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

Transcription

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

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) Y a X 2

3 For non-linear functions Y Y=X α α>1 X Y Y=X α 0< α < 1 X 3

4 First Derivative Y= f (X) = X α dy/dx = f = α X α-1 > 0 Positive Slope: change in Y due to change X is Positive Second Derivative d 2 Y/dX 2 = f = (α-1)α X (α-1)-1 or d 2 Y/dX 2 = f = (α-1)α(y/x 2 ) 4

5 d 2 Y/dX 2 = f = (α-1)α(y/x 2 ) Sign of Second Derivative? d 2 Y/dX 2 = f = 0 if α = 1 constant rate of change d 2 Y/dX 2 = f > 0 if α > 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 α < 1 decreasing rate of change (change Y due to change X is smaller at higher X the change in the slope is negative) 5

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

7 Y B A X*=1 If f (X) < 0 as X will Y If f (X) > 0 as X will Y If f (X) = 0 slope 0 as X no change Y 7

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 8

9 Are these a Max or Min point of the function? 1) examine slope in region near the stationary point Sign of first derivative around a turning point: Before At After Maximum plus zero minus Minimum minus zero plus dy / dx = f (X) is (, 0, +) min dy / dx = f (X) is (+, 0, -) max 9

10 2) or calculate the second derivative look at the change in the slope beyond the stationary point if d 2 Y/dX 2 = f (X) > 0 a minimum the change in the slope is positive beyond the stationary point, so the point is a local minimum if d 2 Y/dX 2 = f (X) < 0 a maximum the change in the slope is negative beyond the stationary point, so the point is a local maximum if d 2 Y/dX 2 = f (X) = 0 indeterminate the change in the slope is zero beyond the stationary point - could be a max, or a min, or an inflection point 10

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

12 To find the Max or Min of a function Y= f(x) 1) First Order Condition (F.O.C.): set slope dy/dx = f (X) = 0 this identifies the stationary point(s) 2) Second Order Condition (S.O.C.): check the sign of the second derivative (gives the change in the slope) d 2 Y/dX 2 = f (X) > 0 a minimum d 2 Y/dX 2 = f (X) < 0 a maximum d 2 Y/dX 2 = f (X) = 0 indeterminate this identifies whether the slope of the function is increasing, decreasing, or does not change after the stationary point(s) 12

13 Find the Maxima and Minima of the following functions: y 2 = x + 2x F.O.C. : slope=0 at stationary point dy dx = 2 x + 2 = 2x = 2 x = 1 0 Y X S.O.C. : check sign of second derivative at x=-1 d 2 dx y 2 = 2 > 0 (slope increases after the stationary point, so must be a minimum at x= -1) 13

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 14

15 Answer.going to take a few slides! The function we want to Maximise is PROFIT. And Profit = Total Revenue Total Cost Find Total Revenue. P = 8-0.5Q inverse demand function TR (Q) = P.Q = 8Q - ½Q 2 TC (Q) = 1 / 3 Q 3-3Q Q N ow write out the profit function MAX Π = TR - TC Π (Q) = -4Q + 2 ½ Q 2 1 / 3 Q 3 15

16 MAX Π (Q) = -4Q + 2 ½ Q 2 1 / 3 Q 3 First Order Condition: dπ/dq = f (Q)= Q Q 2 = 0 (solve quadratic Q 2 + 5Q 4 by applying formula: ) b ± ( b 2 4ac) Q = Optimal Q solves as: Q * =1 and Q * = 4 Second Order Condition: d 2 Π/dQ 2 = f (Q) = 5 2Q Sign? f = 3 > 0 if Q * = 1 (Min) f = - 3 < 0 if Q * = 4 (Max) 2a So profit is max at output Q = 4 16

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 = = +4 Thus At Q = 4, we have MR = MC 17

18 Maximisation and Minimisation Tax Example The (inverse) Supply and Demand Equations of a good are given, respectively, as P- t = 8 + Q S P = 80 3Q D A tax t per unit, imposed on suppliers, is being considered. At what value of t does the government maximise tax revenue in market equilibrium? 18

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 19

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 ¼t 2 t * 20

21 MAX T(t) = 18t ¼t 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 21

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 * ¼t *2 = 324 P e = Q e t * = 53 If t = 0, then tax revenue = 0, Q e = 18, P e = Q e + 8 = 26 22

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) 23

24 Another example Cost Producing Q output given capital K is: 2 C = 8K + Q 2 K (a) if K=20 in Short Run, find the level of Q at which AC is minimised. (b) Show that MC and AC are equal at this point. 24

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 25

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 26

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 27

28 c) What level of K minimises C when Q = 1000? Substitute C = 8 K + 2 K Q = 1000 to C Q 2 2 = 8 K + 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 function ( 1000 ) K 2 more generally, if Q = Q 0, optimal K = ½ Q 0 28

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 29

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. 30

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