Fencing a Community Garden Against a Barn Wall
A community garden committee is laying out a rectangular plot against the straight south wall of a barn. The barn wall forms one side of the plot; the other three sides will be enclosed by 60 meters of fencing. The plot's width x is measured perpendicular to the barn, and its length runs parallel to the wall. For safety and access, the committee requires the width to be at least 12 meters and the area to be at least 400 square meters. A volunteer has proposed making the width 9 meters, arguing that a narrower plot leaves more length and therefore more area. The committee also wants to know whether the 400-square-meter requirement is close to the largest area the fencing can produce.
1. Write the area A as a function of the width x, and state the geometric domain for x before applying the committee's rules.
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The fencing covers two widths and one length, so if the width is x, the length is 60 - 2x. Thus
A(x) = x(60 - 2x) = 60x - 2x^2.
For a genuine rectangle, both dimensions must be positive: x > 0 and 60 - 2x > 0. The second inequality gives x < 30. Therefore the geometric domain is 0 < x < 30. The committee's width rule x >= 12 will be imposed afterward, and it will further restrict this interval.
2. Solve A(x) >= 400 together with x >= 12. Give the possible widths and the corresponding lengths, and check the volunteer's 9 m proposal.
Read the worked answer
Set 60x - 2x^2 >= 400. Dividing by -2 reverses the inequality:
x^2 - 30x + 200 <= 0.
Factor: (x - 10)(x - 20) <= 0, so 10 <= x <= 20. Combining with x >= 12 gives 12 <= x <= 20. The corresponding lengths are 60 - 2x, which range from 36 meters down to 20 meters. At x = 9, the length would be 42 meters and the area would be 9(42) = 378 square meters, which is below 400. The volunteer's claim fails because the area is a quadratic function, not an increasing linear function of length; making the width too small reduces the product.
3. Find the maximum possible area and assess whether the 400-square-meter threshold is close to that maximum.
Read the worked answer
The area function A(x) = -2x^2 + 60x is a downward-opening parabola. Its vertex occurs at x = -b/(2a) = -60/(2(-2)) = 15. Then A(15) = 15(60 - 30) = 15(30) = 450 square meters. Since 15 lies in the feasible interval 12 <= x <= 20, the maximum area is 450 square meters. The 400-square-meter requirement is 50 square meters below the maximum, about 89% of it. So 400 is attainable, but it is not the maximum; the committee can do better by choosing a width between 12 and 20 meters, with the best area at width 15 meters and length 30 meters.