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  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)

3  For non-linear functions

4 Y= f (X) = X 

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)

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

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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

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

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

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

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

13 Find the Maxima and Minima of the following functions:

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

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

16 First Order Condition: d  /dQ = f (Q)= Q – 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

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 Q MC = dTC/dQ = Q 2 – 6Q +12 Evaluate at Q = 4 …. then MC = 16 – = +4 Thus At Q = 4, we have MR = MC

18 Maximisation and Minimisation Tax Example

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

20 Solution To find equilibrium Q, Set Supply equal to Demand…. In equilibrium, QD = QS so Q 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 *

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

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 t * = 53 If t = 0, then tax revenue = 0, Q e = 18, P e = Q e + 8 = 26

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)

24 Another example

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

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 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

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

28 c) What level of K minimises C when Q = 1000? dC/dK = 8 – (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

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( )/ K 2 ) d 2 C/dK 2 = + (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

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.

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