Final answer:
The value of x that maximizes the volume of a box made from a 22 by 27 inches rectangular piece of cardboard is x = 11 inches, as it is the only provided option that is less than half the shortest side, allowing for a positive length and width after the cuts.
Step-by-step explanation:
To determine the value of x that maximizes the volume of the box made from a rectangular piece of cardboard measuring 22 inches by 27 inches, we need to analyze the volume formula in terms of x. Cutting squares of side x from each corner, we get a new length of (27 - 2x) inches, a new width of (22 - 2x) inches, and a height of x inches. The volume V of the resulting box is V = x(27 - 2x)(22 - 2x). To find the maximum volume, we would normally calculate the derivative of the volume function and find its critical points. In high school mathematics, this is typically solved using calculus. However, since we are provided with choices here, we can test which of the given values makes sense particularly whether choice D, x = 11 inches, is feasible. It should be noted that x must be less than half of the shortest side, otherwise the length or width would be negative, which is not possible. The only value that satisfies this condition is x = 11 inches, which indeed maximizes the volume of the box without cancelling the sides.