ECONOMICS OF OPTIMAL
INPUT USE
AAE 575
Paul D. Mitchell
Production Economics Learning Goals
• Economics of identifying optimal input use and
output combinations
• Single input case
• Two/Multiple input case
Economics of Input Use
• Main idea: specify objective of the decision maker and
how choices affect this objective
• Objective: Maximize profit (p)
• Choice(s): Input(s) (x)
• Alternatives: Minimize cost to meet a yield target or
Maximize output given a cost budget
• Can show that these are “dual” to the problem above
• Hold foot still & slide on show or hold shoe steady & slide in foot?
• Need to specify a profit as a function of input choices: p(x)
Model the Decision Problem
• Basic Model: Choose input to maximize profit
• Profit = Revenue – Cost
• How do Revenue and Cost depend on input x?
• Revenue = Price x Yield, but Yield depends on input via
•
•
•
•
the production function, so Revenue = py = pf(x)
Cost: price times how much input used, so cost = rx
p(x) = pf(x) – rx
Textbook presents more general case where price of
output depends on how much produced and input price
depends on how much buy: p(x) = p(f(x))f(x) – r(x)x
The model p(x) = pf(x) – rx assumes perfect competition:
farmer does not affect output or input prices
Economics of Input Use
• Find x to Maximize p(x) = pf(x) – rx
• Use the calculus of optimization
• First Order Condition (FOC): Find x where the
slope of the objective function to zero
• FOC: Means first derivative is zero: p’(x) = 0
• Second Order Condition (SOC): Make sure at the
x where slope is zero is the top of a hill, not the
bottom of a pit or a bench seat
• SOC means second derivative is negative: p’’(x) < 0
Economics of Input Use: FOC
• Find x to Maximize p(x) = pf(x) – rx
pf’(x) – r = 0
pMP – r = 0
• Rearrange:
pf’(x) = r
pMP = r
• pMP is the “Value of the Marginal Product”
(VMP), revenue you would get if sold the MP
• FOC: Increase use of input x until pMP = r, or
until VMP = the cost of the input
• FOC:
Economics of Input Use: SOC
• Find x to Maximize p(x) = pf(x) – rx
• FOC: pf’(x) – r = 0
• SOC: pf’’(x) < 0
• Know p > 0, so SOC becomes: f’’(x) < 0
• If the production function is concave, then we
know the profit maximization SOC is satisfied
• Main Point: FOC and SOC are why we spent time
on MP and curvature of production function
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 revenue generated when
increasing x by one unit
• The rule, keep increasing use of the input x until
the VMP equals the input price (pMP = r), means
keep using x until the revenue the last bit of input
use generates just equals the cost of buying the
last bit of this input
Another Way to Look at Input Use
• Have derived the profit maximizing condition defining
optimal input use as:
pMP = 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!
• This is why economically optimal nitrogen rate (EONR)
and Lauer’s optimal seeding rates use price ratios
What is r/p?
• r/p is the “Relative Price” of input x, how much x
is worth in the market relative to y
• r is $ per unit of x, p is $ per unit of y
• Ratio r/p is units of y per one unit of x
• r/p is how much y 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 y 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 the
Marginal Benefit of N = Marginal Cost of N
1) Output max is
where MP = 0,
x = xymax
20,000
r/p
15,000
y
10,000
5,000
0
0
2
4
6
0
2
4
6
x
8
10
12
10
12
14
16
14
16
3000
2500
MP
2000
1500
1000
500
0
x
8
xopt
xymax
2) Profit Max is
where MP = r/p,
x = xopt
Key Points
• Profit maximizing x is less than output maximizing
x, or xopt < xymax
• Implies profit maximization is not the same as output
maximization
• Profit maximizing x occurs at x levels where MP
is decreasing, meaning will use x so have a
diminishing MP
• Profit maximizing x depends on both r, the price
of x, and p, the output price
• Profit maximizing x is the same whether use VMP
= r or MP = r/p
Calculus of Optimization
• Problem: Choose x to Maximize p(x)
• First Order Condition (FOC)
• Set p’(x) = 0 and solve for x
• May be more than one x where p’(x) = 0
• Call these potential solutions x*
• Identifying x values where the slope of the
objective function (profit) is zero, which occurs at
maximums, minimums, and benches/saddles
Calculus of Optimization
• Second Order Condition (SOC)
• Evaluate p’’(x) at each x* identified
• Condition for a maximum is p’’(x*) < 0
• Condition for a minimum is p’’(x*) > 0
• p’’(x*) is curvature of profit at x*
• Positive curvature (p’’(x*) > 0) is convex (minimum)
• Negative curvature (p’’(x*) < 0) is concave (maximum)
• FOC: finds x values where profit slope is zero, candidate
x’s for profit maximum
• SOC: checks curvature at each candidate x and profit
maximum is curved down (negative)
Simple Example
• In general, p = pf(x) – rx – K
• Suppose your production function is
y = 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 that maximizes p, solve the FOC for x and check
the SOC at this x
•
Simple Example
• p = 10(30 + 5x – 0.4x2) – 2x – 18
• FOC: p’(x) = 10(5 – 0.8x) – 2 = 0
10(5 – 0.8x) = 2
pMP = r
5 – 0.8x = 2/10
MP = r/p
When you solve the FOC, you set VMP = r and/or MP = r/p
and then solve for x
Solution:
5 – 0.8x = 0.2
5 – 0.2 = 0.8x
x = 4.8/0.8 = 6
Simple Example
• First derivative: p’(x) = 10(5 – 0.8x) – 2 = 48 – 8x
• SOC: p’’(x) = –8 < 0 for all x
• Objective function p(x) is globally concave
• Summary
• FOC gave x = 6 as x at which profit slope was zero
• SOC: found that profit was globally concave
• Implication: x = 6 is the single x that maximizes p(x)
• Biologically accurate production functions usually not this
simple, so derivatives get more complex
Hyperbolic Production Function Example
y 
x

