Question

A rectangular piece of sheet metal has a length that is 6 in. less than twice the width. a square piece 3 in. on a side is cut from each corner. the sides are then turned up to form an uncovered box of volume 1014 in. cubed find the length and width of the original piece of metal.

Answer 1

The original width would be 19 and the original length would be 32.

Let w be the width.  Then 2w-6 would be the length.  However, after cutting a 3-inch square from each corner, both the width and length left over to fold into a box would be 6 inches smaller; thus the dimensions would be w-6 and 2w-6-6 or 2w-12.  

Since the section cut out is 3 inches long, 3 will be the height of the box.

Volume is found by multiplying the length, width and height of the box; thus we have:

1014=(w-6)(2w-12)(3)

We multiply the binomials and have:
1014 = [w*2w-12*w-6*2w-6(-12)](3)
1014 = (2w²-12w-12w+72)(3)
1014 = (2w²-24w+72)(3)
1014 = 6w² - 72w + 216

When solving a quadratic equation, we want it set equal to 0.  Subtract 1014 from each side:

1014-1014 = 6w² - 72w + 216 - 1014
0 = 6w² - 72w - 798

We will use the quadratic formula to solve this:

$w=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ \\=\frac{--72\pm \sqrt{(-72)^2-4(6)(-798)}}{2(6)} \\ \\=\frac{72\pm \sqrt{5184--19152}}{12} \\ \\=\frac{72\pm \sqrt{5184+19152}}{12} \\ \\=\frac{72\pm \sqrt{24336}}{12} \\ \\=\frac{72\pm 156}{12} \\ \\=\frac{72+156}{12} \text{ or } \frac{72-156}{12} \\ \\=\frac{228}{12} \text{ or } \frac{-84}{12}=19 \text{ or } -7$

Since we cannot have a negative number for a measurement, 19 has to be the width; then 2(19)-6 = 32 would be the length.