Agricultural Economics & Soil Fertility
Mitscherlich-Baule Fertilizer Yield & Optimum Formula
Calculate Mitscherlich-Baule Crop Yield Y = A_max * (1 - 2^(-(N_0 + N_app) / b_Baule)), Unfertilized Check Yield, Net Agronomic Profit Lift ($/ha), and Profit-Maximizing Economic Optimal Nitrogen Rate.
Price diminishing fertilizer returns with Baule’s halving law and solve exact Economic Optimal Nitrogen!
Eilhard Alfred Mitscherlich (1909) and B. Baule (1918) replaced Justus von Liebig’s rigid linear plateau with the exponential Law of Diminishing Soil Fertility Returns: one "Baule Unit" b_Baule (kg/ha) of available nutrient is the exact quantity that brings yield halfway to Asymptotic Ceiling A_max (1 Baule = 50%, 2 Baules = 75%, 3 Baules = 87.5%, 4 Baules = 93.75%): Y(N_app) = A_max * (1 - 2^(-(N_0 + N_app) / b_Baule)), where N_0 is residual soil-test nitrate! Setting the marginal revenue product dY/dN_app * P_crop equal to Nitrogen Unit Price c_N solves the Profit-Maximizing Economic Optimal Nitrogen Rate (EONR) in closed form: N_opt* = MAX(0, b_Baule * (LN((A_max * P_crop * LN(2)) / (b_Baule * c_N)) / LN(2)) - N_0)—always strictly BELOW biological maximum yield!
Verify on Ceiling_Yield A_max = 12.00 t/ha (B2), Residual_Soil_N N_0 = 50 kg/ha (1.0 Baule unit -> Unfertilized Check Yield = 6.00 t/ha in E3, B3), Planned_Applied_N N_app = 100 kg/ha (Total = 150 kg/ha = 3.0 Baules, B4), Baule_Unit b_Baule = 50 kg/ha (B5), Grain_Price P_crop = $220/t (B6), and Nitrogen_Cost c_N = $1.40/kg N (B7): Predicted Yield at 100 kg/ha applied is 12.00 * (1 - 2^(-3)) = 10.50 t/ha (E2); Net Profit Gain over Unfertilized Check is (10.50 - 6.00) * $220 - 100 * $1.40 = $990 - $140 = +$850.00/ha (E4); and Closed-Form Economic Optimal Nitrogen Rate is 185.41 kg/ha (E5)! Spike fertilizer cost to c_N = $20.00/kg N: EONR cleanly clamps at 0.00 kg/ha!
| Mitscherlich-Baule Metric (A_max=12 t/ha, N0=50, Napp=100, b=50 kg/ha, $220/t, $1.40/kg N) | Formula | Verified Output |
|---|---|---|
| Predicted Crop Yield Y (t/ha) | A_max * (1 - 2^(-(N_0 + N_app) / b)) | 10.50 t/ha (87.5% of Ceiling) |
| Unfertilized Check Yield (t/ha) | A_max * (1 - 2^(-N_0 / b)) | 6.00 t/ha (50.0% @ 1 Baule) |
| Net Profit Over Check ($/ha) | (Y - Y_0) * P_crop - N_app * c_N | +$850.00 / ha |
| Economic Optimal Applied N (kg/ha) | MAX(0, b*LOG2(A*P*LN(2)/(b*c_N)) - N_0) | 185.41 kg/ha |
Closed-Form Economic Optimal Nitrogen Rate (EONR) Formula
Validate A_max > 0, N_0 >= 0, N_app >= 0, b_Baule > 0, P_crop > 0, and c_N > 0!
=LET(a_mx, B2, n_0s, B3, n_ap, B4, b_un, B5, p_cr, B6, c_nt, B7, IF(OR(a_mx <= 0, n_0s < 0, n_ap < 0, b_un <= 0, p_cr <= 0, c_nt <= 0), "Invalid Fertility Model", ROUND(MAX(0, b_un * (LN((a_mx * p_cr * LN(2)) / (b_un * c_nt)) / LN(2)) - n_0s), 2)))Net Agronomic Return Over Fertilizer Cost Formula
Compare Planned Applied N against EONR (E5) to quantify how much profit per hectare under- or over-fertilizing leaves on the table!
=IF(ISTEXT(E5), E5, ROUND((B2 * (1 - 2 ^ (-(B3 + B4) / B5)) - B2 * (1 - 2 ^ (-B3 / B5))) * B6 - B4 * B7, 2))Try the example
Use the original values or change them to test your understanding.