Crop planning with three binding resources and a price the farmer cannot control
A farmer in Bundelkhand has 6.0 ha of irrigated land, 120 man-days of family labour available in the season, a canal allocation of 60 ha-cm of water and ₹45,000 of working capital. Two crops are being considered. Gram (crop A) needs 25 man-days, 12 ha-cm of water and ₹9,000 of capital per hectare, and returns a net profit of ₹28,000 per hectare. Mustard (crop B) needs 15 man-days, 8 ha-cm and ₹6,000 per hectare, and returns ₹18,000 per hectare. Land may be left fallow, inputs cannot be bought beyond the limits stated, and fractional areas are acceptable for planning purposes. The farmer has heard that mustard may fetch a better price next season and wants to know how sensitive the plan is to that price, and what happens to the plan if illness cuts the family labour available to 100 man-days.
1. Formulate the allocation problem and find the profit-maximising plan.
Read the worked answer
Let x, y be hectares of gram and mustard. Constraints: land x + y ≤ 6; labour 25x + 15y ≤ 120; water 12x + 8y ≤ 60, i.e. 3x + 2y ≤ 15; capital 9x + 6y ≤ 45, i.e. the same line 3x + 2y ≤ 15. Water and capital coincide, so capital never binds on its own. Maximise Z = 28x + 18y (₹ thousand).
Vertices: (0, 0) Z = 0; (4.8, 0) Z = 134.4 (labour exactly 120); (0, 6) Z = 108 (labour 90, water 48 ha-cm); intersection of land and water: x + y = 6 and 3x + 2y = 15 gives x = 3, y = 3. At that point water = 15 ha-cm, labour = 25(3) + 15(3) = 120 man-days and capital = 27,000 + 18,000 = ₹45,000 - all three resources bind simultaneously. Z = 84 + 54 = 138, i.e. ₹1,38,000, the maximum.
So the plan is 3 ha gram + 3 ha mustard. The lower-value-per-hectare crop still earns its place because it draws less water and labour per rupee of profit, and land alone cannot produce anything.
2. If mustard returns ₹30,000/ha, does the plan change? Find the price at which the two plans are equally profitable.
Read the worked answer
Recompute with Z = 28x + 30y: (4.8, 0) = 134.4; (3, 3) = 84 + 90 = 174; (0, 6) = 180. The optimum jumps to the corner (0, 6): 6 ha of mustard, using 90 man-days, 48 ha-cm and ₹36,000, for ₹1,80,000.
Switch point: at (3, 3) profit is 84 + 3p; at (0, 6) it is 6p. Setting 84 + 3p = 6p gives p = 28, i.e. ₹28,000/ha. At the 3:3 plan both crops consume the limiting resources in the same 1:1 proportion, so mustard displaces gram exactly when mustard's return equals gram's. Below ₹28,000/ha mustard is a partner crop; above it, gram disappears completely.
Judgement: this is a corner solution with no diversification for its own sake. A 6 ha mustard block carries frost at flowering and aphid risk that the linear model does not price, so a farmer may rationally hold a hectare of gram as insurance even when the arithmetic says otherwise.
3. Labour falls to 100 man-days. Re-solve, and say what the answer implies about hiring labour.
Read the worked answer
Constraints now: x + y ≤ 6, 25x + 15y ≤ 100, 3x + 2y ≤ 15.
Labour and water no longer intersect in the positive quadrant (solving them together gives x = −5), so the optimum lies at the land–labour intersection: x + y = 6 with 25x + 15y = 100, i.e. 25x + 90 − 15x = 100, so x = 1, y = 5. Check feasibility: water = 12 + 40 = 52 ha-cm ≤ 60; capital = 9,000 + 30,000 = ₹39,000 ≤ ₹45,000. Z = 28 + 90 = ₹1,18,000.
Compare: (0, 6) gives ₹1,08,000 but uses only 90 man-days and leaves 10 idle; (4, 0) gives ₹1,12,000. So (1, 5) is best at ₹1,18,000.
Judgement: losing 20 man-days costs ₹20,000 of profit and swings the plan heavily to mustard. Since family labour is the resource the farmer actually controls, the practical rule is that hiring is worthwhile only if the wage bill for those 20 man-days is clearly below the ₹20,000 they recover - and only if the work can be supervised at the right time.