Single Input Production Economics for Farm

ManagementAAE 320

Paul 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 How much nitrogen fertilizer do I use for my

corn? How much corn will I get if I use this much

nitrogen? Application of basic production economics

to farm management

Production Definition: Using inputs to create goods

and services having value to consumers or other producers

Production is what 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 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

(lbs

/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 17,10010 18,40011 19,20012 19,50013 19,60014 19,400

Production Function Mathematically express the

relationship between input(s) and output

Single Input, Single Output Milk = f(TDN) Milk = 50 + 3TDN – 0.2TDN2

Multiple Input, Single Output Milk = f(Corn, Soy) Milk = 50 + 3Corn – 0.2Corn2 + 2Soy

– 0.1Soy2 + 0.4CornSoy

Examples Polynomial: Linear, Quadratic, Cubic

Milk = b0 + b1TDN + b2TDN2

Milk = –2261 + 2.535TDN – 0.000062TDN2

Many functions are 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 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 Example: you harvest 200 bu/ac corn

and applied 100 lbs of nitrogen AP = 200/100 = 2, means on average,

you got 2 bu of corn per pound of nitrogen

Graphics: slope of line between origin and the total product curve

Definitions Marginal Product (MP) = DQ/DX or

derivative dQ/dX: Output Q generated by the last unit of

input used Example: corn yield increases from 199

to 200 bu/ac when you increase nitrogen applied from 99 lbs to 100 lbs

MP = 1/1 = 1, meaning you got 1 bu of corn from last 1 pound of nitrogen

MP: Slope of total product 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

TPMPAP

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 MP0 30 --- ---

25 45 1.8 0.650 75 1.275 105 1.4

100 135 1.35 1.2125 150 0.6150 165 1.1200 170 0.85 0.1250 160 0.64 -0.2

Law of Diminishing Marginal Product

Diminishing MP: Holding all other inputs fixed, as use more and more of one input, eventually the MP will start decreasing, i.e., the returns to increasing that input eventually start to decrease

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 decreases and X decreases means MP increases

Transition We spent time explaining production

functions Q = f(X) and their slope = MP, and AP = Q/X

Now we can ask: How do I decide how much input to use?

How much nitrogen should I use for my corn?

How many seeds should I plant per acre?

Choose each input to maximize farmer profit

Set it up as an economic problem

Intuition MP is the extra output generated when

increasing X by one unit If you add one more pound of N, how

much more corn do you get? The MP How much is this extra corn worth?

Output Price x MP = value of the marginal product or the VMP

What does this last pound of N cost? Input Price x Extra N added = Input Price

x 1

Economics of Input UseHow Much Input to Use?

Mathematical Model: Profit = Revenue – Cost

Profit = 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 Learn this model, we will use it a lot!!!

Economics of Input Use Find X to Maximize profit 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 = r, 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: VMP = p x MP

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 -$4001 800 800 $96 $150 -$4542 1,700 900 $108 $150 -$4963 3,000 1300 $156 $150 -$4904 5,000 2000 $240 $150 -$4005 7,500 2500 $300 $150 -$2506 10,200 2700 $324 $150 -$767 12,800 2600 $312 $150 $868 15,100 2300 $276 $150 $2129 17,100 2000 $240 $150 $302

10 18,400 1300 $156 $150 $308

11 19,200 800 $96 $150 $25412 19,500 300 $36 $150 $14013 19,600 100 $12 $150 $214 19,400 -200 -$24 $150 -$172

Milk Price = $12/cwt or p = $0.12/lbTDN Price = $150

per 1,000 lbsFixed Cost = $400/yrPrice Ratio r/p = $150/$0.12 = 1,250VMP = rOptimal TDN = 10+MP = r/pMaximum 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 0.650 75 1.275 105 1.2

100 135 1.2125 150 0.6150 165 0.6200 170 0.1250 160 -0.1

0 50 100 150 200 250 3000

20406080

100120140160180 Yield

0 50 100 150 200 250 300-0.20

0.20.40.60.8

11.21.4

MP

0 50 100 150 200 250 3000

50100150200250300350 Profit

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 VMP0 30 ---

25 45 0.6 1.250 75 1.2 2.475 105 1.2 2.4

100 135 1.2 2.4125 150 0.6 1.2150 165 0.6 1.2200 170 0.1 0.2250 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 some

function g(X) First Order Condition (FOC) Set g’(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 g’’(X) at each X* identified Condition for a maximum is g’’(X*) <

0 Condition for a minimum is g’’(X*) >

0 g’’(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, g(X) = – 5 + 6X –

X2

FOC: g’(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: g’’(X) = – 2 < 0 Negative, satisfies SOC for a

maximum

0 1 2 3 4 5 6

-8

-6

-4

-2

0

2

4

6

8

g(X)g'(X)

Input X

g(X

) an

d g'

(X)

Example 1: GraphicsSlope = 0g’(X) = 0

Example 2 Maximize, wrt X, g(X) = 10 – 6X + X2

FOC: g’(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: g’’(X) = 2 > 0 Positive, does not satisfy SOC for

maximum

0 1 2 3 4 5 6

-8

-6

-4

-2

0

2

4

6

8

10

12

g(X)g'(X)

Input X

g(X

) an

d g'

(X)

Example 2: GraphicsSlope = 0

g’(X) = 0

What value of X maximizes this function?

Think Break #4Find X to Maximize:

p = 10(30 + 5X – 0.4X2) – 2X – 181) What X satisfies the FOC?2) Does this X satisfy the SOC for a

maximum?

Calculus and Production Economics

In general, p(X) = 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)