Unit 7. Analyses of LR Production and Costs as Functions of Output (Ch. 5, 6, 8)

LR  Max  1. Produce Q where MR = MC 2. Minimize cost of producing Q  optimal input combination

Isoquant The combinations of inputs (K, L) that yield the producer the same level of output. The shape of an isoquant reflects the ease with which a producer can substitute among inputs while maintaining the same level of output.

Typical Isoquant

SR Production in LR Diagram

MRTS and MP MRTS = marginal rate of technical substitution = the rate at which a firm must substitute one input for another in order to keep production at a given level = - slope of isoquant = = the rate at which capital can be exchanged for 1 more (or less) unit of labor MPK = the marginal product of K = MPL = the marginal product of L = Q = MPK K + MPL L Q = 0 along a given isosquant  MPK K + MPL L = 0  = ‘inverse’ MP ratio

Indifference Curve & Isoquant Slopes Indiff Curve Isosquant - slope = MRS = rate at which consumer is willing to exch Y for 1X in order to hold U constant = inverse MU ratio = MUX/MUY For given indiff curve, dU = 0 Derived from diff types of U fns: Cobb Douglas  U = XY Perfect substitutes  U=X+Y Perfect complements  U = min [X,Y] - slope = MRTS = rate at which producer is able to exch K for 1L in order to hold Q constant = inverse MP ratio = MPL/MPK For given isoquant, dQ = 0 Derived from diff types of production fns: Cobb Douglas  Q = LK Perfect substitutes  Q=L+K Perfect complements  Q = min [X,Y]

Cobb-Douglas Isoquants Inputs are not perfectly substitutable Diminishing marginal rate of technical substitution Most production processes have isoquants of this shape

Linear Isoquants Capital and labor are perfect substitutes

Leontief Isoquants Capital and labor are perfect complements Capital and labor are used in fixed-proportions

Budget Line = maximum combinations of 2 goods that can be bought given one’s income = combinations of 2 goods whose cost equals one’s income

Isocost Line = maximum combinations of 2 inputs that can be purchased given a production ‘budget’ (cost level) = combinations of 2 inputs that are equal in cost

Isocost Line Equation TC1 = rK + wL  rK = TC1 – wL  K = Note: slope = ‘inverse’ input price ratio = = rate at which capital can be exchanged for 1 unit of labor, while holding costs constant.

Increasing Isocost

Changing Input Prices

Different Ways (Costs) of Producing q1

Cost Minimization (graph)

LR Cost Min (math)  - slope of isoquant = - slope of isocost line  

SR vs LR Production

Assume a production process: Q = 10K1/2L1/2 Q = units of output K = units of capital L = units of labor R = rental rate for K = $40 W = wage rate for L = $10

Given q = 10K1/2L1/2 Q K L TC=40K+10L 40* 2* 8* 160* 100* 5* 20* 400* 3.2 232 100 2 50 580 * LR optimum for given q

Given q = 10K1/2L1/2, w=10, r=40 Minimum LR Cost Condition  inverse MP ratio = inverse input P ratio  (MP of L)/(MP of K) = w/r  (5K1/2L-1/2)/(5K-1/2L1/2) = 10/40  K/L = ¼  L = 4K

Optimal K for q = 40? (Given L* = 4K*) q = 40 = 10K1/2L1/2  40 = 10 K1/2(4K)1/2  40 = 20K  K* = 2  L* = 8  min SR TC = 40K* + 10L* = 40(2) + 10(8) = 80 + 80 = $160

SR TC for q = 40? (If K = 5) q = 40 = 10K1/2L1/2  40 = 10 (5)1/2(L)1/2  L = 16/5 = 3.2  SR TC = 40K + 10L = 40(5) + 10(3.2) = 200 + 32 = $232

Optimal K for q = 100? (Given L* = 4K*) Q = 100 = 10K1/2L1/2  100 = 10 K1/2(4K)1/2  100 = 20K  K* = 5  L* = 20  min SR TC = 40K* + 10L* = 40(5) + 10(20) = 200 + 200 = $400

SR TC for q = 100? (If K = 2) Q = 100 = 10K1/2L1/2  SR TC = 40K + 10L = 40(2) + 10(50) = 80 + 500 = $580

Two Different costs of q = 100

LRTC Equation Derivation [i.e. LRTC=f(q)]  LRTC = rk* + wL* = r(k* as fn of q) + w(L* as fn of q) To find K* as fn q from equal-slopes condition L*=f(k), sub f(k) for L into production fn and solve for k* as fn q To find L* as fn q from equal-slopes condition L*=f(k), sub k* as fn of q for f(k) deriving L* as fn q

LRTC Calculation Example Assume q = 10K1/2L1/2, r = 40, w = 10 L* = 4K (equal-slopes condition) K* as fn q q = 10K1/2(4K)1/2 = 10K1/22K1/2 = 20K  LR TC = rk* + wL* = 40(.05q)+10(.2q) = 2q + 2q = 4q L* as fn q L* = 4K* = 4(.05 q) L* = .2q

Graph of SRTC and LRTC

Expansion Path  LRTC



Technological Progress

Multiplant Production Strategy Assume: P = output price = 70 - .5qT qT = total output (= q1+q2) q1 = output from plant #1 q2 = output from plant #2  MR = 70 – (q1+q2) TC1 = 100+1.5(q1)2  MC1 = 3q1 TC2 = 300+.5(q2)2  MC2 = q2

Multiplant  Max (#1) MR = MC1 (#2) MR = MC2 (#1) 70 – (q1 + q2) = 3q1 from (#1), q2 = 70 – 4q1 Sub into (#2),  70 – (q1 + 70 – 4q1) = 70 – 4q1  7q1 = 70  q1 = 10, q2 = 30  = TR – TC1 – TC2 = (50)(40) - [100 + 1.5(10)2] - [300 + .5(30)2] = 2000 – 250 – 750 = $1000

 If q1 = q2 = 20?  = TR - TC1 - TC2 = (50)(40) - [100 + 1.5(20)2] - [300 + .5(20)2] = 2000 – 700 – 500 = $800

Multi Plant Profit Max (alternative solution procedure) 1. Solve for MCT as fn of qT knowing cost min  MC1=MC2=MCT  MC1=3q1  q1 = 1/3 - MC1 = 1/3 MCT MC2 = q2 q2 = MC2 = MCT  q1+q2 = qT = 4/3 MCT  MCT = ¾ qT 2. Solve for profit-max qT  MR=MCT  70-qT = ¾ qT  7/4 qT = 70  q*T = 40  MC*T = ¾ (40) = 30

Multi Plant Profit Max (alternative solution procedure) 3. Solve for q*1 where MC1 = MC*T  3q1 = 30  q*1 = 10 4. Solve for q*2 where MC2 = MC*T  q*2 = 30

Graph  Max, 2 Plants (linear MCs)

 Max (?), 2 Plants, nonlinear MCs