Production Economics and Farm Management Essentials
Understanding production economics for effective farm management involves analyzing single and multiple input production functions to optimize input use and output combinations. Explore the concept of production, production functions, and examples such as linear, quadratic, and cubic functions in agricultural settings.
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Single Input Production Economics for Farm Management AAE 320 Paul D. Mitchell
Production Economics Learning Goals Single and Multiple Input Production Functions What are they and how to use them in production economics and farm management Economics to identify optimal input use and output combinations 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 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 Production Function: gives the maximum amount of output that can be produced for the given input(s) Generally two types: 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) 20,000 15,000 Milk (lbs/yr) 10,000 5,000 0 0 5,000 10,000 15,000 TDN (lbs/yr)
Production Function Mathematically express the relationship between input(s) and output Single Input, Single Output Milk = f(TDN) Milk = 50 + 3TDN 0.2TDN2 Multiple Input, Single Output Milk = f(Corn, Soy) Milk = 50 + 3Corn 0.2Corn2 + 2Soy 0.1Soy2 + 0.4CornSoy
Examples Polynomial: Linear, Quadratic, Cubic Milk = b0 + b1TDN + b2TDN2 Milk = 2261 + 2.535TDN 0.000062TDN2 Many functions are used, depending on the process: Cobb-Douglas, von Liebig (plateau), Exponential, Hyperbolic, etc.
Why Production Functions? 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 Input: X, Output: Q Total Product = Output Q Average Product (AP) = Q/X: average output for each unit of the input used Example: you harvest 200 bu/ac corn and applied 100 lbs of nitrogen 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 Marginal Product (MP) = Q/ X or derivative dQ/dX: Output Q generated by the last unit of input used or applied Example: corn yield increases from 199 to 200 bu/ac when you increase nitrogen applied from 99 lbs to 100 lbs 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)MP = 0 when Q at maximum, i.e. slope = 0 1 Q 2 2)AP = MP when AP at maximum, at Q where line btwn origin and Q curve tangent Input X MP AP 3)MP > AP when AP increasing 3 4 4)AP > MP when AP decreasing AP Input X MP
MP and AP: Tabular Form MP = Q/ X = (Q2 Q1)/(X2 X1) AP = Q/X Input TP MP 0 1 2 16 3 29 4 44 5 55 6 60 7 62 8 62 9 61 10 59 AP 0 6 6 6.0 8.0 9.7 11.0 11.0 MP: 6 = (6 0)/(1 0) 10 13 15 11 AP: 8.0 = 16/2 MP: 5 = (60 55)/(6 5) AP: 8.9 = 62/7 5 10.0 2 0 -1 -2 8.9 7.8 6.8 5.9
Same Data: Graphically 55 TP MP AP 35 15 -5 0 2 4 6 8 10
Think Break #2 N 0 Yield 30 45 75 105 135 150 165 170 160 AP --- 1.8 MP --- 0.6 1.2 Fill in the missing numbers in the table for Nitrogen and Corn Yield 25 50 75 1.4 1.35 Remember the Formulas MP = Q/ X = (Q2 Q1)/(X2 X1) AP = Q/X 100 125 150 200 250 1.2 0.6 1.1 0.85 0.64 0.1 -0.2
Law of Diminishing Marginal Product 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 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 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 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? How much nitrogen should I use for my corn? 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 Additional Revenues Extra yield = 200 bu 199 bu = 1 bu/acre x $3.00/bu = $3/acre (Value of the MP) Costs Reduced None Costs Additional Costs $0.50/lb x 1 lb of N/acre = $0.50/acre (Input price) Revenues Reduced None Total Benefits Total Benefits Total Costs = Net Gain $3/acre Total Costs $0.50/acre $2.50/acre
Intuition MP is the extra output generated when increasing X by one unit 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? Output Price x MP = value of the marginal product or the VMP (top left partial budget) What does this last pound of N cost? 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 = pQ rX K = pf(X) rX K = profit Q = output p = output price r = input price f(X) = production function Learn this model, we will use it a lot!!! X = input K = fixed cost
