Single Input Production Economics for Farm Management AAE

Single Input Production Economics for Farm Management AAE 320 Paul D. Mitchell

Production Economics Learning Goals n Single and Multiple Input Production Functions n n Economics to identify optimal input use and output combinations n n n What are they and how to use them in production economics and farm management How much nitrogen fertilizer do I use for my corn? How much corn will I get if I use this much nitrogen? Application of basic production economics to farm management

Production n n Definition: Using inputs to create goods and services having value to consumers or other producers Production is what farms do! Using land, labor, time, machinery, animals, seeds, fertilizer, water, etc. to grow crops, livestock, milk, eggs, etc. Can further process outputs: flour, cheese, ham Can produce services: dude ranch, bed and breakfast, orchard/pumpkin farm/hay rides, etc. selling the “fall country experience”

Production Function n n Production Function: gives the maximum amount of output that can be produced for the given input(s) Generally two types: n n Tabular Form (Production Schedule) Mathematical Function

TDN (1000 lbs/yr) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Milk (lbs/yr) 0 800 1, 700 3, 000 5, 000 7, 500 10, 200 12, 800 15, 100 17, 100 18, 400 19, 200 19, 500 19, 600 19, 400 Tabular Form A table listing the maximum output for each given input level TDN = total digestible nutrition (feed)

Production Function n n Mathematically express the relationship between input(s) and output Single Input, Single Output n n n Milk = f(TDN) Milk = 50 + 3 TDN – 0. 2 TDN 2 Multiple Input, Single Output n n Milk = f(Corn, Soy) Milk = 50 + 3 Corn – 0. 2 Corn 2 + 2 Soy – 0. 1 Soy 2 + 0. 4 Corn. Soy

Examples n Polynomial: Linear, Quadratic, Cubic n n n Milk = b 0 + b 1 TDN + b 2 TDN 2 Milk = – 2261 + 2. 535 TDN – 0. 000062 TDN 2 Many functions are used, depending on the process: Cobb-Douglas, von Liebig (plateau), Exponential, Hyperbolic, etc.

Why Production Functions? n n n More convenient & easier to use than tables Estimate via regression with the tables of data from experiments Increased understanding of production process: identify important factors and how important factor each is Allows use of calculus for optimization Common activity of agricultural research scientists

Definitions n n Input: X, Output: Q Total Product = Output Q Average Product (AP) = Q/X: average output for each unit of the input used n Example: you harvest 200 bu/ac corn and applied 100 lbs of nitrogen n AP = 200/100 = 2, means on average, you got 2 bu of corn per pound of nitrogen Graphics: slope of line between origin and the total product curve

Definitions n n Marginal Product (MP) = DQ/DX or derivative d. Q/d. X: n Output Q generated by the last unit of input used or applied n Example: corn yield increases from 199 to 200 bu/ac when you increase nitrogen applied from 99 lbs to 100 lbs n MP = 1/1 = 1, meaning you got 1 bu of corn from last 1 pound of nitrogen applied MP: Slope of total product curve

Graphics Output Q 1 Q 2 1) MP = 0 when Q at maximum, i. e. slope = 0 2) AP = MP when AP at maximum, at Q where line btwn origin and Input X Q curve tangent MP AP 3 3) MP > AP when AP increasing 4 MP 4) AP > MP when AP AP decreasing Input X

MP and AP: Tabular Form Input TP MP AP 0 0 1 6 6 6. 0 2 16 10 8. 0 3 29 13 9. 7 4 44 15 11. 0 5 55 11 11. 0 6 60 5 10. 0 7 62 2 8. 9 8 62 0 7. 8 9 61 -1 6. 8 10 59 -2 5. 9 MP = DQ/DX = (Q 2 – Q 1)/(X 2 – X 1) AP = Q/X MP: 6 = (6 – 0)/(1 – 0) AP: 8. 0 = 16/2 MP: 5 = (60 – 55)/(6 – 5) AP: 8. 9 = 62/7

Same Data: Graphically

Think Break #2 n n Fill in the missing numbers in the table for Nitrogen and Corn Yield Remember the Formulas MP = DQ/DX = (Q 2 – Q 1)/(X 2 – X 1) AP = Q/X N Yield AP MP --0. 6 1. 2 0 25 50 30 45 75 --1. 8 75 100 125 150 200 250 105 135 150 165 170 160 1. 4 1. 35 1. 1 0. 85 0. 64 1. 2 0. 6 0. 1 -0. 2

Law of Diminishing Marginal Product n n n Diminishing MP: Holding all other inputs fixed, as you use more and more of one input, eventually the MP will start decreasing, i. e. , the returns to increasing that input eventually start to decrease For example, as you make more and more feed available for a cow, the extra milk produced eventually starts to decrease Main Point: X increase means MP decreases and X decreases means MP increases

Law of Diminishing Marginal Product n n Happens all the time in biological, physical and social systems: eventually the marginal product (MP) will start decreasing Farm example: Acres farmed is your input, total corn produced your output n As you add more and more acres, eventually your MP (the per acre yields you get for the new acres) will decrease: less time to manage, more travel time to get to fields, poorer quality land available, …

