Fencing Two Pens Against a Stone Wall
An agricultural extension officer is helping a farmer near Deir Alla lay out two identical rectangular pens for sheep. The pens will share a long straight stone wall, 50 m long, as one side. A single fence line perpendicular to the wall divides the rectangle into two equal pens, and the two short outer sides are also fenced. The farmer has 90 m of fencing. One short side must contain a 3 m gate, and the gate is not made of fencing. For tractor access, the depth of the enclosure measured perpendicular to the wall must be at least 8 m and at most 20 m. The farmer wants to maximise the total enclosed area, but he also insists that each of the two pens must have at least 360 m² so that the animals are not crowded. The extension officer must decide whether these two goals can be met at the same layout, and if so, what dimensions to recommend.
1. Define a variable for the depth, write the fencing equation, and state the interval of depths allowed by the wall and access constraints.
Read the worked answer
Let (x) be the depth perpendicular to the wall, and let (L) be the length along the wall. The divider is perpendicular to the wall, so each pen is (L/2) by (x).
Fencing used: two short outer sides contribute (2x), the gate removes 3 m, the divider contributes (x), and the long side opposite the wall contributes (L). Hence [L+2x+x-3=90 \Rightarrow L+3x=93.] So (L=93-3x).
The wall is only 50 m long, so (L\le 50): [93-3x\le 50 \Rightarrow x\ge 43/3\approx 14.33\text{ m}.] The access constraint gives (8\le x\le 20). Combining these, the feasible interval is [14.33\le x\le 20.] The total area is (A=Lx=x(93-3x)), and each pen has area (\frac{L}{2}x=\frac{x(93-3x)}{2}).
2. Find the depth and length that maximise total area, and check whether each pen then has at least 360 m².
Read the worked answer
The area is [A(x)=x(93-3x)=-3x^2+93x.] This is a downward-opening parabola. Its vertex is at [x=-\frac{b}{2a}=-\frac{93}{2(-3)}=15.5\text{ m}.] This lies inside ([14.33,20]), so it is the maximum. Then [L=93-3(15.5)=46.5\text{ m}.] The wall constraint is satisfied because (46.5<50). The total area is [A=46.5\times 15.5=720.75\text{ m}^2.] Each pen has area [\frac{46.5}{2}\times 15.5=23.25\times 15.5=360.375\text{ m}^2.] This is at least 360 m², so both goals are met at the same layout, but only just. The maximum is genuine because (A''(x)=-6<0), and the endpoint values are smaller.
3. The farmer raises the minimum to 380 m² per pen. Show that no layout satisfies all constraints, and state the largest minimum area per pen that can be achieved.
Read the worked answer
Each pen has area [a(x)=\frac{x(93-3x)}{2}.] This is also a downward-opening parabola, with the same vertex at (x=15.5). Its maximum value is [a(15.5)=\frac{15.5(93-46.5)}{2}=\frac{15.5\times 46.5}{2}=360.375\text{ m}^2.] Since (360.375<380), no depth in the feasible interval can give each pen 380 m². The constraints are incompatible.
To reach 380 m² per pen, the farmer must relax at least one constraint: use more fencing, shorten the gate, allow a longer wall, or accept a smaller depth/area requirement. Under the given conditions, the largest achievable minimum area per pen is about (360.4\text{ m}^2), at (x=15.5) m and (L=46.5) m.