AAE 320 Paul Mitchell pdmitchellwiscedu 6083201162 Learning Goals Economics of identifying optimal input use when have two inputs to choose at the same time Multiple Input Production Function ID: 1028435
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1. Multiple Input Production Economics for Farm ManagementAAE 320Paul Mitchellpdmitchell@wisc.edu 608-320-1162
2. Learning GoalsEconomics of identifying optimal input use when have two inputs to choose at the same timeMultiple Input Production FunctionBoth in tabular form and using calculusHow to use the new optimality conditionKey concept: Tradeoff or Substitution between inputs
3. Multiple Input ProductionMost agricultural production processes have more than one input, e.g., capital, labor, land, machinery, fertilizer, herbicides, insecticides, tillage, water, etc.How do you decide how much of each input to use when you are choosing more than one?Will derive the Equal Margin Principle and show its use to answer this questionWill derive it using calculus, so you see where it comes fromApplies whether you use a function or not, so can apply Equal Margin Principle to the tabular form of the multiple input production scheduleMultivariate calculus requires use of Partial Derivatives, so let’s review
4. Partial DerivativesDerivative of function that has more than one variableWhat’s the derivative of Q = f(x,y)? Depends on which variable you are talking about.Remember a derivative is the slopeDerivative of Q = f(x,y) with respect to x is the slope of the function in the x directionDerivative of Q = f(x,y) with respect to y is the slope of the function in the y direction
5. Partial DerivativesThink of a hill; its elevation Q is a function of the location in latitude (x) and in longitude (y): Q = f(x,y)At any spot on the hill, defined by a latitude-longitude pair (x,y), the hill will have a slope in the x direction and in the y directionSlope in x direction: dQ/dx = fx(x,y)Slope in y direction: dQ/dy = fy(x,y)XYQ
6. Partial DerivativesNotation: if have Q = f(x,y)First Partial DerivativesdQ/dx = fx(x,y) and dQ/dy = fy(x,y)Second (Own) Partial Derivatived2Q/dx2 = fxx(x,y) and d2Q/dy2 = fyy(x,y)Second Cross Partial Derivatived2Q/dxdy = fxy(x,y)
7. Partial DerivativesPartial derivatives are the same as regular derivatives, just treat the other variables as constantsQ = f(x,y) = 2 + 3x + 6y – 2x2 – 3y2 – 5xyWhen taking the derivative with respect to x, treat y as a constant and vice versaQ = f(x,y) = 2 + 3x + 6y – 2x2 – 3y2 – 5xy fx(x,y) = 3 – 4x – 5yQ = f(x,y) = 2 + 3x + 6y – 2x2 – 3y2 – 5xy fy(x,y) = 6 – 6y – 5x
8. Think Break #6Give the 1st and 2nd derivatives [fx(x,y), fy(x,y), fxx(x,y), fyy(x,y), fxy(x,y)] of each function:f(x,y) = 7 + 5x + 2y – 5x2 – 4y2 – 11xyf(x,y) = – 5 – 2x + y – x2 – 3y2 + 2xy
9. Equal Margin PrincipleGiven production function Q = f(x,y), find (x,y) to maximize p(x,y) = pf(x,y) – rxx – ryy – KFOC’s: dp/dx = 0 and dp/dy = 0 and solve for pair (x,y)dp/dx = pfx(x,y) – rx = 0dp/dy = pfy(x,y) – ry = 0Just p x MPx = rx and p x MPy = ryJust MPx = rx/p and MPy = ry/pThese still hold, but we have moreSame as before
10. Equal Margin PrincipleProfit Maximization again impliesp x MPx = rx and p x MPy = ryNote p x MPx = rx depends on y and p x MPy = ry depends on xThese are two equations and both must be satisfied, so rearrange by making the ratio
11. Equal Margin PrincipleEqual Margin Principle: expressed mathematically in two ways1) MPx/rx = MPy/ryRatio of MPi/ri must be equal for all inputs2) MPx/MPy = rx/ryRatio of MP’s must equal input price ratio
12. Intuition for MPx/rx = MPy/ry Corn ExampleMPi is bu of corn from the last lb of N fertilizer (bu/lb)ri is $ per lb of N fertilizer ($/lb)Units for MPi/ri (bu/lb) / ($/lb) = bu/$ MPi/ri is bu of corn from the last $ spent on N fertilizerMPi/ri is how many bushels of corn you get from the last dollar spent on N fertilizerMPx/rx = MPy/ry means use each input so that the last dollar spent on each input gives the same extra output