Transition n n n We spent time explaining production functions Q = f(X) and their slope = MP, and AP = Q/X Now we can ask: How do I decide how much input to use? n How much nitrogen should I use for my corn? n How many seeds should I plant per acre? Choose each input to maximize farmer profit Set it up as an economic problem First as partial budget and then to calculus

I currently apply 99 pounds of Nitrogen per acre to corn. Should I apply 100 pounds? Benefits Costs Additional Revenues Additional Costs Extra yield = 200 bu – 199 bu $0. 50/lb x 1 lb of N/acre = = 1 bu/acre x $3. 00/bu = $0. 50/acre (Input price) $3/acre (Value of the MP) Costs Reduced None Total Benefits Revenues Reduced None $3/acre Total Costs $0. 50/acre Total Benefits – Total Costs = Net Gain $2. 50/acre

Intuition n n MP is the extra output generated when increasing X by one unit n If you add one more pound of N, how much more corn do you get? The MP (top left partial budget) How much is this extra corn worth? n Output Price x MP = value of the marginal product or the VMP (top left partial budget) What does this last pound of N cost? n Input Price x Extra N added = Input Price x 1 (top right partial budget) It’s a partial budget analysis for adding 1 pound of N and the only cost is buying the extra input

Economics of Input Use How Much Input to Use? Mathematical Model: Profit = Revenue – Cost Profit = price x output – input cost – fixed cost p = p. Q – r. X – K = pf(X) – r. X – K p = profit Q = output X = input p = output price r = input price f(X) = production function K = fixed cost n Learn this model, we will use it a lot!!!

Economics of Input Use n n n Find X to Maximize profit p = pf(X) – r. X – K Calculus: Set first derivative of p with respect to X equal to 0 and solve for X, the “First Order Condition” (FOC) FOC: pf’(X) – r = 0 p x MP – r = 0 Rearrange: pf’(X) = r p x MP is the “Value of the Marginal Product” (VMP), what would get if sold the MP FOC: Increase use of input X until p x MP = r, i. e. , until VMP = r, the price of the input

Intuition n Remember, MP is the extra output generated when increasing X by one unit The value of this MP is the output price p times the MP, or the extra income you get when increasing X by one unit: VMP = p x MP Optimal Rule: keep increasing use of the input X until VMP equals the input price (p x MP = r) n n Keep increasing X until the income the last bit of input generates just equals the cost of buying the last bit of input It’s like a partial budget analysis 1 pound at a time

Milk Cow Example p r X TDN Q Milk MP 0 0 0 $0 $150 -$400 1 800 $96 $150 -$454 2 1, 700 900 $108 $150 -$496 3 3, 000 1300 $156 $150 -$490 4 5, 000 2000 $240 $150 -$400 5 7, 500 2500 $300 $150 -$250 6 10, 200 2700 $324 $150 -$76 7 12, 800 2600 $312 $150 $86 8 15, 100 2300 $276 $150 $212 9 17, 100 2000 $240 $150 $302 VMP = r 10 18, 400 1300 $156 $150 $308 Optimal TDN = 10+ 11 19, 200 800 $96 $150 $254 12 19, 500 300 $36 $150 $140 13 19, 600 100 $12 $150 $2 14 19, 400 -200 -$24 $150 -$172 VMP price TDN profit Milk Price = $12/cwt or p = $0. 12/lb TDN Price = $150 per 1, 000 lbs Fixed Cost = $400/yr Price Ratio r/p = $150/$0. 12 = 1, 250 MP = r/p Maximum Production

1) Output max is where MP = 0 r/p 2) Profit Max is where MP = r/p Q TDN MP TDN

Another Way to Look at Input Use n n n Have derived the profit maximizing condition defining optimal input use as: p x MP = r or VMP = r Rearrange this condition to get an alternative: MP = r/p Keep increasing use of the input X until its MP equals the price ratio r/p Both give the same answer! Price ratio version useful to understand effect of price changes

MP=r/p: What is r/p? n n n r/p is the “Relative Price” of input X, how much X is worth in the market relative to Q r is $ per unit of X, p is $ per unit of Q Ratio r/p is units of Q per one unit of X r/p is how much Q the market place would give you if you traded in one unit of X r/p is the cost of X if you were buying X in the market using Q in trade

MP = r/p Example: Cow feed n n r = $/ton of TDN (feed), p = $/cwt of milk, so r/p = ($/ton)/($/cwt) = cwt/ton, or the hundredweight of milk you buy if “traded in” one ton of feed MP = cwt of milk from by the last ton of feed Condition MP = r/p means: Find TDN amount that gives the same conversion between TDN and milk in the production process as in the market, or find the TDN amount set the Marginal Benefit of TDN = Marginal Cost of TDN

Milk Cow Example: Key Points n n Profit maximizing TDN is less than output maximizing TDN, which implies profit maximization ≠ output maximization Profit maximizing TDN occurs at TDN levels where MP is decreasing, meaning will use TDN so have a diminishing MP Profit maximizing TDN depends on both the TDN price and the milk price Profit maximizing TDN same whether use VMP = r or MP = r/p

