1 / 13

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  ∞.

dulcea
Download Presentation

Appendix 2A Differential Calculus Techniques in Management

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related