ECONOMICS OF OPTIMAL INPUT USEAAE 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 ()• 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: (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• (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: (x) = p(f(x))f(x) – r(x)x

• The model (x) = pf(x) – rx assumes perfect competition: farmer does not affect output or input prices

Economics of Input Use• Find x to Maximize (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: ’(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: ’’(x) < 0

Economics of Input Use: FOC• Find x to Maximize (x) = pf(x) – rx• FOC: 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

Economics of Input Use: SOC• Find x to Maximize (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

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

x

y

MP

1)Output max is where MP = 0, x = xymax

2)Profit Max is where MP = r/p, x = xopt

r/p

x xopt xymax

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 (x)• First Order Condition (FOC)• Set ’(x) = 0 and solve for x• May be more than one x where ’(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 ’’(x) at each x* identified• Condition for a maximum is ’’(x*) < 0• Condition for a minimum is ’’(x*) > 0• ’’(x*) is curvature of profit at x*

• Positive curvature (’’(x*) > 0) is convex (minimum)• Negative curvature (’’(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, = 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

= 10(30 + 5x – 0.4x2) – 2x – 18• To find x that maximizes , solve the FOC for x and check

the SOC at this x

Simple Example• = 10(30 + 5x – 0.4x2) – 2x – 18• FOC: ’(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: ’(x) = 10(5 – 0.8x) – 2 = 48 – 8x• SOC: ’’(x) = –8 < 0 for all x• Objective function (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 (x)

• Biologically accurate production functions usually not this simple, so derivatives get more complex

Hyperbolic Production Function Example

2

2 22

2

0.5 0.5

0.5 0.5

0.5 2 2

0.5

( )1 /

( ) ( )FOC: '( )

( )

( )( ) /

1

( / ) ( / )

21 SOC: ''( )

( / )

x x xy x p rx

x x x

x x rx

x p

rx

x p r p

x xr p r p

px x

r p

30

( )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 x1,x2 = pf(x1,x2) – r1x1 – r2x2 – K

• FOC’s: d/dx1 = 0 and d/dx2 = 0 and solve for pair (x1,x2)

• d/dx = pf1(x1,x2) – r1 = 0• d/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)

1 1 2 1

2 1 2 2

pf (x ,x ) r=

pf (x ,x ) r1 1 2 2 1 2

1 2

pf (x ,x ) pf (x ,x )=

r r

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 PrincipleGraphical 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

x1

x2

Isoquant y = y0

-r1/r2

x1*

x2*

= -MP1/MP2

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 (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 – x2

2 – 2x1x2

price p = 10

price r1 = 2

price r2 = 3

Multiple Input Production with Calculus

• Max 10(7 + 9x1 + 8x2 – 2x12 – x2

2 – 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 = 040x1 = 88 – 20x2

x1 = (88 – 20x2)/40 = 2.2 – 0.5x2

2) Substitute this x1 into FOC2 and solve for x2

77 – 20x2 – 20x1 = 077 – 20x2 – 20(2.2 – 0.5y) = 077 – 20x2 – 44 + 10x2 = 033 – 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

2 2

21 1 2 11 12

2 221 22

22 1 2

1 11

22 11 22 12

0

( ) 0

x x x

x x x

H

H

H

1 11

2 22 11 22 12

Since ( )

0

( ( ) ) 0

ij ijpf

pf

p f f f

H

H

Second Order Conditions• FOC1: 10(9 – 4x1 – 2x2) – 2 = 0

• FOC2: 10(8 – 2x2 – 2x1) – 3 = 0

• Second Derivatives

1) 11 = pf11 = 10(– 4) = – 40

2) 22 = pf11 = 10(– 2) = – 20

3) 12 = pf11 = 10(– 2) = – 20

SOC’s

1) xx = – 40 < 0 and yy = – 20 < 0

2) xxyy – (xy)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