from the corners of the sheet metal has a side that is 3 inches, and that the cutting of the corners allows the sides of the sheet metal to be turned up to form the box, the height of the box is just the length of one of the sides of the squares we cut out. So the height (H) of the box is 3 inches.
Now we have our 3 dimensions for the box, the height is 3, the length is W and the width is W - 6. We also know the total volume of the box is 273 cubic inches, and this is also equal to height times length times width. Remember that the length and width of the box are different from the length and width of the sheet metal, and we have redefined length and width in terms of only the width. So we can write:
V = H * L * W
V = 3 * (W) * (W - 6)
273 = 3(W) * (W - 6)
273 = 3W^2 -18W
We can subract 273 from both sides to get:
0 = 3W^2 -18W - 273
Next, divide all terms by 3 to get:
0 = W^2 -6W -91
This is a quadratic equation that can be solved by factoring or by using the quadratic formula. If we factor it, we are looking for 2 numbers that add up to -6 and multiply to -91. After trial and error, we can see that -13 and 7 fit, so our factored form of the equation becomes:
0 = (W - 13)(W + 7)
The solutions of this equation are W = 13 and W = -7. Since our width must be a positive number, since a negative width makes no sense, we can ignore the -7 and say that our width is 13 inches.
Remember, that we defined L and W to mean the length and width of the original piece of sheet metal, and then modified them to become the dimensions of the box. So 13 inches is the width of the sheet metal, and since L = W + 6, the length of the sheet metal is then 19 inches.
The dimensions of the sheet metal is 19 x 13 inches.