The Fence Problem: Maximizing Area with Limited Resources
A homeowner wants to enclose a rectangular garden using 40 meters of fencing. One side of the garden will be against a barn and requires no fencing. The homeowner wants to maximize the area of the garden. They consider two options: (1) using the barn as one entire side, so fencing covers three sides, or (2) using only part of the barn as one side, with fencing on the other three sides but not necessarily using the full length of the barn. The barn is 30 meters long. The homeowner also wants the garden to be at least 5 meters wide (the side parallel to the barn) to accommodate a path.
1. Let x be the length of the side parallel to the barn (in meters) and y be the length of each of the two sides perpendicular to the barn. Write the area A as a function of x, and state the domain of x given the constraints.
Read the worked answer
The fencing is used for three sides: the two perpendicular sides (each of length y) and the side parallel to the barn (length x). So 2y + x = 40, hence y = (40 - x)/2. The area is A = x*y = x(40 - x)/2 = -0.5x^2 + 20x. Constraints: x ≥ 0, y ≥ 0 ⇒ x ≤ 40. Also, the garden must be at least 5 m wide: width is y? Wait, careful: the side parallel to the barn is x, and the width (the side perpendicular) is y. The problem says 'at least 5 meters wide (the side parallel to the barn)' - that means x ≥ 5? Actually, it says 'the side parallel to the barn' is the width, so x ≥ 5. Also, the barn is 30 m long, so x ≤ 30 (since the garden side parallel to the barn cannot exceed the barn's length if it uses the barn as one side? But if the garden is not using the barn as a side, then the barn constraint might not apply. However, the problem states the garden will be against the barn, so the side parallel to the barn must be at most the barn's length, so x ≤ 30. Thus domain: 5 ≤ x ≤ 30.
2. Find the value of x that maximizes the area, and determine whether this value is feasible given the constraints. If not, what is the maximum area achievable?
Read the worked answer
The area function A(x) = -0.5x^2 + 20x is a downward-opening parabola. Its vertex occurs at x = -b/(2a) = -20/(2*(-0.5)) = 20. So the unconstrained maximum is at x = 20. Check constraints: 5 ≤ 20 ≤ 30, so it is feasible. Then y = (40-20)/2 = 10. The maximum area is A(20) = 20*10 = 200 m². Since the vertex is within the domain, this is the maximum. The garden should be 20 m parallel to the barn and 10 m perpendicular.
3. Suppose the homeowner insists that the garden use the entire length of the barn (so x = 30). What is the area then? Compare with the maximum found in part (b) and explain the difference in terms of the function's behavior.
Read the worked answer
If x = 30, then y = (40-30)/2 = 5. Area = 30*5 = 150 m². This is less than the maximum of 200 m². The function A(x) = -0.5x^2 + 20x has its vertex at x=20; for x > 20, the function decreases. Since 30 > 20, the area is smaller. The difference is 50 m². This illustrates that the maximum occurs at the vertex, and forcing x to be larger reduces the area because the fencing is allocated more to the parallel side and less to the perpendicular sides, which reduces the product.