Unformatted text preview: Single Input Production Economics for Farm Management
AAE 320 Paul D. Mitchell Production Economics: Topics Single and Multiple Input Production Functions 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: flour, cheese, ham 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 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 Milk (lbs/yr) 15,000 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(s) Single Input, Single Output Multiple Input, Single Output Milk = f(TDN) Multiple Input, Multiple Output Implicit Function Milk = f(TDN, Labor, Equipment) F(Milk, Meat, TDN, Labor, Equipment) = 0 Examples Polynomial: Linear, Quadratic, Cubic Milk = b0 + b1TDN + b2TDN2 Milk = 2261 + 2.535TDN 0.000062TDN2 Are many functions used, depending on the process: CobbDouglas, 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 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 Output Q Graphics
Q 1) MP = 0 when Q at maximum, i.e. slope = 0 MP AP 2) AP = MP when AP at maximum, at Q where line btwn origin and Input X Q curve tangent 3) MP > AP when AP increasing 4) AP > MP when AP AP decreasing Input X MP MP and AP: Tabular Form
Input TP MP 0 1 2 3 4 5 6 7 8 9 10 0 6 16 29 44 55 60 62 62 61 59 6 10 13 6.0 8.0 9.7 AP MP = Q/ X = (Q2 Q1)/(X2 X1) 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 15 11.0 11 11.0 5 10.0 2 0 -1 -2 8.9 7.8 6.8 5.9 Same Data: Graphically
55 35 TP MP AP 15 -5 0 2 4 6 8 10 Think Break #1 Fill in the missing numbers in the table for Nitrogen and Corn Yield Remember the Formulas N Yield 0 25 50 75 100 125 150 200 250 30 45 75 105 135 150 165 170 160 AP 1.8 1.4 1.35 1.1 0.85 0.64 MP 0.6 1.2 1.2 0.6 0.1 -0.1 MP = Q/ X = (Q2 Q1)/(X2 X1) AP = Q/X 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 = profit Q = output X = input p = output price r = input price f(X) = production function K = 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 = 0 p x MP r = 0 Rearrange: pf'(X) = r p 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 corn and N 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
TDN 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Milk 0 800 1,700 3,000 5,000 7,500 MP 0 800 900 VMP price TDN $0 $96 $108 $150 $150 $150 $150 $150 $150 $150 $150 $150 $150 $150 $150 $150 $150 $150 profit -$400 -$454 -$496 -$490 -$400 -$250 -$76 $86 $212 $302 $308 $254 $140 $2 -$172 Milk Price = $12/cwt, or $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 1300 $156 2000 $240 2500 $300 10,200 2700 $324 12,800 2600 $312 15,100 2300 $276 17,100 2000 $240 18,400 1300 $156 19,200 19,500 19,600 800 300 100 $96 $36 $12 -$24 19,400 -200 20,000 15,000 1) Output max is where MP = 0 2) Profit Max is where MP = r/p
0 2 4 6 8 10 12 14 16 Q 10,000
5,000 0 3000 2500 MP 2000 1500 1000 500 0 0 2 4 6 8 10 12 14 16 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 #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 Yield MP 0 25 50 75 100 125 150 200 250 30 45 0.6 75 1.2 105 1.2 135 1.2 150 0.6 165 0.6 170 0.1 160 -0.1 VMP 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 on the Table? N 0 25 50 75 100 125 150 200 250 Yield MP 30 45 75 105 135 150 165 170 0.6 1.2 1.2 1.2 0.6 0.6 0.1 VMP 1.2 2.4 2.4 2.4 1.2 1.2 0.2 -0.2 160 -0.1 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 = b0 + b1X + b2X2 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) Constant Function Quick Review of Derivatives
If Q = f(X) = K, then f'(X) = 0 Q = f(X) = 7, then f'(X) = 0 If Q = f(X) = aXb, then f'(X) = abXb1 Q = f(X) = 7X = 7X1, then f'(X) = 7(1)X11 = 7 Q = f(X) = 3X2, then f'(X) = 3(2)X21 = 6X Q = f(X) + g(X), then dQ/dX = f'(X) + g'(X) Q = 3 + 5X 0.1X2, dQ/dX = 5 0.2X Power Function Sum of Functions Think Break #3 1. 1. 2. 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(b0 + b1X + b2X2) 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 X2 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
15 10 5 f(x) and f'(x) 0 0 -5 -10 -15 x 1 2 3 4 5 6 f(x) f'(x)
Slope = 0 f'(X) = 0 Example 2 Maximize, wrt X, f(X) = 10 6X + X2 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 What value of X maximizes this function?
12 8 f(x) and f'(x) Example 2: Graphics Slope = 0 f'(X) = 0
4 0 0 -4 -8 x 1 2 3 4 5 6 f(x) f'(x) Think Break #4
Find 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, = 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? 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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