1 x / 
FO C : p '( x ) 
 
2
(   x)

0.5
 
 (   x)   ( x)
r
(   x)
 


0.5
  (r / p )

  x
 rx
r
p
 (   x)
2
r/ p
  x
0.5
2
 x
2
p

p ( x)  p

 x
0.5
(r / p )
x
2

 x

0.5

1  
x


0.5
  (r / p )

2
2

2 p 
 1  S O C : p ''( x )  
0
3
(   x)

Multiple Input Production
• Most agricultural production processes have more than
one input, e.g., N, P, K fertilizer, plus herbicides,
insecticides, tillage, water, etc.
• How do you decide how much of each input to use when
you have more than one input?
• Derive the Equal Margin Principle and show its use to
answer this question
Equal Margin Principle
• Given production function y = f(x1,x2), find (x1,x2) to
maximize p(x1,x2) = pf(x1,x2) – r1x1 – r2x2 – K
• FOC’s: dp/dx1 = 0 and dp/dx2 = 0 and solve for pair
(x1,x2)
• dp/dx = pf1(x1,x2) – r1 = 0
• dp/dy = pf2(x1,x2) – r2 = 0
• Just pMP1 = r1 and pMP2 = r2
• Just MP1 = r1/p and MP2 = r2/p
• These still hold, but we also have more
Equal Margin Principle
• Profit Maximization again implies
• pf1(x1,x2) = r1 and pf2(x1,x2) = r2
• Note that pf1(x1,x2) = r1 depends on x2 and
pf2(x1,x2) = r2 depends on x1
• Have two equations and both must be satisfied, so
rearrange (make ratio)
p f 1 ( x 1 ,x 2 )
p f 2 ( x 1 ,x 2 )
=
r1
pf 1 ( x 1 ,x 2 )
r2
r1
=
pf 2 ( x 1 ,x 2 )
r2
Equal Margin Principle
• Equal Margin Principle is expressed mathematically in two
ways
1) MP1/r1 = MP2/r2
2) MP1/MP2 = r1/r2
• Ratio of MPi/ri must be equal for all inputs
• Ratio of MP’s must equal input price ratio
Intuition: Corn Example
• MPi is bu of corn per lb of N fertilizer (bu/lb)
• ri is $ per lb of N fertilizer ($/lb)
• MPi/ri is bu corn per $ spent on N fertilizer
(bu/lb)/ ($/lb) = bu/$
• MPi/ri is how many bushels of corn you get for the last dollar
spent on N fertilizer
• MP1/r1 = MP2/r2 means use inputs so that the last dollar
spent on each input gives the same extra output
Intuition: Corn Example
• MP1 is bu of corn from last lb of N fert. (bu/lb N)
• MP2 is bu of corn from last lb of P fert. (bu/lb P)
• MP1/MP2 = (lbs P/lbs N) is how much P need if
cut N by 1 lb and want to keep output constant
• Ratio of marginal products is the substitution
rate between N and P in the production
process when hold output constant
Intuition: Corn Example
• r1 is $ per lb of N fertilizer ($/lb N)
• r2 is $ per lb of P fertilizer ($/lb P)
• r1/r2 is ($/lb N)/($/lb P) = lbs P/lbs N, or the
substitution rate between N and P in the
market
• MP1/MP2 = r1/r2 means use inputs so that the
substitution rate between inputs in the
production process is the same as the
substitution rate between inputs in the market
Marginal Rate of Technical Substitution
• The ratio of marginal products (MP1/MP2) is the
substitution rate between inputs in the production
process when holding output fixed (i.e., the slope
of the isoquant)
• MP1/MP2 is called the Marginal Rate of Technical
Substitution (MRTS): the input substitution rate at
the margin for the production technology (the
slope of the isoquant)
• If cut x1 by one unit, how much must you increase
x2 to keep output the same
• Optimality condition MP1/MP2 = r1/r2 means set
substitution rates equal
Equal Margin Principle Intuition
• MP1/r1 = MP2/r2 means use inputs so the last dollar spent