Think Break #3 n n Yield N lbs/A bu/A MP Fill in the VMP column 0 30 --in the table using 25 45 0. 6 $2/bu for the corn price. What is the profit maximizing N fertilizer rate if the N fertilizer price is $0. 2/lb? 50 75 100 125 150 200 250 75 1. 2 105 1. 2 135 1. 2 150 0. 6 165 0. 6 170 0. 1 160 -0. 1 VMP

Why We Need Calculus n n n What do you do if the relative price ratio r/p for the input is not on the table? What do you do if the VMP is not on the table? If you have the production function Q = f(X), then you can use calculus to derive an equation for the MP = f’(X) With an equation for MP, you can “fill in the gaps” in the tabular form of the production schedule

Calculus and AAE 320 n n n I will keep the calculus simple!!! Production Functions will be Quadratic Equations: Q = f(X) = a + b. X + c. X 2 First derivative = slope of production function = Marginal Product 3 different notations for derivatives dy/dx (Newton), f′(x) and fx(x) (Leibniz) 2 nd derivatives: d 2 y/dx 2, f′’(x), fxx (x)

Quick Review of Derivatives n Constant Function n Power Function n n If Q = f(X) = K, then f’(X) = 0 Q = f(X) = 7, then f’(X) = 0 If Q = f(X) = a. Xb, then f’(X) = ab. Xb-1 Q = f(X) = 7 X 1, then f’(X) = 7(1)X 1 -1 = 7 Q = f(X) = 3 X 2, then f’(X) = 3(2)X 2 -1 = 6 X Sum of Functions n n Q = f(X) + g(X), then d. Q/d. X = f’(X) + g’(X) Q = 3 + 5 X – 0. 1 X 2, d. Q/d. X = 5 – 0. 2 X

Think Break #4 n 1. 2. 3. What are the 1 st and 2 nd derivatives with respect to X of the following functions? Q = 4 + 15 X – 7 X 2 p = 2(5 – X – 3 X 2) – 8 X – 15 p = p(a + b. X + c. X 2) – r. X – K

Calculus of Optimization n n n Problem: Choose X to Maximize some function g(X) First Order Condition (FOC) Set g’(X) = 0 and solve for X May be more than one Call these potential solutions X* Identifying X values where the slope of the objective function is zero, which occurs at maximums and minimums

Calculus of Optimization n n n Second Order Condition (SOC) Evaluate g’’(X) at each X* identified Condition for a maximum is g’’(X*) < 0 Condition for a minimum is g’’(X*) > 0 g’’(X) is function's curvature at X Positive curvature is convex (minimum) Negative curvature is concave (maximum)

Calculus of Optimization: Intuition n n FOC: finding the X values where the objective function's slope is zero, candidates for minimum/maximum SOC: checks the curvature at each candidates identified by FOC Maximum is curved down (negative) Minimum is curved up (positive)

Example 1 n n n Choose X to maximize g(X) = – 5 + 6 X – X 2 FOC: g’(X) = 6 – 2 X = 0 FOC satisfied when X = 3 Is this a maximum or a minimum or an inflection point? How do you know? Check the SOC: g’’(X) = – 2 < 0 Negative, satisfies SOC for a maximum

Example 1: Graphics Slope = 0 8 g’(X) = 0 6 g(X) and g'(X) 4 2 g(X) 0 0 1 2 3 -2 -4 -6 -8 Input X 4 5 6 g'(X)

Example 2 n n n Choose X to maximize g(X) = 10 – 6 X + X 2 FOC: g’(X) = – 6 + 2 X = 0 FOC satisfied when X = 3 Is this a maximum or a minimum or an inflection point? How do you know? Check the SOC: g’’(X) = 2 > 0 Positive, does not satisfy SOC for maximum

Example 2: Graphics What value of X maximizes this function? Slope = 0 12 10 8 g’(X) = 0 g(X) and g'(X) 6 4 g(X) 2 g'(X) 0 -2 0 1 2 3 -4 -6 -8 Input X 4 5 6

Think Break #5 Choose X to Maximize: p = 10(30 + 5 X – 0. 4 X 2) – 2 X – 18 1) What X satisfies the FOC? 2) Does this X satisfy the SOC for a maximum?

Calculus and Production Economics n n In general, p(X) = pf(X) – r. X – K Suppose your production function is Q = f(X) = 30 + 5 X – 0. 4 X 2 Suppose output price is 10, input price is 2, and fixed cost is 18, then p = 10(30 + 5 X – 0. 4 X 2) – 2 X – 18 To find X to maximize p, solve the FOC and check the SOC

Calculus and Production Economics p = 10(30 + 5 X – 0. 4 X 2) – 2 X – 18 n FOC: 10(5 – 0. 8 X) – 2 = 0 10(5 – 0. 8 X) = 2 p x MP =r 5 – 0. 8 X = 2/10 MP = r/p When you solve the FOC, you set VMP = r and/or MP = r/p n

Summary Single Input Production Function n n Condition to find optimal input use: VMP = r or MP = r/p What does this condition mean? What does it look like graphically? Know how to use condition to find optimal input use n 1) With a production schedule (table) n 2) With a production function (calculus)