Assume Statement 1 alone. Since the length of one side is at minimum 4 feet, or equivalently,  yards, its area is at least the square of this, or  square yards. Since the length of one side is at most 6 feet, or, equivalently,  yards, its area is at most the square of this, or 4 square yards. Since the area must be equal to a whole number of yards, we can only narrow the area down to 2, 3, or 4 square yards.

Assume Statement 2 alone. Since the length of one side is at minimum 44 inches, or equivalently,  yards, its area is at least the square of this, or  square yards. Since the length of one side is at most 54 inches, or, equivalently,  yards, its area is at most the square of this, or   square yards. Since the area must be equal to a whole number of yards, the only possible area for the sheet is 2 square yards.

Assume both statements are true. Since, as known from Statement 2, the weight of the bag of oranges is given by a whole number of pounds, then the weight in ounces must be a multiple of 16. From Statement 1, this number is between 110 and 130 inclusive. However, there are two multiples of 16 - 112 and 128 - that fall in this range. Therefore, the question cannot be answered for certain.

Assume Statement 1 alone. One yard is equal to 36 inches, so one square yard is equal to  square inches. The maximum area of 3,000 square inches is equal to  square yards; since the area is a whole number of square yards less than this, we can narrow the area down to 1 or 2 square yards. Without further information, we cannot narrow this down further.

Assume Statement 2 alone. One yard is equal to 3 feet, so one square yard is equal to  square feet. The maximum area of 15 square feet is equivalent to  square yards; since the area is a whole number of square yards less than this, the only possible area is one square yard. 

If Statement 1 alone is assumed, since the length of the rod is given by a whole number of feet, then the length in inches must be a multiple of 12 - 12 inches, 24 inches, etc. However, no other information can be found.

If Statement 2 alone is assumed, there are eleven possible lengths of the rod - 40 inches, 41 inches, and so on up to 50 inches.

Now assume both statements. The length of the rod in inches must be a multiple of 12 between 40 and 50; this narrows it down to one possibility, 48 inches. 

Assume Statement 1 alone. Since one meter is equal to 100 centimeters, the minimum and maximum lengths in meters are  meters and  meters. The squares of these lengths provide the minimum and maximum areas of the sheet:

From the main body of the problem, the area in square meters must be a whole number, so its area must be seven meters.

Assume Statement 2 alone. One meter is equal to 1,000 millimeters, so the minumum and maximum lengths are   meters and  meters. These are the same boundaries derived from Statement 1, so again, the area can be determined to be seven meters.