Optimizing a Wheat Field with a Fixed Fence Length
A farmer in Balkh province has 120 meters of fencing and wants to enclose a rectangular wheat plot against a straight irrigation canal, so only three sides need fencing (the canal forms the fourth side). Let the two sides perpendicular to the canal each have length x meters, and the side parallel to the canal have length y meters. The farmer wants the enclosed area A to be as large as possible, but he also insists that the plot be at least 10 meters wide in the direction parallel to the canal (so y >= 10) and that each perpendicular side be at least 5 meters (x >= 5). The fencing constraint is 2x + y = 120.
1. Express the area A as a function of x, state the feasible interval for x implied by the constraints, and find the value of x that maximizes A.
Read the worked answer
From 2x + y = 120 we get y = 120 - 2x. Area A = x*y = x(120 - 2x) = 120x - 2x^2.
Constraints: x >= 5 and y >= 10 means 120 - 2x >= 10, so 2x <= 110, x <= 55. Thus x is in [5, 55].
A(x) = -2x^2 + 120x is a downward parabola with vertex at x = -b/(2a) = -120/(2*(-2)) = 30. Since 30 lies in [5, 55], the maximum occurs at x = 30.
2. Compute the maximum area and the corresponding dimensions, and verify the answer is feasible.
Read the worked answer
At x = 30: y = 120 - 2(30) = 60. A = 30 * 60 = 1800 square meters.
Check constraints: x = 30 >= 5 (ok), y = 60 >= 10 (ok), and 2(30) + 60 = 120 (fencing used exactly). So the maximum feasible area is 1800 m^2 with perpendicular sides 30 m and canal-parallel side 60 m.
3. Suppose the farmer instead insists the plot be square (x = y). Would the area be larger or smaller than the optimum, and by how much?
Read the worked answer
If x = y, then 2x + x = 120, so 3x = 120, x = 40. Area = 40 * 40 = 1600 m^2. This is smaller than 1800 m^2. The loss is 1800 - 1600 = 200 m^2.
Reason: for a fixed perimeter the rectangle maximizing area is the square, but here the perimeter is not fixed at 4x + nothing - the canal side is free, so the optimum is not the square; it is the rectangle where the fenced parallel side equals the sum of the two perpendicular sides (y = 2x).