13. Intuition for MPx/MPy = rx/ry Corn ExampleMPx is bu of corn from last lb of N fertilizer (bu/lb N)MPy is bu of corn from last 1k seeds planted (bu/1k seeds)MPx/MPy = (1k seeds/lbs N) is how many seeds/ac need to increase if cut N by 1 lb/ac and want to keep yield the sameRatio of marginal products is the substitution rate between Nitrogen and Seeds in the production processHave not talked about substitution between inputs, but the Ratio of the marginal products is the slope of the tradeoff curve between inputs
14. Intuition for MPx/MPy = rx/ry Corn Examplerx is $ per lb of N fertilizer ($/lb N)ry is $ per 1k seeds ($/1k seeds)rx/ry is ($/lb N)/($/1k seeds) = 1k seeds/lbs N, or the substitution rate between Nitrogen and Seeds in the marketMPx/MPy = rx/ry 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
15. Marginal Rate of Technical SubstitutionThe ratio of marginal products (MPx/MPy) is the substitution rate between inputs in the production processSlope of the tradeoff curveMPx/MPy is called the Marginal Rate of Technical Substitution (MRTS): the input substitution rate at the margin for the production technologyMRTS: If you cut X by one unit, how much must you increase Y to keep output the sameOptimality condition MPx/MPy = rx/ry means set substitution rates equal
16. Equal Margin Principle IntuitionMPx/rx = MPy/ry means use inputs so the last dollar spent on each input gives the same extra output at the marginCompare to p x MP = r or VMP = rMPx/MPy = rx/ry means use inputs so the substitution rate at the margin between inputs is the same in the production process as in the marketplaceCompare to MP = r/p
17. Equal Margin PrincipleGraphical Analysis via IsoquantsIsoquant (“equal-quantity”) plot or function representing all combinations of two inputs producing the same output quantityIntuition: Isoquants are the two dimensional “contour lines” of the three dimensional production “hill”First look at in Theory, then a Table
18. Isoquants in TheoryInput XInput YOutput Q = Q ~
19. Capital ($1,000)Labor (persons/year)12345678910250.5123814.5222527.526501.53814.521.527.5293029.528753814.52227.530313130.529.51005122227.5303131.531.531.531125814.527.52930.531.532323232150814.527.5303131.53232.532.532.517582227.5303131.53232.533332006.5222527.530313232.53333.5Swine Feeding Operation: Output (hogs sold/year)
20. Capital ($1,000)Labor (persons/year)12345678910250.5123814.5222527.526501.53814.521.527.5293029.528753814.52227.530313130.529.51005122227.5303131.531.531.531125814.527.52930.531.532323232150814.527.5303131.53232.532.532.517582227.5303131.53232.533332006.5222527.530313232.53333.5Swine feeding operation: Output (hogs sold/year)
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22. Isoquants and MPx/MPy Input XInput YOutput Q = Q~DXDYIsoquant Slope = DY/DX = Substitution rate between X and Y at the margin If reduce X by the amount DX, then must increase Y by the amount DY to keep output fixed
23. DYDXIsoquants and MPx/MPy Input XInput YOutput Q = Q~Isoquant slope = dY/dX dY/dX = – (dQ/dX)/(dQ/dY) dY/dX = – MPx/MPy = – Ratio of MP’s Isoquant Slope = – MRTSNeed minus sign since marginal products are positive and slope is negative
24. Isoquants and MPx/MPy = rx/ryIsoquant slope (– MPx/MPy) is (minus) the substitution rate between X and Y at the marginThus define minus the isoquant slope as the “Marginal Rate of Technical Substitution” (MRTS)Price ratio rx/ry is the substitution rate between X and Y (at the margin) in the marketplaceEconomically optimal use of inputs sets these substitution rates equal, or MPx/MPy = rx/ry
25. Isoquants and MPx/MPy = rx/ryInput XInput YOutput Q = Q~Slope = – MPx/MPy = – rx/ry Find the point of tangency between the isoquant and the input price ratio Kind of like the MP = r/p: tangency between the production function and input-output price ratio line
26. Soybean Meal (lbs)Corn (lbs)10376.815356.320339.025326.030315.035307.540300.645294.650289.255284.460280.265276.470272.9Soybean meal and corn needed for 125 lb feeder pigs to gain 125 lbs Which feed ration do you use?
