1. business and industrial
2. company
3. earnings

Single Input Production
Economics for Farm Management
AAE 320
Paul D. Mitchell
Production Economics
Learning Goals
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Single and Multiple Input Production Functions
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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
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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
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Production Function: gives the maximum
amount of output(s) that can be produced
for the given input(s)
Generally two types:
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Tabular Form (Production Schedule)
Mathematical Function
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)
TDN (1000
lbs/yr)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15,000
10,000
5,000
0
0
5,000
10,000
TDN (lbs/yr)
15,000
Production Function
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Mathematically express the relationship
between input(s) and output(s)
Single Input, Single Output
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Multiple Input, Single Output
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Milk = f(TDN)
Milk = f(TDN, Labor, Equipment)
Multiple Input, Multiple Output
Implicit Function
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F(Milk, Meat, TDN, Labor, Equipment) = 0
Examples
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Polynomial: Linear, Quadratic, Cubic
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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?
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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
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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
Output Q
Graphics
Q
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) MP > AP when
AP increasing
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 = (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
Same Data: Graphically
55
TP
MP
AP
35
15
-5 0
2
4
6
8
10
Think Break #1
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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
0
30
25
45
50
75
75 105
100 135
125 150
150 165
200 170
250 160
AP
MP
--1.8
--0.6
1.2
1.4
1.35
1.1
0.85
0.64
1.2
0.6
0.1
-0.2
Law of Diminishing Marginal Product
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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
p = pQ – rX – K = pf(X) – rX – K
p = profit
Q = output
X = input
p = output price
r = input price
f(X) = production function
K = fixed cost
Economics of Input Use
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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
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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
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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?
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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
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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
r
X
TDN
Q
Milk
MP
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
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
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
MP
2000
1500
1000
500
TDN
0
0
2
4
6
8
10
12
14
16
Milk Cow Example: Key Points
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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
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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/A MP
0
25
50
30
45
75
--0.6
1.2
75
100
125
150
105
135
150
165
1.2
1.2
0.6
0.6
200
170
0.1
250
160 -0.1
VMP
Using MP = r/p Price Changes
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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
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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
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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
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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
25
50
75
30
45
75
105
--0.6
1.2
1.2
1.2
2.4
2.4
100
125
150
135
150
165
1.2
0.6
0.6
2.4
1.2
1.2
200
250
170 0.1
160 -0.2
0.2
-0.4
Why We Need Calculus
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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
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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
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Constant Function
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Power Function
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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) = 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
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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
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1.
2.
3.
What are the 1st and 2nd derivatives with
respect to X of the following functions?
Q = 4 + 15X – 7X2
p = 2(5 – X – 3X2) – 8X - 15
p = p(b0 + b1X + b2X2) – rX – K
Calculus of Optimization
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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
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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
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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
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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
Slope = 0
15
f’(X) = 0
f(x) and f'(x)
10
5
f(x)
0
0
1
2
3
-5
-10
-15
x
4
5
6
f'(x)
Example 2
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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
What value of X maximizes this function?
Slope = 0
12
8
f(x) and f'(x)
f’(X) = 0
4
f(x)
f'(x)
0
0
1
2
3
-4
-8
x
4
5
6
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
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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
=r
5 – 0.8X = 2/10
MP = r/p
When you solve the FOC, you set VMP = r
and/or MP = r/p
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Summary
Single Input Production Function
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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
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1) with a production schedule (table)
2) with a production function (calculus)