Managerial Economics Lecture: Optimization Technique Date:

Maximizing profit Manager determine the quantity of output to be produced Quantity of sales to maximize profit

Optimization Techniques Methods for maximizing or minimizing an objective function Examples – Consumers maximize utility by purchasing an optimal combination of goods – Firms maximize profit by producing and selling an optimal quantity of goods – Firms minimize their cost of production by using an optimal combination of inputs

Optimization Techniques Concept of the Derivative The derivative of Y with respect to X is equal to the limit of the ratio ΔY/ΔX as ΔX approaches zero dy/dx=LimΔY/ΔX x→0

Geometric Relationships A marginal value is positive, zero, and negative, respectively, when a total curve slopes upward, is horizontal, and slopes downward A marginal value may be negative, but an average value can never be negative

Steps in Optimization Define an objective function of one or more choice variables Define the constraint on the values of the objective function Determine the values of the choice variables that maximize or minimize the objective function while satisfying the constraint

New Management Tools for Optimization Benchmarking (tool for improving productivity and quality) Total Quality Management (constantly improving the quality of products and the firm’s processes to deliver more value to customers; e.g. Six Sigma) Reengineering (radical redesign of all the firm’s processes to achieve major gains) Learning Organization (values continuing learning, both individual and collective)

Management Tools for Optimization Broad-banding (elimination of multiple salary grades to foster movement among jobs within the firm and lower cost) Direct Business Model (eliminating the time and cost of third-party distribution)

Management Tools for Optimization Networking (forming of temporary strategic alliances among firms as per their core competence) Performance Management (holding executives and their subordinates accountable for delivering the desired results)

Other Management Tools for Optimization Pricing Power (ability of a firm to raise prices faster than the rise in its costs and vice-versa) Small-World Model (linking well-connected individuals from each level of the organization to one another to improve flow of information and the operational efficiency)

Other Management Tools for Optimization Strategic Development (continuous review of strategic decisions) Virtual Integration (treating suppliers and customers as if they were part of the company which reduces the need for inventories) Virtual Management (ability of a manager to simulate consumer behavior using computer models)

Univariate Optimization Given objective function Y = f(X) Find X such that dY/dX = 0 Second derivative rules: If d2Y/dX2 > 0, then X is a minimum If d2Y/dX2 < 0, then X is a maximum

Example 1 Given the following total revenue (TR) function, determine the quantity of output (Q) that will maximize total revenue: TR = 100Q – 10Q2 dTR/dQ = 100 – 20Q = 0 Q* = 5 and d2TR/dQ2 = -20 < 0

Example 2 Given the following total revenue (TR) function, determine the quantity of output (Q) that will maximize total revenue: TR = 45Q – 0.5Q2 dTR/dQ = 45 – Q = 0 Q* = 45 and d2TR/dQ2 = -1 < 0

Example 3 Given the following marginal cost function (MC), determine the quantity of output that will minimize MC: MC = 3Q2 – 16Q + 57 dMC/dQ = 6Q - 16 = 0 Q* = 2.67 and d2MC/dQ2 = 6 > 0

Example 4 Given TR = 45Q – 0.5Q2 TC = Q3 – 8Q2 + 57Q + 2 Determine Q that maximizes profit (π): π = 45Q – 0.5Q2 – (Q3 – 8Q2 + 57Q + 2)

Example 4: Solution Method 1 dπ/dQ = 45 – Q – 3Q2 + 16Q – 57 = 0 – Q – 3Q2 = 0 Method 2 MR = dTR/dQ = 45 – Q MC = dTC/dQ = 3Q2 – 16Q + 57 Set MR = MC: 45 – Q = 3Q2 – 16Q + 57 Use quadratic formula: Q* = 4

Multivariate Optimization Objective function Y = f(X1, X2,...,Xk) Find all Xi such that ∂Y/∂Xi = 0 Partial derivative: ∂Y/∂Xi = dY/dXi while all Xj (where j ≠ i) are held constant

Example 5 Determine the values of X and Y that maximize the following profit function: π = 80X – 2X2 – XY – 3Y Y Solution ∂π/∂X = 80 – 4X – Y = 0 ∂π/∂Y = – X – 6Y = 0 Solve simultaneously X = and Y = 13.91

Constrained Optimization Substitution Method – Substitute constraints into the objective function and then maximize the objective function Lagrangian Method – Form the Lagrangian function by adding the Lagrangian variable and constraint to the objective function and then maximize the Lagrangian function

Example 6 Use the substitution method to maximize the following profit function: π = 80X – 2X2 – XY – 3Y Y Subject to the following constraint: X + Y = 12

Example 6: Solution Substitute X = 12 – Y into profit: π = 80(12 – Y) – 2(12 – Y)2 – (12 – Y)Y – 3Y Y π = – 4Y2 + 56Y Solve as univariate function: dπ/dY = – 8Y + 56 = 0 Y = 7 and X = 5

Example 7 Use the Lagrangian method to maximize the following profit function: π = 80X – 2X2 – XY – 3Y Y Subject to the following constraint: X + Y = 12

Example 7: Solution Form the Lagrangian function L = 80X – 2X2 – XY – 3Y Y + λ(X + Y – 12) Find the partial derivatives and solve simultaneously dL/dX = 80 – 4X –Y + λ = 0 dL/dY = – X – 6Y λ = 0 dL/dλ = X + Y – 12 = 0 Solution: X = 5, Y = 7, and λ = -53

Interpretation of the Lagrangian Multiplier, λ Lambda, λ, is the derivative of the optimal value of the objective function with respect to the constraint – In Example 7, λ = -53, so a one-unit increase in the value of the constraint (from -12 to -11) will cause profit to decrease by approximately 53 units