Fencing a Vegetable Plot Against a Stone Wall
A gardener is fencing a rectangular vegetable plot and will use an existing straight stone wall as one full side of the rectangle, so only three sides require fencing. The gardener has exactly 60 feet of fencing and intends to use all of it. The stone wall is 24 feet long, and the plot's side lying along the wall may not be longer than the wall itself. Let x be the length in feet of each of the two sides perpendicular to the wall, and let y be the length of the side along the wall. Dimensions need not be whole numbers. The gardener wants the plot with the greatest possible area subject to these constraints and asks whether extending the stone wall would help.
1. Express the plot's area A as a function of x alone, and state the complete feasible domain for x given the wall-length restriction.
Read the worked answer
The three fenced sides give 2x + y = 60, so y = 60 − 2x.
Area: A(x) = x·y = x(60 − 2x) = −2x² + 60x.
Constraints:
- y ≤ 24 → 60 − 2x ≤ 24 → x ≥ 18
- y > 0 → x < 30
So the feasible domain is 18 ≤ x ≤ 30. The lower endpoint x = 18 is forced by the wall, not by the fencing - without it the algebra would allow x as small as 0. Every x in this interval uses all 60 feet of fencing, with y shrinking from 24 to 0.
2. Find the dimensions and area of the largest feasible plot. Be explicit about what happens to the parabola's vertex.
Read the worked answer
A(x) = −2x² + 60x is a downward parabola. The unconstrained vertex is at x = −b/(2a) = −60/(2·(−2)) = 15, giving y = 30 and area A = 15(30) = 450 sq ft.
But x = 15 lies outside the feasible domain (18 ≤ x ≤ 30) because it would require y = 30 > 24, i.e., more wall than exists. On [18, 30] the parabola is strictly decreasing (vertex at 15 is to the left of the interval), so the maximum occurs at the left endpoint x = 18.
At x = 18: y = 60 − 36 = 24 ft, area = 18 × 24 = 432 sq ft.
Endpoint check: x = 30 gives y = 0 and area 0. The optimum is therefore 18 ft by 24 ft, with the 24-ft side along the wall.
For contrast, fully fencing a rectangle with 60 ft of fencing gives at most 15 × 15 = 225 sq ft; using the wall nearly doubles the attainable area.
3. How long would the stone wall need to be for the unconstrained optimum to become feasible, and would making it longer than that increase the maximum area?
Read the worked answer
The unconstrained optimum needs y = 60 − 2(15) = 30 ft of wall. So a wall of at least 30 ft makes x = 15 feasible, and the maximum area rises to 450 sq ft.
Beyond 30 ft, extra length does nothing: the optimum already uses only 30 ft of wall, and the constraint y ≤ (wall length) stops binding. With a 40-ft wall the answer is still 15 ft by 30 ft, 450 sq ft.
General pattern: for a fixed perimeter P on three sides of a rectangle, the free optimum puts P/2 along the wall and P/4 on each perpendicular side. The wall constraint binds exactly when the available wall is shorter than P/2 - here 30 ft > 24 ft, which is why the constrained answer (18 × 24) fell short of 450.