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

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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 Application of basic production economics to farm management This will take a few weeks

Production Definition: Using inputs to create goods and services having value to consumers or other producers Production is what firms/farms do! Using land, labor, time, chemicals, animals, 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(s) that can be produced for the given input(s) Generally two types: Tabular Form (Production Schedule) Mathematical Function

Tabular Form A table listing the maximum output for each given input level TDN = total digestible nutrition (feed) TDN (1000 lbs/yr) Milk (lbs/yr) 00 1800 21,700 33,000 45,000 57,500 610,200 712,800 815,100 917,100 1018,400 1119,200 1219,500 1319,600 1419,400

Production Function Mathematically express the relationship between input(s) and output(s) Single Input, Single Output Milk = f(TDN) Multiple Input, Single Output Milk = f(TDN, Labor, Equipment) Multiple Input, Multiple Output Implicit Function F(Milk, Meat, TDN, Labor, Equipment) = 0

Examples Polynomial: Linear, Quadratic, Cubic Milk = b 0 + b 1 TDN + b 2 TDN 2 Milk = -2261 + 2.535TDN – 0.000062TDN 2 Are many functions 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 methods 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 AP: slope of line btwn origin and TP curve Marginal Product (MP) = Q/ X or derivative dQ/dX: output generated by the last unit of input used MP: Slope of TP curve

Graphics Input X Output Q MP AP Q MP AP 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 Q curve tangent 3)MP > AP when AP increasing 4)AP > MP when AP decreasing

MP and AP: Tabular Form InputTPMPAP 00 1666.0 216108.0 329139.7 4441511.0 5551111.0 660510.0 76228.9 86207.8 9616.8 1059-25.9 MP: 6 = (6 – 0)/(1 – 0) AP: 8.0 = 16/2 MP: 5 = (60 – 55)/(6 – 5) AP: 8.9 = 62/7 MP = Q/ X = (Q 2 – Q 1 )/(X 2 – X 1 ) AP = Q/X

Same Data: Graphically

Think Break #1 Fill in the missing numbers in the table for Nitrogen and Corn Yield Remember the Formulas MP = Q/ X = (Q 2 – Q 1 )/(X 2 – X 1 ) AP = Q/X NYieldAPMP 030 --- 25451.80.6 50751.2 751051.4 1001351.351.2 1251500.6 1501651.1 2001700.850.1 2501600.64-0.2

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

Economics of Input Use How Much Input to Use? Mathematically: Profit = Revenue – Cost Profit = price x output – input cost – fixed cost = pQ – rX – K = pf(X) – rX – K = profitQ = outputX = input p = output pricer = input price f(X) = production functionK = fixed cost

Economics of Input Use Find X to Maximize = pf(X) – rX 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 = 0p x MP – r = 0 Rearrange:pf(X) = rp x MP = 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 = the price of the input

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 The rule, keep increasing use of the input X until VMP equals the input price (p x MP = r), means keep using X until the income the last bit of input generates just equals the cost of buying the last bit of input

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: N fertilizer r = \$/lb of N, p = \$/bu of corn, so r/p = (\$/lb)/(\$/bu) = bu/lb, or the bushels of corn received if traded in one pound of N MP = bushels of corn generated by the last pound of N Condition MP = r/p means: Find N rate that gives the same conversion between N and corn in the production process as in the market, or find N rate to set the Marginal Benefit of N = Marginal Cost of N

Milk Cow Example TDNMilkMPVMPprice TDNprofit 000\$0\$150-\$400 1800 \$96\$150-\$454 21,700900\$108\$150-\$496 33,0001300\$156\$150-\$490 45,0002000\$240\$150-\$400 57,5002500\$300\$150-\$250 610,2002700\$324\$150-\$76 712,8002600\$312\$150\$86 815,1002300\$276\$150\$212 917,1002000\$240\$150\$302 1018,4001300\$156\$150\$308 1119,200800\$96\$150\$254 1219,500300\$36\$150\$140 1319,600100\$12\$150\$2 1419,400-200-\$24\$150-\$172 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 VMP = r Optimal TDN = 10+ MP = r/p Maximum Production X Q r

