Fixed Perimeter, Changing Area: The 24-Centimeter Rectangle Problem
Mr. Chen gives fourth graders 24 centimeter-long tiles and asks them to build rectangles with a perimeter of 24 cm using whole-centimeter side lengths. Groups record dimensions: 1 × 11, 2 × 10, 3 × 9, 4 × 8, 5 × 7, and 6 × 6. One group concludes that 6 × 6 has the greatest area because it is 'the most square.' Another group insists 1 × 11 must have the greatest area because 11 is the largest side. A third group notices that area increases from 11 to 36 square centimeters as the sides get closer together. Mr. Chen wants students to justify the pattern with multiplication and to connect it to the perimeter formula. The class has grid paper and can make a table of length, width, perimeter, and area.
1. Evaluate the two competing claims using the data. What is the correct conclusion and what misconception does each claim reflect?
Read the worked answer
The areas are 11, 20, 27, 32, 35, and 36 square centimeters, so 6 × 6 has the greatest area. The first group's conclusion is correct, but 'most square' is an intuitive observation, not yet a justification.
The second group confuses the largest single dimension with the largest area. Area is a product of two dimensions, not a sum or a maximum side length. A 1 × 11 rectangle uses the longest side but has the smallest area because the other factor is only 1. This is a multiplicative-reasoning error: the student focuses on one factor instead of the product. The table should be displayed so students can see that the area rises as the factors become closer in value.
2. Explain, using the perimeter constraint, why area is maximized at the square for whole-number dimensions. Include an algebraic or tabular argument appropriate for upper elementary.
Read the worked answer
The perimeter constraint gives 2(L + W) = 24, so L + W = 12 and W = 12 - L. The area is L(12 - L) = 12L - L². For whole-number side lengths, L can be 1 through 6; the paired widths are 11 through 6, producing the areas in the table.
A pattern argument is accessible to upper elementary students: because L + W is fixed at 12, the product is greatest when the two factors are as close as possible. Moving 1 unit from the width to the length changes the product by 1: (6 - k)(6 + k) = 36 - k². For example, 5 × 7 = 35 and 4 × 8 = 32. The square, 6 × 6, is the closest pair of whole-number factors, so it gives the maximum area.
3. A student says, 'If I double all sides of a 3 × 9 rectangle, the perimeter doubles and the area doubles.' Assess this statement and propose a representation to correct it.
Read the worked answer
The statement is half right. Doubling all sides doubles the perimeter: the original perimeter is 24 cm and the doubled rectangle is 6 × 18 with perimeter 48 cm. But the area does not double. The original area is 27 square centimeters, and the doubled area is 108 square centimeters, which is four times as large.
A grid representation makes the scaling visible. Draw the 3 × 9 rectangle, then draw the 6 × 18 rectangle around it. The larger rectangle can be partitioned into four copies of the original (a 2-by-2 array), showing why the area factor is 2 × 2 = 4. The key idea is that area is measured in square units, so scaling both dimensions by a factor of 2 scales area by 2². Perimeter is measured in linear units, so it scales by 2.