27. Soyb MealCornMRTS10376.815356.34.1020339.03.4625326.02.6030315.02.2035307.51.5040300.61.3845294.61.2050289.21.0855284.40.9660280.20.8465276.40.7670272.90.70 Can’t use MRTS = – MPx/MPy since DQ = 0 on an isoquantUse MRTS = – DY/DXDX = (SoyM2 – SoyM1) DY = (Corn2 – Corn1) MRTS = As increase Soybean Meal, how much can you decrease Corn and keep output constant?= – (294.6 – 300.6)/(45 – 40)= – (284.4 – 289.2)/(55 – 50)Interpretation: If increase soybean meal 1 lb, decrease corn by 1.20 or 0.96 lbs and keep same gain on hogs
28. Soy Meal (lbs)Corn(lbs)MRTSP ratio10376.82.2015356.34.102.2020339.03.462.2025326.02.602.2030315.02.202.2035307.51.502.2040300.61.382.2045294.61.202.2050289.21.082.2055284.40.962.2060280.20.842.2065276.40.762.2070272.90.702.20Economically Optimal input use is where MRTS = input price ratio, or – DY/DX = rx/ry(Same as MPx/MPy = rx/ry)Soybean Meal Price $176/ton = $0.088/lbCorn Price $2.24/bu = $0.04/lbRatio: 0.088/0.04 = 2.20 lbs corn/lbs soyKeep straight which is X and which is Y!!!
29. IntuitionWhat does – DY/DX = rx/ry mean?Ratio – DY/DX is how much less Y you need if you increase X, or the MRTSCross multiply to get – DYry = DXrx– DYry = cost savings from decreasing YDXrx = cost increase from increasing XKeep sliding down the isoquant (decreasing Y and increasing with output staying the same) as long as the cost saving exceeds the cost increaseLike a partial budget analysis of the cost changes (revenue doesn’t change, as output doesn’t change)
30. Main Point on MRTSWhen you have information on inputs needed to generate the same output, you can identify the optimal input combinationTheory: (MRTS) MPx/MPy = rx/ryPractice: (isoquant slope) – DY/DX = rx/ryNote how the input in the numerator and denominator changes between theory and practiceThe numbers in a “table” like I have created actually have underneath them a 2-input production function
31. Think Break #7Grain(lbs)Hay(lbs)MRTSpriceratio825135090011302.9397593510507702.2011256251.93120052512754451.07The table is rations of grain and hay that put 300 lbs of gain on 900 lb steers.1) Fill in the missing MRTS2) If grain is $0.04/lb and hay is $0.03/lb, what is the economically optimal feed ration?
