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 (  ) 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 = 0pMP – r = 0 Rearrange:pf’(x) = rpMP = 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

x y MP 1)Output max is where MP = 0, x = x ymax 2)Profit Max is where MP = r/p, x = x opt r/p x x opt x ymax

Key Points Profit maximizing x is less than output maximizing x, or x opt < x ymax 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) = x – 0.4x 2 Suppose output price is 10, input price is 2, and fixed cost is 18, then  = 10(30 + 5x – 0.4x 2 ) – 2x – 18 To find x that maximizes , solve the FOC for x and check the SOC at this x

Simple Example  = 10(30 + 5x – 0.4x 2 ) – 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 = 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

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(x 1,x 2 ), find (x 1,x 2 ) to maximize  x 1,x 2  = pf(x 1,x 2 ) – r 1 x 1 – r 2 x 2 – K FOC’s: d  /dx 1 = 0 and d  /dx 2 = 0 and solve for pair (x 1,x 2 ) d  /dx = pf 1 (x 1,x 2 ) – r 1 = 0 d  /dy = pf 2 (x 1,x 2 ) – r 2 = 0 Just pMP 1 = r 1 and pMP 2 = r 2 Just MP 1 = r 1 /p and MP 2 = r 2 /p These still hold, but we also have more

Equal Margin Principle Profit Maximization again implies pf 1 (x 1,x 2 ) = r 1 and pf 2 (x 1,x 2 ) = r 2 Note that pf 1 (x 1,x 2 ) = r 1 depends on x 2 and pf 2 (x 1,x 2 ) = r 2 depends on x 1 Have two equations and both must be satisfied, so rearrange (make ratio)

Equal Margin Principle Equal Margin Principle is expressed mathematically in two ways 1) MP 1 /r 1 = MP 2 /r 2 2) MP 1 /MP 2 = r 1 /r 2 Ratio of MP i /r i must be equal for all inputs Ratio of MP’s must equal input price ratio

Intuition: Corn Example MP i is bu of corn per lb of N fertilizer (bu/lb) r i is $ per lb of N fertilizer ($/lb) MP i /r i is bu corn per $ spent on N fertilizer (bu/lb)/ ($/lb) = bu/$ MP i /r i is how many bushels of corn you get for the last dollar spent on N fertilizer MP 1 /r 1 = MP 2 /r 2 means use inputs so that the last dollar spent on each input gives the same extra output

Intuition: Corn Example MP 1 is bu of corn from last lb of N fert. (bu/lb N) MP 2 is bu of corn from last lb of P fert. (bu/lb P) MP 1 /MP 2 = (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 r 1 is $ per lb of N fertilizer ($/lb N) r 2 is $ per lb of P fertilizer ($/lb P) r 1 /r 2 is ($/lb N)/($/lb P) = lbs P/lbs N, or the substitution rate between N and P in the market MP 1 /MP 2 = r 1 /r 2 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 (MP 1 /MP 2 ) is the substitution rate between inputs in the production process when holding output fixed (i.e., the slope of the isoquant) MP 1 /MP 2 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 x 1 by one unit, how much must you increase x 2 to keep output the same Optimality condition MP 1 /MP 2 = r 1 /r 2 means set substitution rates equal

Equal Margin Principle Intuition MP 1 /r 1 = MP 2 /r 2 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 MP 1 /MP 2 = r 1 /r 2 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: MP 1 /MP 2 = r 1 /r 2 Remember: -MP 1 /MP 2 is called the Marginal Rate of Technical Substitution (MRTS) = slope of the isoquant Optimality condition for two inputs: find the input combination (x 1,x 2 ) where slope of isoquant = price ratio

Graphics x1x1 x2x2 Isoquant y = y 0 -r 1 /r 2 x1*x1* x2*x2* = -MP 1 /MP 2

Multiple Input Production with Calculus Use calculus with production function to find the optimal input combination General problem we’ve seen: Find (x 1,x 2 ) to maximize  (x 1,x 2 ) = pf(x 1,x 2 ) – r 1 x 1 – r 2 x 2 – K Will get FOC’s, one for each choice variable, and SOC’s are more complicated Assume quadratic production function y = 7 + 9x 1 + 8x 2 – 2x 1 2 – x 2 2 – 2x 1 x 2 price p = 10 price r 1 = 2 price r 2 = 3

Multiple Input Production with Calculus Max 10(7 + 9x 1 + 8x 2 – 2x 1 2 – x 2 2 – 2x 1 x 2 ) – 2x 1 – 3x 2 – 8 FOC1: 10(9 – 4x 1 – 2x 2 ) – 2 = 0 FOC2: 10(8 – 2x 2 – 2x 1 ) – 3 = 0 Find the (x 1,x 2 ) pair that satisfies these two equations Same as finding where the two lines from the FOC’s intersect

Solve FOC1 and FOC2 for (x 1,x 2 ) 1) Solve FOC1 for x 1 88 – 40x 1 – 20x 2 = 0 40x 1 = 88 – 20x 2 x 1 = (88 – 20x 2 )/40 = 2.2 – 0.5x 2 2) Substitute this x 1 into FOC2 and solve for x 2 77 – 20x 2 – 20x 1 = 0 77 – 20x 2 – 20(2.2 – 0.5y) = 0 77 – 20x 2 – x 2 = 0 33 – 10x 2 = 0, or 10x 2 = 33, or x 2 = 3.3 3) Calculate x 1 : x 1 = 2.2 – 0.5x 2 = 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: f 11 < 0, f 22 < 0 (both > 0 for a minimum) 2) Another condition: f 11 f 22 – (f 12 ) 2 > 0 Main point: if concave production function, SOC satisfied Typically express SOC as condition on the determinants of the Hessian matrix of 2 nd derivatives: determinant of the principal minors alternate in sign, with odd 0

Second Order Conditions FOC1: 10(9 – 4x 1 – 2x 2 ) – 2 = 0 FOC2: 10(8 – 2x 2 – 2x 1 ) – 3 = 0 Second Derivatives 1)  11 = pf 11 = 10(– 4) = – 40 2)  22 = pf 11 = 10(– 2) = – 20 3)  12 = pf 11 = 10(– 2) = – 20 SOC’s 1)  xx = – 40 < 0 and  yy = – 20 < 0 2)  xx  yy – (  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