Optimising a Fenced Paddock with a Fixed Budget
A farmer has 120 m of fencing to enclose a rectangular paddock against a straight river, so only three sides need fencing (two widths and one length parallel to the river). Let the width perpendicular to the river be x metres and the length parallel to the river be y metres. The farmer wants to maximise area A = xy. A neighbour suggests making the paddock a square with side 40 m, arguing that squares always maximise area. The farmer also has a budget constraint: fencing costs $15 per metre and the budget is $1800, exactly matching 120 m.
1. Set up the constraint and objective, and find the dimensions that maximise area. Does the square suggestion hold?
Read the worked answer
Constraint: 2x + y = 120, so y = 120 - 2x. Objective: A(x) = x(120 - 2x) = 120x - 2x^2, with 0 < x < 60. Differentiate: A'(x) = 120 - 4x. Setting A' = 0 gives x = 30, y = 120 - 60 = 60. A''(x) = -4 < 0, so it is a maximum. Maximum area = 30*60 = 1800 m^2. The neighbour's square (40 by 40) uses 2(40) + 40 = 120 m and gives area 1600 m^2, which is less. The square maximises area only when all four sides are fenced with a fixed perimeter; with one side free, the optimum is not square. The judgement is that the shape of the constraint changes the optimum.
2. The fencing cost rises to $18 per metre but the budget stays at $1800. Re-solve for the optimal dimensions and explain the effect.
Read the worked answer
Budget allows 1800/18 = 100 m of fencing. Constraint: 2x + y = 100, so y = 100 - 2x. A(x) = 100x - 2x^2, A'(x) = 100 - 4x = 0 gives x = 25, y = 50. Maximum area = 1250 m^2. The optimal shape keeps the same ratio (y = 2x: length twice the width), because the structure of the problem is unchanged; only the scale shrinks. This is a useful general insight: for a three-sided rectangle against a wall, the optimal length is always twice the width regardless of the total fencing available. Recognising scale-invariance versus shape-invariance is the judgement being tested, and it prevents students from re-deriving from scratch each time.
3. The farmer now wants area at least 1500 m^2 with the original 120 m of fencing. Find the range of x that achieves this and interpret the endpoints.
Read the worked answer
Need 120x - 2x^2 >= 1500, i.e. 2x^2 - 120x + 1500 <= 0, or x^2 - 60x + 750 <= 0. Discriminant = 3600 - 3000 = 600, so x = (60 +/- sqrt(600))/2 = 30 +/- 5*sqrt(6) ≈ 30 +/- 12.247. Thus x is between about 17.75 m and 42.25 m. At each endpoint the area is exactly 1500 m^2; inside the interval the area exceeds 1500, peaking at x = 30 (area 1800). The endpoints correspond to the same area achieved by two different shapes, one tall and narrow, one short and wide. Interpreting both endpoints, not just the maximum, is the judgement; students often report only the peak and miss the feasible band.