Fencing a Plot Against a Wall
A homeowner is fencing a rectangular vegetable plot against the back wall of her garage, so only three sides need fence: two equal sides perpendicular to the wall and one side parallel to it. She has bought exactly 60 feet of fencing and intends to use all of it. She wants the plot as large as possible, but she also insists the plot be at least 400 square feet so her rows fit. Later she remembers the gate will consume 4 feet of that fencing, leaving only 56 feet for the three sides, and she will not buy more.
1. Let x be the length of each side perpendicular to the wall. Express the area as a function of x and find the dimensions that maximize it.
Read the worked answer
The parallel side uses what is left: 60 − 2x.
Area: A(x) = x(60 − 2x) = −2x² + 60x
Complete the square: A = −2(x² − 30x) = −2[(x − 15)² − 225] = −2(x − 15)² + 450.
This is a downward parabola with vertex at x = 15. Maximum area = 450 sq ft, with perpendicular sides of 15 ft each and a parallel side of 60 − 30 = 30 ft.
Check: 15 × 30 = 450, and 15 + 15 + 30 = 60 feet of fence. A square 20 × 20 would need 60 ft too but would enclose only 400 sq ft - the wall side is what makes the non-square shape better.
2. What range of x keeps the plot at least 400 square feet?
Read the worked answer
Set A(x) ≥ 400:
−2x² + 60x ≥ 400 x² − 30x + 200 ≤ 0 (x − 10)(x − 20) ≤ 0
So 10 ≤ x ≤ 20.
Interpretation: perpendicular sides between 10 and 20 ft, with the parallel side between 40 ft (when x = 10) and 20 ft (when x = 20). At either endpoint the area is exactly 400 sq ft (10 × 40 and 20 × 20), so both endpoints are acceptable if 'at least 400' is the rule. The maximum, x = 15, sits comfortably inside the range.
3. With the 4-foot gate taken out, redo the maximization and say what the homeowner should do.
Read the worked answer
Now only 56 ft of fence is available for three sides: parallel side = 56 − 2x.
A(x) = x(56 − 2x) = −2x² + 56x = −2(x − 14)² + 392
Maximum = 392 sq ft at x = 14 ft, parallel side 56 − 28 = 28 ft.
Judgement: the best possible plot now falls 8 sq ft short of her 400 sq ft requirement. The shortfall has to be resolved by changing something, not by arithmetic: shorten the gate to about 2 ft (giving 58 ft of fence, which allows 14.5 × 29 = 420 sq ft), buy a few more feet of fencing, or accept 392 sq ft. Simply building the 14 × 28 layout and calling it 'close enough' silently violates the stated constraint.