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

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

3. 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

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

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

8. Average and Marginal Revenue
Total Revenue
Average and Marginal Revenue

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

13. 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

14. 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

15. Profit Maximization

18. 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

20. New Management Tools Benchmarking Total Quality Management
Reengineering
The Learning Organization

21. Other Management Tools
Broadbanding
Direct Business Model
Networking
Performance Management

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

23. Chapter 2 Appendix

24. 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.

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

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

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

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

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

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

34. 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.

35. 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.

36. 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

37. 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

38. 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

39. 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)

40. Example 4: Solution Method 1 Method 2 Use quadratic formula: Q* = 4
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

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

42. 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

43. 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 = 13.92

45. 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

46. 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

47. 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 + 672
Solve as univariate function:
dπ/dY = – 8Y + 56 = 0
Y = 7 and X = 5

48. 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

49. Example 7: Solution Form the Lagrangian function
L = 80X – 2X2 – 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

50. 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