on each input gives the same extra output at the margin
• Compare to p x MP = r or VMP = r
• MP1/MP2 = r1/r2 means use inputs so the substitution rate
at the margin between inputs is the same in the
production process as in the market place
• Compare to MP = r/p
Equal Margin Principle
Graphical Analysis via Isoquants
• Remember: Isoquant (“equal-quantity”) plot or function
representing all combinations of two inputs producing the
same output quantity
• Intuition: Isoquants are the two dimensional “contour lines” of the
three dimensional production “hill”
• Optimality Condition: MP1/MP2 = r1/r2
• Remember: -MP1/MP2 is called the Marginal Rate of
Technical Substitution (MRTS) = slope of the isoquant
• Optimality condition for two inputs: find the input
combination (x1,x2) where slope of isoquant = price ratio
Graphics
x2
Isoquant y = y0
-r1/r2 = -MP1/MP2
x2*
x1*
x1
Multiple Input Production with Calculus
• Use calculus with production function to find the optimal
input combination
• General problem we’ve seen: Find (x1,x2) to maximize
p(x1,x2) = pf(x1,x2) – r1x1 – r2x2 – K
• Will get FOC’s, one for each choice variable, and SOC’s
are more complicated
• Assume quadratic production function
y = 7 + 9x1 + 8x2 – 2x12 – x22 – 2x1x2
price p = 10
price r1 = 2
price r2 = 3
Multiple Input Production with Calculus
• Max 10(7 + 9x1 + 8x2 – 2x12 – x22 – 2x1x2) – 2x1 – 3x2 – 8
• FOC1: 10(9 – 4x1 – 2x2) – 2 = 0
• FOC2: 10(8 – 2x2 – 2x1) – 3 = 0
• Find the (x1,x2) pair that satisfies these two equations
• Same as finding where the two lines from the FOC’s
intersect
Solve FOC1 and FOC2 for (x1,x2)
1) Solve FOC1 for x1
88 – 40x1 – 20x2 = 0
40x1 = 88 – 20x2
x1 = (88 – 20x2)/40 = 2.2 – 0.5x2
2) Substitute this x1 into FOC2 and solve for x2
77 – 20x2 – 20x1 = 0
77 – 20x2 – 20(2.2 – 0.5y) = 0
77 – 20x2 – 44 + 10x2 = 0
33 – 10x2 = 0, or 10x2 = 33, or x2 = 3.3
3) Calculate x1:
x1 = 2.2 – 0.5x2 = 2.2 – 0.5(3.3) = 0.55
Second Order Conditions
• SOC’s are more complicated with multiple inputs, must
look at curvature in each direction, plus the “cross”
direction (to ensure do not have a saddle point).
1) Own second derivatives of production function must be
negative for a maximum: f11 < 0, f22 < 0
(both > 0 for a minimum)
2) Another condition: f11f22 – (f12)2 > 0
Main point: if concave production function, SOC satisfied
Typically express SOC as condition on the determinants of
the Hessian matrix of 2nd derivatives: determinant of the
principal minors alternate in sign, with odd < 0, even > 0
  2p
 2
 x1

H 
  2p

  x 2  x1
2
 p 

 x1  x 2
p
   11

2

 p
 p 21

2
 x2 
H 1  p 11  0
H 2  p 11p 22  (p 12 )  0
2
S ince p ij  pf ij ( )
H 1  pf 11  0
H 2  p ( f 11 f 22  ( f 12 ) )  0
2
2
p 12 

p 22 
Second Order Conditions
• FOC1: 10(9 – 4x1 – 2x2) – 2 = 0
• FOC2: 10(8 – 2x2 – 2x1) – 3 = 0
• Second Derivatives
1) p11 = pf11 = 10(– 4) = – 40
2) p22 = pf11 = 10(– 2) = – 20
3) p12 = pf11 = 10(– 2) = – 20
SOC’s
1) pxx = – 40 < 0 and pyy = – 20 < 0
2) pxxpyy – (pxy)2 > 0
(– 40)(– 20) – (– 20)2 = 800 – 400 = 400 > 0
Solution x = 0.55 and y = 3.3 is a maximum
• Biologically accurate production functions usually not this
simple, so calculus gets more complicated