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Appendix 2A Differential Calculus Techniques in Management

Appendix 2A Differential Calculus Techniques in Management. A function with one decision variable, X, can be written as Y = f(X) The marginal value of Y, with a small change in X, is M y = D Y/ D X For a very small change in X, the derivative is written: dY/dX = limit D Y/ D X D X  ∞.

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Appendix 2A Differential Calculus Techniques in Management

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  1. Appendix 2ADifferential Calculus Techniques in Management • A function with one decision variable, X, can be written as Y = f(X) • The marginal value of Y, with a small change in X, is My = DY/DX • For a very small change in X, the derivative is written: dY/dX = limit DY/DX DX  ∞

  2. Marginal = Slope = Derivative D Y • The slope of line C-D is DY/DX • The marginal at point C is My 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 DY DX C X

  3. The optimum can be the highest or the lowest • Finding the maximum flying range for a plane is an example of 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!

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

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

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

  7. Applications of Calculus in Managerial Economics • 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 (F.O.C) for a maximum is that the derivative at that point is zero. • If  = 50·Q – Q2, then d/dQ = 50 – 2·Q (using the rules of differentiation). • Hence, Q = 25 will maximize profits where 50 – 2·Q = 0.

  8. More Applications of Calculus • 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 (F.O.C) for a minimum is that the derivative at that point is zero. • If C = 5·Q2 - 60·Q, then dC/dQ = 10·Q – 60. • Hence, Q = 6 will minimize cost where 10•Q – 60 = 0.

  9. More Examples • Competitive Firm: Maximize Profits • where  = TR - TC = P•Q – TC(Q) • Use first order condition (F,O.C): d/dQ = P - dTC/dQ = 0. • Decision Rule: P = MC. TC a function of Q Problem 1Problem 2 • Max = 100•Q - Q2 • 100 -2•Q = 0 implies Q = 50 and  = 2,500 • Max= 50 + 5•X2 • So, 10•X = 0 implies Q = 0 and= 50

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

  11. Partial Differentiation • 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.

  12. Problem: • Sales are a function of advertising in newspapers and magazines ( N, M) • Max S = 200N + 100M -10N2 -20M2 +20NM • Differentiate with respect to N and M and set equal to zero. S/N = 200 – 20N + 20M= 0 S/M = 100 – 40M + 20N = 0 • solve for N & M and Sales

  13. Solution: 2 equations & 2 unknowns • 200 – 20N + 20M= 0 • 100 – 40M + 20N = 0 • Adding them, the -20N and +20N cancel, so we get 300 – 20M = 0, or M* =15 • Plug into one of them: 200 – 20N + 300 = 0, hence N* = 25 • To find Sales, plug into equation: S = 200N + 100M -10N2 -20M2 +20NM = 3,250

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