Riverside Pens: When the Vertex Lies Outside the Feasible Interval
A farming cooperative in Nankoku has bought 60 m of fencing and wants to enclose a rectangular holding pen on the bank of a straight irrigation channel, using the channel edge as one side of the rectangle so that no fence is needed there. The pen is to be divided into three equal strips by two internal fences running perpendicular to the channel; each internal fence costs the same per metre as the outer fence. A drainage easement forces the pen to be at least 8 m deep measured away from the channel, and the cooperative also insists that every strip be at least 4 m wide along the channel. The fencing comes in continuous rolls, so the dimensions need not be whole metres. The treasurer sketches a parabola and claims the best pen is about 7.5 m deep, which the site manager disputes.
1. Let x be the depth of the pen in metres (perpendicular to the channel) and y the length along the channel. Write the area as a function of x and state the exact interval of x values allowed by the two constraints.
Read the worked answer
Fencing is used on two ends and two internal dividers, each of length x, plus the one side of length y parallel to the channel:
4x + y = 60, so y = 60 - 4x.
Area A(x) = xy = x(60 - 4x) = -4x² + 60x.
Constraints:
- Depth at least 8 m: x ≥ 8.
- Each of the three strips at least 4 m wide along the channel: y/3 ≥ 4, i.e. y ≥ 12, so 60 - 4x ≥ 12, giving x ≤ 12.
- y > 0 would only require x < 15, which is weaker.
Hence the feasible set is 8 ≤ x ≤ 12, a closed interval, so by the extreme value theorem a maximum exists and occurs either at the vertex (if inside) or at an endpoint.
2. Find the maximum possible area and the corresponding dimensions, and explain carefully why the vertex of the parabola is not used.
Read the worked answer
Complete the square: A(x) = -4x² + 60x = -4(x - 7.5)² + 225.
The vertex is at x = 7.5 with value 225, but 7.5 does not lie in [8, 12]: a pen 7.5 m deep would violate the drainage easement. On the feasible interval the function is strictly decreasing, since A'(x) = 60 - 8x ≤ 60 - 64 = -4 < 0 for all x ≥ 8.
So the maximum is at the left endpoint x = 8: A(8) = 8 × 28 = 224 m², with y = 28 m, and each strip 28/3 ≈ 9.3 m wide (comfortably above the 4 m minimum).
A useful check: the other endpoint gives A(12) = 12 × 12 = 144 m², so the binding constraint really is the depth, not the strip width. Writing '225 m²' would be the classic error of reporting an infeasible optimum.
3. The easement is later withdrawn. Quantify what the cooperative gains, and comment on how sensitive the answer is to small errors in the measured 8 m depth.
Read the worked answer
With the easement removed the feasible interval becomes 0 < x < 15, the vertex is admissible, and the optimum is x = 7.5, y = 30, A = 225 m². The strip rule is still satisfied: 30/3 = 10 m per strip.
The gain is 225 - 224 = 1 m², about 0.44%. Because the optimum sits only 0.5 m inside the old boundary and a parabola is flat near its vertex, the objective is insensitive: A(7) = 7 × 32 = 224 and A(8) = 224 as well, so even a half-metre error in the easement measurement changes the achievable area by about 1 m².
Judgement: fighting the easement is not worth legal or surveying costs; the real risk to the plan is arithmetic (forgetting that the channel side needs no fence), not the constraint.