Three-Sided Fencing and the Quadratic Maximum
Students are building a rectangular vegetable garden against the straight side of a barn. They have 36 feet of fencing for the other three sides. The barn forms the fourth side, so no fencing is needed there. Working in pairs, students must find the dimensions that give the greatest area. One pair sets up 2w + l = 36 and tests w = 8, 9, and 10. Another pair assumes all four sides are fenced and concludes that a 9 ft by 9 ft square is best, with area 81 ft². A third pair writes A = w(36 − 2w) but stops because they do not know how to find the maximum without graphing. The teacher wants to use the three approaches to compare models and justify the maximum.
1. Calculate the area for w = 8, 9, and 10 using the correct three-side model. What do the results suggest, and how can the class justify the maximum without calculus?
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With the barn as one side, the fencing equation is 2w + l = 36, so l = 36 − 2w and A = w(36 − 2w). For w = 8, l = 20 and A = 160 ft². For w = 9, l = 18 and A = 162 ft². For w = 10, l = 16 and A = 160 ft². The areas are equal at w = 8 and w = 10 and larger at w = 9, which suggests a symmetric maximum at w = 9. To justify without calculus, expand A = 36w − 2w² and complete the square: A = −2(w² − 18w) = −2[(w − 9)² − 81] = 162 − 2(w − 9)². Since 2(w − 9)² is never negative, A is at most 162, attained when w = 9. A table of values around 9 also shows the pattern, but completing the square gives a proof. The domain is 0 < w < 18 because l must be positive.
2. Explain the error in the 9 ft by 9 ft conclusion and describe how the teacher could use a diagram to redirect that pair.
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The 9 ft by 9 ft pair used the four-side perimeter 4s = 36, giving s = 9 and area 81 ft². But the barn replaces one side, so fencing only three sides changes both the equation and the best shape. A 9 ft by 9 ft garden would use only 27 ft of the 36 ft of fencing, leaving 9 ft unused. The teacher can sketch the barn as a fixed line and label the three fenced sides; then ask the pair to keep width 9 and use the leftover 9 ft to extend the length to 18 ft. The area doubles from 81 ft² to 162 ft². Comparing 9 × 9 = 81 with 9 × 18 = 162 makes the error concrete: the square is optimal only when all four sides are fenced. The teacher should then ask which model matches the constraint 2w + l = 36 and have the pair test w = 8 and w = 10 to see the symmetry.