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Managerial Economics in a Global Economy

Managerial Economics in a Global Economy. Chapter 2 Optimization Techniques and New Management Tools. OPTIMIZATION

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Managerial Economics in a Global Economy

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  1. Managerial Economicsin a Global Economy Chapter 2 Optimization Techniquesand New Management Tools

  2. OPTIMIZATION Managerial economics is concerned with the ways in which managers should make decisions in order to maximize the effectiveness or performance of the organizations they manage. To understand how this can be done we must understand the basic optimization techniques. Functional relationships: relationships can be expressed by graphs:

  3. This form can be expressed in an equation: • Q = f ( P ) (1) • Though useful, it does not tell us how Q responds to P, but this equation do. • Q = 200 - 5 p (2) Marginal Analysis • The marginal value of a dependent variable is defined as the change in this dependent variable associated with a 1-unit change in a particular independent variable. e.g.

  4. Note: Total profit is maximized when marginal profit shifts from positive to negative.

  5. Slope of ray from the origin: • Rise / Run • Profit / Q = average profit • Maximizing average profit doesn’t maximize total profit Average Profit = Profit / Q PROFITS MAX C B profits quantity Q

  6. Q1 is breakeven (zero profit) • maximum marginal profits occur at theinflection point (Q2) • Max average profit at Q3 • Max total profit at Q4 where marginal profit is zero • So the best place to produce is where marginal profits = 0. Marginal Profits = /Q (Figure 2.1) max profits Q4 Q3 Q2 Q1 Q average profits marginal profits Q

  7. A function with one decision variable, X, can be written as: Y = f(X) • The marginal value of Y, with a small increase of X, is My = DY/DX • For a very small change in X, the derivative is written: dY/dX = limit DY/DX DX  B Differential Calculus in Management 2005 South-Western Publishing

  8. The slope of line C-D is DY/DX • The marginal at point C is DY/DX • The slope at point C is the rise (DY) over the run (DX) • The derivative at point C is also this slope Marginal = Slope = Derivative D Y DY DX C X

  9. Finding the maximum flying range for the Stealth Bomber is an optimization problem. • Calculus teaches that when the first derivative is zero, the solution is at an optimum. • The original Stealth Bomber study showed that a controversial flying V-wing design optimized the bomber's range, but the original researchers failed to find that their solution in fact minimized the range. • It is critical that managers make decision that maximize, not minimize, profit potential! Optimum Can Be Highest or Lowest

  10. Constant Y = c dY/dX = 0 Y = 5 Functions dY/dX = 0 • A Line Y = c • X dY/dX = c Y = 5X dY/dX = 5 • Power Y = cXbdY/dX = b•c•X b-1Y = 5X2 Functions dY/dX = 10X Quick Differentiation Review Name Function Derivative Example

  11. SumofY = G(X) + H(X) dY/dX = dG/dX + dH/dX Functions exampleY = 5X + 5X2 dY/dX = 5 + 10X • Product ofY= G(X) • H(X) Two FunctionsdY/dX = (dG/dX)H + (dH/dX)G exampleY = (5X)(5X2 ) dY/dX = 5(5X2 ) + (10X)(5X) = 75X2 Quick Differentiation Review

  12. Quotient of Two Y = G(X) / H(X) Functions dY/dX = (dG/dX)•H - (dH/dX)•G H2 Y = (5X) / (5X2) dY/dX = 5(5X2) -(10X)(5X) (5X2)2 = -25X2 / 25X4 = - X-2 • Chain RuleY = G [ H(X) ] dY/dX = (dG/dH)•(dH/dX)Y = (5 + 5X)2 dY/dX = 2(5 + 5X)1(5) = 50 + 50X Quick Differentiation Review

  13. USING DERIVATIVES TO SOLVE MAXIMIZATION AND MINIMIZATION PROBLEMS • Maximum or minimum points occur only if the slope of the curve equals zero. • Look at the following graph

  14. Max of x y Slope = 0 value of x x 10 20 dy/dx Value of Dy/dx when y is max 0 Value of dy/dx which Is the slope of y curve x 20 10 Note that this is not sufficient for maximization or minimization problems.

  15. y Max value of y Min value of y x Dy/dx value of dy/dx d2y/dx2 > 0 d2y/dx2 < 0 x

  16. Graphs of an original third-order function and its first and second derivatives. (what if the second order = 0)

  17. Optimization Rules

  18. maximization problem: • A profit function might look like an arch, rising to a peak and then declining at even larger outputs. A firm might sell huge amounts at very low prices, but discover that profits are low or negative. • At the maximum, the slope of the profit function is zero. The first order condition for a maximum is that the derivative at that point is zero. If  = 50Q - Q2, then d/dQ = 50 - 2·Q, using the rules of differentiation. Hence, Q = 25 will maximize profits where 50 - 2Q = 0. Applications of Calculus in Managerial Economics

  19. minimization problem: Cost minimization • supposes that there is a least cost point to produce. An average cost curve might have a U-shape. At the least cost point, the slope of the cost function is zero. The first order condition for a minimum is that the derivative at that point is zero. If TC = 5Q2 – 60Q, then dC/dQ = 10Q - 60. Hence, Q = 6 will minimize cost Where: 10Q - 60 = 0. More Applications of Calculus

  20. Competitive Firm: Maximize Profits • where  = TR - TC = P • Q - TC(Q) • Use our first order condition: d/dQ = P - dTC/dQ = 0. • Decision Rule: P = MC. More Examples a function of Q • Max = 100Q - Q2 First order = 100 -2Q = 0 implies Q = 50 and;  = 2,500

  21. If the second derivative is negative, then it’s a maximum • If the second derivative is positive, then it’s a minimum Second Order Condition: one variable Problem 1 Problem 2 Max = 100Q - Q2 First derivative 100 -2Q = 0 second derivative is: -2 implies Q =50 is a MAX Max= 50 + 5X2 First derivative 10X = 0 second derivative is: 10 implies Q = 10 is a MIN

  22. Extra examples

  23. Economic relationships usually involve several independent variables. • A partial derivative is like a controlled experiment- it holds the “other” variables constant • Suppose price is increased, holding the disposable income of the economy constant as in Q = f (P, I ) • then Q/Pholds income constant. Partial Differentiation

  24. Sales are a function of advertising in newspapers and magazines ( X, Y) Max S = 200X + 100Y -10X2 -20Y2 +20XY • Differentiate with respect to X and Y and set equal to zero. S/X = 200 - 20X + 20Y= 0 S/Y = 100 - 40Y + 20X = 0 • solve for X & Y and Sales Problem:

  25. 200 - 20X + 20Y= 0 • 100 - 40Y + 20X = 0 • Adding them, the -20X and +20X cancel, so we get 300 - 20Y = 0, or Y =15 • Plug into one of them: • 200 - 20X + 300 = 0, hence X = 25 • To find Sales, plug into equation: • S = 200X + 100Y -10X2 -20Y2 +20XY = 3,250 Solution: 2 equations & 2 unknowns

  26. Marginal profit • Average profit • Marginal cost • Marginal revenue • Marginal analysis • Optimization • Derivative of Y with respect to X (dy/dx) • Differentiation • Second derivative • Partial derivative • Constrained optimization • Lagrangian multiplier method • Lagrangian function Key Terms

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