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)