32. SummaryCan identify economically optimal input combination using tabular data and pricesRequires input data on the isoquant, i.e., different feed rations that generate the same amount of gainAgain: we will use calculus to fill in gaps in the tabular form of isoquants
33. Types of SubstitutionPerfect substitutes: soybean meal & canola meal, corn & sorghum, wheat and barleyImperfect substitutes: corn & soybean mealNon-substitutes/perfect complements: tractors and drivers, wire and fence posts
34. Types of Input SubstitutionPerfect substitutes: soybean meal & canola meal, corn & sorghum, wheat and barleyImperfect substitutes: corn & soybean mealNon-substitutes/perfect complements: tractors and drivers, wire and fence posts
35. Perfect SubstitutesMRTS (slope of isoquant) is constantExamples: canola meal and soybean meal, or corn and sorghum in a feed rationConstant conversion between two inputs2 lbs of canola meal = 1 lb of soybean meal: – DCnla/DSoyb = 21.2 bu of sorghum = 1 bu corn – DSrgm/DCorn = 1.2
36. Economics of Perfect SubstitutesMRTS = rx/ry still applies MRTS = k: – DY/DX = – DCnla/DSoyb = 2If rx/ry < k, use only input xsoybean meal < twice canola meal pricesoyb meal $200/cwt, cnla meal $150/cwtIf rx/ry > k, use only input ysoybean meal > twice canola meal pricesoyb meal $200/cwt, cnla meal $90/cwtIf rx/ry = k, use any x-y combination
37. Canola MealSoybean Meal21rx/ry < 2Use only Soybean Mealrx/ry > 2Use only Canola MealPerfect Substitutes: Graphicsrx/ry = 2Use any combination of Canola or Soybean Meal
38. Economics of Imperfect SubstitutesInput XInput YThis the case we already did!
39. Perfect Complements, or Non-SubstitutesNo substitution is possibleThe two inputs must be used togetherWithout the other input, neither input is productive, they must be used togetherTractors and drivers, fence posts and wire, chemical reactions, digger and shovelInputs used in fixed proportions 1 driver for 1 tractor
40. Economics of Perfect ComplementsMRTS is undefined, the price ratio rx/ry does not identify the optimal combinationLeontief Production FunctionQ = min(aX,bY):Marginal products of inputs = 0Either an output quota or a cost budget, with the fixed input proportions defining how much of the inputs to use
41. TractorsDriversrx/ry < 2rx/ry > 2Perfect Complements: GraphicsPrice ratio does not matter
42. Multiple Input Production with CalculusUse calculus with production function to find the optimal input combinationGeneral problem we’ve seen: Find (x,y) to maximize p(x,y) = pf(x,y) – rxx – ryy – KWill get FOC’s, one for each choice variable, and SOC’s are more complicatedEconomic Problem: How much of the input x and input y do I use to maximize net returns?
43. Multiple Input Production with CalculusIf price of output p = 10, price of x (rx) = 2 and price of y (ry) = 3How much x and y do I use to maximize net returns?Set up: Max 10(7 + 9x + 8y – 2x2 – y2 – 2xy) – 2x – 3y – 8FOC1: 10(9 – 4x – 2y) – 2 = 0FOC2: 10(8 – 2y – 2x) – 3 = 0Find the (x,y) pair that satisfies these two equationsSame as finding where the two lines from the FOC’s intersect
44. FOC1: 10(9 – 4x – 2y) – 2 = 0FOC2: 10(8 – 2y – 2x) – 3 = 01) Solve FOC1 for x 88 – 40x – 20y = 0 40x = 88 – 20y x = (88 – 20y)/40 = 2.2 – 0.5y2) Substitute this x into FOC2 and solve for y 77 – 20y – 20x = 0 77 – 20y – 20(2.2 – 0.5y) = 0 77 – 20y – 44 + 10y = 0 33 – 10y = 0, or 10y = 33, or y = 3.33) Calculate x: x = 2.2 – 0.5y = 2.2 – 0.5(3.3) = 0.55Solve these two equations for the two unknowns (x, y) Substitution Method Solve 1st equation for x as function of ySubstitute this equation into the 2nd equation for x and solve for ySubstitute this value for y into the 1st equation to calculate xSubstitute x and y into f(x,y) to get Q and calculate costs and net returns
