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Single Input Production Economics for Farm
Management
AAE 320Paul D. Mitchell
Production EconomicsLearning 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 FormA table listing the maximum output for each given input levelTDN = total digestible nutrition (feed)
0
5,000
10,000
15,000
20,000
0 5,000 10,000 15,000
TDN (lbs/yr)
Milk
(lb
s/yr
)
TDN (1000 lbs/yr)
Milk (lbs/yr)
0 01 8002 1,7003 3,0004 5,0005 7,5006 10,2007 12,8008 15,1009 17,100
10 18,40011 19,20012 19,50013 19,60014 19,400
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 OutputImplicit Function
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: 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) = DQ/DX or
derivative dQ/dX: output generated by the last unit of input used
MP: Slope of TP curve
Graphics
Input X
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 FormInput 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: 6 = (6 – 0)/(1 – 0)AP: 8.0 = 16/2
MP: 5 = (60 – 55)/(6 – 5)AP: 8.9 = 62/7
MP = DQ/DX = (Q2 – Q1)/(X2 – X1)
AP = Q/X
Same Data: Graphically
-5
15
35
55
0 2 4 6 8 10
TP
MP
AP
Think Break #1
Fill in the missing numbers in the table for Nitrogen and Corn Yield
Remember the Formulas
MP = DQ/DX = (Q2 – Q1)/(X2 – X1)
AP = Q/X
N Yield AP MP
0 30 --- ---25 45 1.8 0.6
50 75 1.2
75 105 1.4
100 135 1.35 1.2
125 150 0.6
150 165 1.1
200 170 0.85 0.1
250 160 0.64 -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 UseHow Much Input to Use?
Mathematically: Profit = Revenue – CostProfit = price x output – input cost – fixed
cost p = pQ – rX – K = pf(X) – rX – Kp = profit Q = output X = inputp = output price r = input pricef(X) = production function K = fixed
cost
Economics of Input Use Find X to Maximize p = pf(X) – rX 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 = 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 theMarginal Benefit of N = Marginal Cost of N
Milk Cow ExampleTDN Milk MP VMP price TDN profit
0 0 0 $0 $150 -$400
1 800 800 $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
10 18,400 1300 $156 $150 $308
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
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
0
5,000
10,000
15,000
20,000
0 2 4 6 8 10 12 14 16
0
500
1000
1500
2000
2500
3000
0 2 4 6 8 10 12 14 16
TDN
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
Yieldbu/A MP VMP
0 30 ---25 45 0.6
50 75 1.2
75 105 1.2
100 135 1.2
125 150 0.6
150 165 0.6
200 170 0.1
250 160 -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?
N Yield MP VMP
0 30 ---25 45 0.6 1.2
50 75 1.2 2.4
75 105 1.2 2.4
100 135 1.2 2.4
125 150 0.6 1.2
150 165 0.6 1.2
200 170 0.1 0.2
250 160 -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 = 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)
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 #3
What are the 1st and 2nd derivatives with respect to X of the following functions?
1. Q = 4 + 15X – 7X2
2. p = 2(5 – X – 3X2) – 8X - 153. p = 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
0
5
10
15
0 1 2 3 4 5 6
x
f(x) a
nd f'
(x)
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
Example 2: Graphics
-8
-4
0
4
8
12
0 1 2 3 4 5 6
x
f(x
) a
nd
f'(x
)
f(x)
f'(x)
Slope = 0
f’(X) = 0
What value of X maximizes this function?
Think Break #4
Find X to Maximize:p = 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, p = 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 p = 10(30 + 5X – 0.4X2) – 2X – 18
To find X to maximize p, solve the FOC and check the SOC
Calculus and Production Economics
p = 10(30 + 5X – 0.4X2) – 2X – 18 FOC: 10(5 – 0.8X) – 2 = 0
10(5 – 0.8X) = 2 p x MP = r5 – 0.8X = 2/10 MP = r/p
When you solve the FOC, you set VMP = r and/or MP = r/p
SummarySingle 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)