Economics of Input Use Find X to Maximize profit = pf(X) rX K Calculus: Set first derivative of with respect to X equal to 0 and solve for X, the First Order Condition (FOC) FOC: pf (X) 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 p x MP r = 0 p x MP = r
Intuition 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) 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 r Q X TDN 0 1 2 3 4 5 6 7 8 9 10 Milk 0 800 1,700 3,000 5,000 7,500 10,200 2700 $324 12,800 2600 $312 15,100 2300 $276 17,100 2000 $240 18,400 1300 $156 MP 0 800 900 1300 $156 2000 $240 2500 $300 VMP price TDN $0 $96 $108 profit -$400 -$454 -$496 -$490 -$400 -$250 -$76 $86 $212 $302 $308 Milk Price = $12/cwt or p = $0.12/lb $150 $150 $150 $150 $150 $150 $150 $150 $150 $150 $150 TDN Price = $150 per 1,000 lbs Fixed Cost = $400/yr Price Ratio r/p = $150/$0.12 = 1,250 VMP = r Optimal TDN = 10+ 11 12 13 14 19,200 19,500 19,600 19,400 800 300 100 -200 $96 $36 $12 -$24 $150 $150 $150 $150 $254 $140 $2 -$172 MP = r/p Maximum Production
1)Output max is where MP = 0 20,000 r/p 15,000 2)Profit Max is where MP = r/p Q 10,000 5,000 TDN 0 0 2 4 6 8 10 12 14 16 3000 2500 2000 MP 1500 1000 500 TDN 0 0 2 4 6 8 10 12 14 16
Another Way to Look at Input Use 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? 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 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 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 Yield bu/A MP 30 45 75 105 135 150 165 170 160 lbs/A 0 25 50 75 100 125 150 200 250 VMP Fill in the VMP column in the table using $2/bu for the corn price. What is the profit maximizing N fertilizer rate if the N fertilizer price is $0.2/lb? --- 0.6 1.2 1.2 1.2 0.6 0.6 0.1 -0.1
Why We Need Calculus 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 I will keep the calculus simple!!! Production Functions will be Quadratic Equations: Q = f(X) = a + bX + cX2 First derivative = slope of production function = Marginal Product 3 different notations for derivatives dy/dx (Newton), f (x) and fx(x) (Leibniz) 2nd derivatives: d2y/dx2, f (x), fxx (x)
Quick Review of Derivatives Constant Function If Q = f(X) = K, then f (X) = 0 Q = f(X) = 7, then f (X) = 0 Power Function If Q = f(X) = aXb, then f (X) = abXb-1 Q = f(X) = 7X = 7X1, then f (X) = 7(1)X1-1 = 7 Q = f(X) = 3X2, then f (X) = 3(2)X2-1 = 6X Sum of Functions Q = f(X) + g(X), then dQ/dX = f (X) + g (X) Q = 3 + 5X 0.1X2, dQ/dX = 5 0.2X
Think Break #4 What are the 1st and 2nd derivatives with respect to X of the following functions? Q = 4 + 15X 7X2 = 2(5 X 3X2) 8X 15 = p(a + bX + cX2) rX K 1. 2. 3.
Calculus of Optimization 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 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 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 Choose X to maximize g(X) = 5 + 6X X2 FOC: g (X) = 6 2X = 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 4 2 g(X) and g'(X) g(X) g'(X) 0 0 1 2 3 4 5 6 -2 -4 -6 -8 Input X
Example 2 Choose X to maximize g(X) = 10 6X + X2 FOC: g (X) = 6 + 2X = 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 6 g(X) and g'(X) 4 g(X) g'(X) 2 0 0 1 2 3 4 5 6 -2 -4 -6 -8 Input X
Think Break #5 Choose X to Maximize: = 10(30 + 5X 0.4X2) 2X 18 1) What X satisfies the FOC? 2) Does this X satisfy the SOC for a maximum?
Calculus and Production Economics In general, (X) = pf(X) rX K Suppose your production function is Q = f(X) = 30 + 5X 0.4X2 Suppose output price is 10, input price is 2, and fixed cost is 18, then = 10(30 + 5X 0.4X2) 2X 18 To find X to maximize , solve the FOC and check the SOC
Calculus and Production Economics = 10(30 + 5X 0.4X2) 2X 18 FOC: 10(5 0.8X) 2 = 0 10(5 0.8X) = 2 p x MP = r 5 0.8X = 2/10 MP = r/p When you solve the FOC, you set VMP = r and/or MP = r/p
Summary Single Input Production Function 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 1) With a production schedule (table) 2) With a production function (calculus)