45. Second Order ConditionsSOC’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 must be negative for a maximum: pxx < 0, pyy < 0 (both > 0 for a minimum)2) Another condition: pxxpyy – (pxy)2 > 0
46. FOC1: 10(9 – 4x – 2y) – 2 = 0FOC2: 10(8 – 2y – 2x) – 3 = 0Second Derivatives 1) pxx: 10(– 4) = – 40 2) pyy: 10(– 2) = – 20 3) pxy: 10(– 2) = – 20SOC’s 1) pxx = – 40 < 0 and pyy = – 20 < 0 2) pxxpyy – (pxy)2 > 0 (– 40)(– 20) – (– 20)2 = 800 – 400 > 0 Solution x = 0.55 and y = 3.3 is a maximum
47. Milk ProductionM = – 25.9 + 2.56G + 1.05H – 0.00505G2 – 0.00109H2 – 0.00352GHM = milk production (lbs/week)G = grain (lbs/week)H = hay (lbs/week)Milk $14/cwt, Grain $3/cwt, Hay $1.50/cwtWhat are the profit maximizing inputs? (Source: Heady and Bhide 1983)
48. Milk Profit Functionp(G,H) = 0.14(– 25.9 + 2.56G + 1.05H – 0.00505G2 – 0.00109H2 – 0.00352GH) – 0.03G – 0.015HFOC’s pG = 0.14(2.56 – 0.0101G – 0.00352H) – 0.03 = 0 pH = 0.14(1.05 – 0.00218H – 0.00352G) – 0.015 = 0Alternative: jump to optimality conditions MPG = rG/p and MPH = rH/p MPG = 2.56 – 0.0101G – 0.00352H = rG/p = 0.214 MPH = 1.05 – 0.00218H – 0.00352G = rH/p = 0.107
49. Solve FOC1 for H: 2.56 – 0.0101G – 0.00352H = 0.214 2.346 – 0.0101G = 0.00352H H = (2.346 – 0.0101G)/0.00352 H = 666 – 2.87GSubstitute this H into FOC2 and solve for G: 1.05 – 0.00218H – 0.00352G = 0.107 0.943 – 0.00218(666 – 2.87G) – 0.00352G = 0 0.943 – 1.45 + 0.00626G – 0.00352G = 0 0.00274G – 0.507 = 0 G = 0.507/0.00274 = 185 lbs/week
50. Substitute this G into the equation for H H = 666 – 2.87G H = 666 – 2.87(185) H = 666 – 531 = 135 lbs/weekCheck SOCs pGG = – (0.14)0.0101 = – 0.001414 < 0 pHH = – (0.14)0.00218 = – 0.000305 < 0 pGH = – (0.14)0.00352 = – 0.000493 pGGpHH – (pGH)2 = 1.88 x 10-7 > 0SOCs satisfied for a maximum
51. Calculate Milk Output using this feed rationM = – 25.9 + 2.56G + 1.05H – 0.00505G2 – 0.00109H2 – 0.00352GHM = – 25.9 + 2.56(185) + 1.05(135) – 0.00505(185)2 – 0.00109(135)2 – 0.00352(185)(135) = 309 lbs/week 309 x 52 = 16,068 lbs/year (It’s 1983!)Calculate Profit (Net Returns) using this feed rationProfit = 0.14(309) – 0.03(185) – 0.015(135) = $41.20/week
52. Think Break #8Repeat the previous problem, but now use a milk price of Milk $10/cwtM = – 25.9 + 2.56G + 1.05H – 0.00505G2 – 0.00109H2 – 0.00352GHPrices: Grain $3/cwt, Hay $1.50/cwtWhat are the profit maximizing inputs?
53. Think Break #8 DiscussionPrice decreased 29%, from $14/cwt to $10/cwtMilk output decreased 0.9%, from 309 to 306 lbs/wkOptimal grain use decreased 1.4%, from 185 to 182.5 lbs/wkOptimal hay use decreased 12.4%, from 135 to 118.2 lbs/wkNet Returns decreased 43%, from $41.20 to $23.35/wkEconomically optimal response to a large price decrease is a small production decrease and a large profit decrease
54. Think Break #8 DiscussionThe standard agricultural problem: output is non-responsive (inelastic) to price changes, but net returns is responsiveQPSDDPDQ
55. SummaryOptimality Conditions for multiple inputs have several expressions, all imply Equal Margin PrincipleMRTS, isoquant, and optimal input use are all connected, know the graphicsKnow the different types of input substitutionBe able to identify optimal input useTabular formUsing calculus