Max Profit vs. Max Production

Max Profit vs. Max Production

Thursday, September 26, 2019

UNL report examines economics of N fertilizer usage 

In these challenging times for the agriculture economy, it’s important to carefully evaluate the return on investment of nitrogen fertilizer at the individual producer level. According to a recently published article from Cornhusker Economics, excessive use of nitrogen does little or nothing to increase revenues but can inflate costs, thereby reducing profit. How does a grower calculate the difference between maximum profit and maximum production? Many factors must be considered, including the price of corn, the price of fertilizer, and the expected yield gains from application of N at increasing levels on specific varieties of corn.

Read the full report from University of Nebraska Lincoln

Summary

There is a level of N where physical production is maximized and going beyond that point makes no yield difference and no economic sense, since to do so means incurring expense (buying N) with no hope of a return (no yield response). The economic optimum or breakeven point is when each unit (lb) of N returns enough revenue to at least cover the costs of applying that unit. Using data from the 2018 cycle, researchers determined that for D60-69 corn, the economic optimum was between 176 and 177 lbs of applied N per acre. As corn and fertilizer prices shift, this economic optimum won’t remain constant in future growing seasons, but it is a useful measure as corn growers evaluate their growing practices.

The UNL report concludes:
  • Having individual area of field N response functions would enable the implementation of precision farming.
  • Having N response functions for individual areas and/or fields would save applying excessive N fertilizer that makes farmers no money and may become a contaminant in the local ground or surface water, thus increasing the likelihood for future regulation.
  • Low commodity prices call for the use of less N than high prices and that N expense plays a role in the economically optimal quantity of N to be applied.