Principles and Worldwide Applications, 7th Edition Managerial Economics Principles and Worldwide Applications, 7th Edition Dominick Salvatore & Ravikesh Srivastava

Optimization Techniques and New Chapter 2: Optimization Techniques and New Management Tools

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

Expressing Economic Relationships Equations: TR = 100Q - 10Q2 Tables: Graphs:

Total, Average, and Marginal Revenue TR = PQ AR = TR/Q MR = TR/Q

Average and Marginal Revenue Total Revenue Average and Marginal Revenue

Total, Average, and Marginal Cost AC = TC/Q MC = TC/Q

Geometric Relationships The slope of a tangent to a total curve at a point is equal to the marginal value at that point The slope of a ray from the origin to a point on a total curve is equal to the average value at that point

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 is above, equal to, and below an average value, respectively, when the slope of the average curve is positive, zero, and negative

Profit Maximization

Steps in Optimization Define an objective mathematically as a function of one or more choice variables Define one or more constraints on the values of the objective function and/or the choice variables Determine the values of the choice variables that maximize or minimize the objective function while satisfying all of the constraints

New Management Tools Benchmarking Total Quality Management Reengineering The Learning Organization

Other Management Tools Broadbanding Direct Business Model Networking Performance Management

Other Management Tools Pricing Power Small-World Model Strategic Development Virtual Integration Virtual Management

Chapter 2 Appendix

Concept of the Derivative The derivative of Y with respect to X is equal to the limit of the ratio Y/X as X approaches zero.

Rules of Differentiation Constant Function Rule: The derivative of a constant, Y = f(X) = a, is zero for all values of a (the constant).

Rules of Differentiation Power Function Rule: The derivative of a power function, where a and b are constants, is defined as follows.

Rules of Differentiation Sum-and-Differences Rule: The derivative of the sum or difference of two functions, U and V, is defined as follows.

Rules of Differentiation Product Rule: The derivative of the product of two functions, U and V, is defined as follows.

Rules of Differentiation Quotient Rule: The derivative of the ratio of two functions, U and V, is defined as follows.

Rules of Differentiation Chain Rule: The derivative of a function that is a function of X is defined as follows.

Optimization with Calculus 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.

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 Determine Q that maximizes profit (π): 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 Method 2 Use quadratic formula: Q* = 4 dπ/dQ = 45 – Q - 3Q2 + 16Q – 57 = 0 -12 + 15Q - 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

Quadratic Formula Write the equation in the following form: aX2 + bX + c = 0 The solutions have the following form:

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 – 3Y2 + 100Y Solution ∂π/∂X = 80 – 4X – Y = 0 ∂π/∂Y = -X – 6Y + 100 = 0 Solve simultaneously X = 16.52 and Y = 13.92

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 variables and constraints 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 – 3Y2 + 100Y 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 – 3Y2 + 100Y π = – 4Y2 + 56Y + 672 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 – 3Y2 + 100Y Subject to the following constraint: X + Y = 12

Example 7: Solution Form the Lagrangian function L = 80X – 2X2 – XY – 3Y2 + 100Y + (X + Y – 12) Find the partial derivatives and solve simultaneously dL/dX = 80 – 4X –Y +  = 0 dL/dY = – X – 6Y + 100 +  = 0 dL/d = X + Y – 12 = 0 Solution: X = 5, Y = 7, and  = -53

Interpretation of the Lagrangian Multiplier,  Lambda, , is the derivative of the optimal value of the objective function with respect to the constraint In Example 7,  = -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 Actual decrease is 66.5 units