1 / 31

Topic 6: Optimization I

Topic 6: Optimization I. Maximisation and Minimisation Jacques (4th Edition): Chapter 4.6 & 4.7. For a straight line. Y=a+bX Y= f (X) = a + bX First Derivative dY/dX = f  = b constant slope b Second Derivative d 2 Y/dX 2 = f  = 0

duante
Télécharger la présentation

Topic 6: Optimization I

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. Topic 6: Optimization I Maximisation and Minimisation Jacques (4th Edition): Chapter 4.6 & 4.7

  2. For a straight line • Y=a+bX • Y= f (X) = a + bX • First Derivative • dY/dX = f  = b constant slope b • Second Derivative • d2Y/dX2 = f  = 0 • constant rate of change - the change in the slope is zero - (i.e. change in Y due to change in X does not depend on X)

  3. For non-linear functions

  4. Y= f (X) = X

  5. d2Y/dX2 = f = (-1)(Y/X2) Sign of Second Derivative? • d2Y/dX2 = f = 0 if  = 1 constant rate of change • d2Y/dX2 = f  > 0 if  > 1 increasing rate of change (change Y due to change X is bigger at higher X – the change in the slope is positive) • d2Y/dX2 = f  < 0 if  < 1 decreasing rate of change (change Y due to change X is smaller at higher X – the change in the slope is negative)

  6. Maximisation and Minimisation • Stationary Points • •Second-order derivatives • •Applications

  7. Definition • Stationary points are the turning points or critical points of a function • Slope of tangent to curve is zero at stationary points • Stationary point(s) at A & B: • where f (X) = 0

  8. Are these a Max or Min point of the function?

  9. 2) or calculate the second derivative…… look at the change in the slope beyond the stationary point

  10. e.g. inflection point f =0 & f =0

  11. To find the Max or Min of a function Y= f(X)

  12. Find the Maxima and Minima of the following functions:

  13. Example 1: Profit Maximisation • Question. • A firm faces the demand curve P=8-0.5Q and total cost function TC=1/3Q3-3Q2+12Q. Find the level of Q that maximises total profit and verify that this value of Q is where MC=MR

  14. Answer….going to take a few slides! The function we want to Maximise is PROFIT…. And Profit = Total Revenue – Total Cost

  15. First Order Condition: d/dQ = f (Q)= - 4 + 5Q – Q2 = 0 (solve quadratic – Q2 + 5Q – 4 by applying formula: ) Optimal Q solves as: Q*=1 and Q*= 4 Second Order Condition: d2/dQ2 = f (Q) = 5 – 2Q Sign ? f  = 3 > 0 if Q*= 1 (Min) f = - 3 < 0 if Q*= 4 (Max) So profit is max at output Q = 4

  16. Continued….. Verify that MR = MC at Q = 4: TR (Q) = 8Q - ½Q2 MR = dTR/dQ = 8 – Q Evaluate at Q = 4 …..then MR = 4 TC (Q) = 1/3Q3 - 3Q2 + 12Q MC = dTC/dQ = Q2 – 6Q +12 Evaluate at Q = 4 …. then MC = 16 – 24 +12 = +4 Thus At Q = 4, we have MR = MC

  17. Maximisation and MinimisationTax Example

  18. What do we want to maximise? Tax revenue in equilibrium….. This will be equal to the tax rate t multiplied by the equilibrium quantity So first we need to find the equilibrium quantity

  19. Solution To find equilibrium Q, Set Supply equal to Demand…. In equilibrium, QD = QS so Q + 8 + t= 80 – 3Q Now Solve for Q Qe = 18 – ¼ t  Now we can write out our objective function… Tax Revenue T = t.Qe = t(18 – ¼ t) MAX T(t) = 18t – ¼ t2 t*

  20. MAX T(t) = 18t – ¼ t2t* • First Order Condition for max: set the slope (or first derivative) = 0 dT/dt = 18 – ½ t = 0 t* = 36 Second Order Condition for max: check sign of second derivative d2T/dt2 = -½ < 0 at all values of x Thus, tax rate of 36 will Maximise tax revenue in equilibrium

  21. Now we can compute out the equilibrium P and Q and the total tax revenue when t = 36 At t* = 36 Qe = 18 – ¼ t*= 9 Tax Revenue T = t*.Qe = 18t* – ¼ t*2 = 324 Pe = Qe + 8 + t*= 53 If t = 0, then tax revenue = 0, Qe = 18 , Pe = Qe + 8 = 26

  22. Is the full burden of the tax passed on to consumers? Ex-ante (no tax) Pe = 26 Ex-post (t* =36) Pe = 53 The tax is t*= 36, but the price increase is only 27 (75% paid by consumer)

  23. Another example

  24. Solution Substituting in K = 20 to our C function: C = (8*20) + (2/20)Q2 = 160 + 0.1Q2  AC = C/Q = 160/Q + 0.1Q and MC = dC/dQ = 0.2Q First Order Condition: set first derivative (slope)=0 AC is at min when dAC/dQ = 0 So dAC/dQ = - 160/Q2 + 0.1 = 0 And this solves as Q2 = 1600 Q = 40

  25. Second Order Condition Second Order Condition: check sign at Q = 40 If d2AC/dQ2 >0  min. Since dAC/dQ = - 160/Q2 + 0.1 Then d2AC/dQ2 = + 320/Q3 Evaluate at Q = 40, d2AC/dQ2 = 320 / 403 >0  min AC at Q = 40

  26. b) Now show MC = AC when Q = 40: AC = C/Q = 160/Q + 0.1Q  AC at Q=40: 160/40 + (0.1*40) = 8 and MC = dC/dQ = 0.2Q So MC at Q = 40: 0.2*40 = 8  MC = AC at min AC when Q=40

  27. c) What level of K minimises C when Q = 1000? dC/dK = 8 – (2(10002 )/ K2 )= 0 Solving  8K2 = 2.(1000)2  K2 = ¼ .(1000)2  optimal K* = ¼.(1000) =½(1000)=500 if Q = 1000, optimal K* = 500 more generally, if Q = Q0, optimal K = ½ Q0

  28. Second Order Condition: check sign of second derivative at K = 500 If d2C/dK2 >0  min. Since dC/dK = 8 – (2(10002 )/ K2 ) d2C/dK2 = + (2.(10002).2K )/ K4 >0 for all values of K>0 and so C are at a min when K = 500 The min cost producing Q=1000 occurs when K = 500 Subbing in value k = 500 we get C = 8000 or more generally, min cost producing Q0 occurs when K = Q0/2 and so C = 4Q0 + 4Q0 = 8Q0

  29. Topic 6: Maximisation and Minimisation • Second DeIdentifying the max and min of various functions • Identifying the max and min of various functions – sketch graphs • Finding value of t that maximises tax revenues, given D and S functions • Identifying all local max and min of various functions. Identifying profit max output level. • Differentiate various functions.

  30. Maximisation and Minimisation • Second-order derivatives • Stationary Points • Optimisation I • Applications

More Related