2. 1 Topic 6: Optimization I Maximisation and Minimisation Jacques (4th Edition): Chapter 4.6 & 4.7

3. 2 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 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)

4. 3  For non-linear functions

5. 4 Y= f (X) = X 

6. 5  d 2 Y/dX 2 = f  = (  -1)  (Y/X 2 ) Sign of Second Derivative?  d 2 Y/dX 2 = f  = 0 if  = 1 constant rate of change  d 2 Y/dX 2 = 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)  d 2 Y/dX 2 = 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)

7. 6 Maximisation and Minimisation  Stationary Points  Second-order derivatives  Applications

8. 7

9. 8 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

10. 9 Are these a Max or Min point of the function?

11. 10 2) or calculate the second derivative…… look at the change in the slope beyond the stationary point

12. 11 e.g. inflection point f =0 & f  =0

13. 12 To find the Max or Min of a function Y= f(X)

14. 13 Find the Maxima and Minima of the following functions:

15. 14 Example 1: Profit Maximisation  Question.  A firm faces the demand curve P=8-0.5Q and total cost function TC=1/3Q 3 -3Q 2 +12Q. Find the level of Q that maximises total profit and verify that this value of Q is where MC=MR

16. 15 Answer…. going to take a few slides! The function we want to Maximise is PROFIT…. And Profit = Total Revenue – Total Cost

17. 16 First Order Condition: d  /dQ = f (Q)= - 4 + 5Q – Q 2 = 0 (solve quadratic – Q 2 + 5Q – 4 by applying formula: ) Optimal Q solves as: Q * =1 and Q * = 4 Second Order Condition: d 2  /dQ 2 = 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

18. 17 Continued….. Verify that MR = MC at Q = 4: TR (Q) = 8Q - ½Q 2 MR = dTR/dQ = 8 – Q Evaluate at Q = 4 …..then MR = 4 TC (Q) = 1 / 3 Q 3 - 3Q 2 + 12Q MC = dTC/dQ = Q 2 – 6Q +12 Evaluate at Q = 4 …. then MC = 16 – 24 +12 = +4 Thus At Q = 4, we have MR = MC

19. 18 Maximisation and Minimisation Tax Example

20. 19 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

21. 20 Solution To find equilibrium Q, Set Supply equal to Demand…. In equilibrium, QD = QS so Q + 8 + t = 80 – 3Q Now Solve for Q Q e = 18 – ¼ t Now we can write out our objective function… Tax Revenue T = t.Q e = t(18 – ¼ t) MAX T(t) = 18t – ¼ t 2 t *

22. 21 MAX T(t) = 18t – ¼ t 2 t*  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 d 2 T/dt 2 = -½ < 0 at all values of x Thus, tax rate of 36 will Maximise tax revenue in equilibrium

23. 22 Now we can compute out the equilibrium P and Q and the total tax revenue when t = 36 At t * = 36 Q e = 18 – ¼ t * = 9 Tax Revenue T = t *.Q e = 18t * – ¼ t *2 = 324 P e = Q e + 8 + t * = 53 If t = 0, then tax revenue = 0, Q e = 18, P e = Q e + 8 = 26

24. 23 Is the full burden of the tax passed on to consumers? Ex-ante (no tax) P e = 26 Ex-post (t * =36) P e = 53 The tax is t * = 36, but the price increase is only 27 (75% paid by consumer)

25. 24 Another example

26. 25 Solution Substituting in K = 20 to our C function: C = (8*20) + (2/20)Q 2 = 160 + 0.1Q 2  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/Q 2 + 0.1 = 0 And this solves as Q 2 = 1600  Q = 40

27. 26 Second Order Condition Second Order Condition: check sign at Q = 40 If d 2 AC/dQ 2 >0  min. Since dAC/dQ = - 160/Q 2 + 0.1 Then d 2 AC/dQ 2 = + 320/Q 3 Evaluate at Q = 40, d 2 AC/dQ 2 = 320 / 40 3 >0  min AC at Q = 40

28. 27 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

29. 28 c) What level of K minimises C when Q = 1000? dC/dK = 8 – (2(1000 2 )/ K 2 )= 0 Solving  8K 2 = 2.(1000) 2  K 2 = ¼.(1000) 2  optimal K* =  ¼.(1000) =½(1000)=500  if Q = 1000, optimal K* = 500 more generally, if Q = Q 0, optimal K = ½ Q 0

30. 29 Second Order Condition: check sign of second derivative at K = 500 If d 2 C/dK 2 >0  min. Since dC/dK = 8 – (2(1000 2 )/ K 2 ) d 2 C/dK 2 = + (2.(1000 2 ).2K )/ K 4 >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 Q 0 occurs when K = Q 0 /2 and so C = 4Q 0 + 4Q 0 = 8Q 0

31. 30 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.

32. 31 Maximisation and Minimisation  Second-order derivatives  Stationary Points  Optimisation I  Applications