4. Use a 1-meter long iron wire to enclose a rectangle with both length and width being integer centimeters, there are $\qquad$ different ways to do so. The maximum area of the rectangle is $\qquad$ square centimeters.
【Answer】 25 ways, 625 square centimeters
【Key Points】Perimeter of a Rectangle, Maximum Value Problem
【Analysis】
1 meter $=100$ centimeters, which is the perimeter of the rectangle, so the length + width of the rectangle $=100 \div 2=50$ centimeters;
Different ways to enclose: $50=49+1=48+2=47+3=\ldots=25+25$, a total of 25 ways;
Since the sum of the length and width is constant, the product, which is the area of the rectangle, is maximized when their difference is the smallest. Therefore, the maximum area of the rectangle is $25 \times 25=625$ square centimeters.