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 - 10Q 2 Tables: Graphs:

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

Total, Average, and Marginal Cost

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

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 d 2 Y/dX 2 > 0, then X is a minimum. If d 2 Y/dX 2 < 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 d 2 Y/dX 2 > 0, then X is a minimum. If d 2 Y/dX 2 < 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 – 10Q 2 dTR/dQ = 100 – 20Q = 0 Q* = 5 and d 2 TR/dQ 2 = -20 < 0

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

Example 3 Given the following marginal cost function (MC), determine the quantity of output that will minimize MC: MC = 3Q 2 – 16Q + 57 dMC/dQ = 6Q - 16 = 0 Q* = 2.67 and d 2 MC/dQ 2 = 6 > 0

Example 4 Given –TR = 45Q – 0.5Q 2 –TC = Q 3 – 8Q Q + 2 Determine Q that maximizes profit ( π): –π = 45Q – 0.5Q 2 – (Q 3 – 8Q Q + 2)

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

Quadratic Formula Write the equation in the following form: aX 2 + bX + c = 0 The solutions have the following form:

Multivariate Optimization Objective function Y = f(X 1, X 2,...,X k ) Find all X i such that ∂ Y/ ∂ X i = 0 Partial derivative: –∂ Y/ ∂ X i = dY/dX i while all X j (where j ≠ i) are held constant

Example 5 Determine the values of X and Y that maximize the following profit function: –π = 80X – 2X 2 – XY – 3Y Y Solution –∂π / ∂ X = 80 – 4X – Y = 0 –∂π / ∂ Y = -X – 6Y = 0 –Solve simultaneously –X = 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 – 2X 2 – 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 = – 4Y Y Solve as univariate function: –d π /dY = – 8Y + 56 = 0 –Y = 7 and X = 5

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

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

New Management Tools Benchmarking Total Quality Management Reengineering The Learning Organization

Other Management Tools Broadbanding Direct Business Model Networking Pricing Power Small-World Model Virtual Integration Virtual Management