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

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 #2 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? N lbs/A Yield bu/AMPVMP 030 --- 25450.6 50751.2 751051.2 1001351.2 1251500.6 1501650.6 2001700.1 250160-0.1

Using MP = r/p Price Changes Can use MP = r/p to find optimal X Can also use MP = r/p to examine effect of price changes: what happens to profit maximizing X if output price and/or input price change? Use MP = r/p and the Law of Diminishing MP Output price p increases r/p decreases Input price r increases r/p increases X increases MP decreases X decreases MP increases

Optimal X for Output Price Change Output price p increases r/p decreases Need to change use of X so that the MP equals this new, lower, price ratio r/p Law of Diminishing MP implies that to decrease MP, use more X Intuition: p increase means output more valuable, so use more X to increase output Everything reversed if p decreases

Optimal X for Input Price Change Input price r increases r/p increases Need to change use of X so that the MP equals this new, higher, price ratio r/p Law of Diminishing MP implies that to increase MP, use less X Intuition: r increase means input more costly, so use less X Everything reversed if r increases

Think Break #2 Example Corn price = \$2.00/bu N price = \$0.20/lb Optimal N where VMP = r, or VMP = 0.20 Alternative: MP = r/p, or MP = 0.2/2 = 0.1 What if p = \$2.25/bu and r = \$0.30/lb, r/p = 0.133? Relative price of N has increased, so reduce N, but where is it on the Table? NYieldMPVMP 030 --- 25450.61.2 50751.22.4 751051.22.4 1001351.22.4 1251500.61.2 1501650.61.2 2001700.10.2 250160-0.2-0.4

Why We Need Calculus What do you do if the relative price ratio of 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 = b 0 + b 1 X + b 2 X 2 First derivative = slope of production function = Marginal Product 3 different notations for derivatives dy/dx (Newton), f(x) and f x (x) (Leibniz) 2 nd derivatives: d 2 y/dx 2, f(x), f xx (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) = aX b, then f(X) = abX b-1 Q = f(X) = 7X = 7X 1, then f(X) = 7(1)X 1-1 = 7 Q = f(X) = 3X 2, then f(X) = 3(2)X 2-1 = 6X Sum of Functions Q = f(X) + g(X), then dQ/dX = f(X) + g(X) Q = 3 + 5X – 0.1X 2, dQ/dX = 5 – 0.2X

Think Break #3 What are the 1 st and 2 nd derivatives with respect to X of the following functions? 1. 1. Q = 4 + 15X – 7X 2 = 2(5 – X – 3X 2 ) – 8X - 15 = p(b 0 + b 1 X + b 2 X 2 ) – rX – K

Calculus of Optimization Problem: Choose X to Maximize f(X) First Order Condition (FOC) Set f(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 f(X) at each X* identified Condition for a maximum is f(X * ) < 0 Condition for a minimum is f(X * ) > 0 f(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 Maximize, wrt X, f(X) = – 5 + 6X – X 2 FOC: f(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: f(X) = – 2 < 0 Negative, satisfies SOC for a maximum

Example 1: Graphics Slope = 0 f(X) = 0

Example 2 Maximize, wrt X, f(X) = 10 – 6X + X 2 FOC: f(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: f(X) = 2 > 0 Positive, does not satisfy SOC for maximum

Example 2: Graphics Slope = 0 f(X) = 0 What value of X maximizes this function?

Think Break #4 Find X to Maximize: = 10(30 + 5X – 0.4X 2 ) – 2X – 18 1) What X satisfies the FOC? 2) Does this X satisfy the SOC for a maximum?

Calculus and Production Economics In general, = pf(X) – rX – K Suppose your production function is Q = f(X) = 30 + 5X – 0.4X 2 Suppose output price is 10, input price is 2, and fixed cost is 18, then = 10(30 + 5X – 0.4X 2 ) – 2X – 18 To find X to maximize, solve the FOC and check the SOC

Calculus and Production Economics = 10(30 + 5X – 0.4X 2 ) – 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? Effect of price changes Know how to use condition to find optimal input use 1) with a production schedule (table) 2) with a production function (calculus)

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