Topic 6 Optimization I Maximisation and Minimisation Jacques

Topic 6: Optimization I Maximisation and Minimisation Jacques (4 th Edition): Chapter 4. 6 & 4. 7 1

For a straight line o o Y=a+b. X Y= f (X) = a + b. X First Derivative d. Y/d. X = f = b constant slope b o o Second Derivative d 2 Y/d. X 2 = f = 0 n 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) 2

o For non-linear functions 3

Y= f (X) = X 4

o o d 2 Y/d. X 2 = f = ( -1) (Y/X 2) Sign of Second Derivative? d 2 Y/d. X 2 = f = 0 if = 1 constant rate of change d 2 Y/d. X 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/d. X 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) 5

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

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Definition o o o 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: o 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 o Question. A firm faces the demand curve P=8 -0. 5 Q and total cost function TC=1/3 Q 3 -3 Q 2+12 Q. o 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 /d. Q = f (Q)= - 4 + 5 Q – Q 2 = 0 (solve quadratic – Q 2 + 5 Q – 4 by applying formula: ) Optimal Q solves as: Q*=1 and Q*= 4 Second Order Condition: d 2 /d. Q 2 = f (Q) = 5 – 2 Q 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) = 8 Q - ½Q 2 MR = d. TR/d. Q = 8 – Q Evaluate at Q = 4 …. . then MR = 4 TC (Q) = 1/3 Q 3 - 3 Q 2 + 12 Q MC = d. TC/d. Q = Q 2 – 6 Q +12 Evaluate at Q = 4 …. then MC = 16 – 24 +12 = +4 Thus At Q = 4, we have MR = MC 17

Maximisation and Minimisation Tax 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 – 3 Q 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) = 18 t – ¼ t 2 t* 20

MAX t* T(t) = 18 t – ¼ t 2 o First Order Condition for max: set the slope (or first derivative) = 0 d. T/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

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 = 18 t* – 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)Q 2 = 160 + 0. 1 Q 2 AC = C/Q = 160/Q + 0. 1 Q and MC = d. C/d. Q = 0. 2 Q First Order Condition: set first derivative (slope)=0 AC is at min when d. AC/d. Q = 0 So d. AC/d. Q = - 160/Q 2 + 0. 1 = 0 And this solves as Q 2 = 1600 Q = 40 25

Second Order Condition: check sign at Q = 40 If d 2 AC/d. Q 2 >0 min. Since d. AC/d. Q = - 160/Q 2 + 0. 1 Then d 2 AC/d. Q 2 = + 320/Q 3 Evaluate at Q = 40, d 2 AC/d. Q 2 = 320 / 403 >0 min AC at Q = 40 26

b) Now show MC = AC when Q = 40: AC = C/Q = 160/Q + 0. 1 Q AC at Q=40: 160/40 + (0. 1*40) = 8 and MC = d. C/d. Q = 0. 2 Q 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? d. C/d. K = 8 – (2(10002 )/ K 2 )= 0 Solving 8 K 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 28

Second Order Condition: check sign of second derivative at K = 500 If d 2 C/d. K 2 >0 min. Since d. C/d. K = 8 – (2(10002 )/ K 2 ) d 2 C/d. K 2 = + (2. (10002). 2 K )/ 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 = 4 Q 0 + 4 Q 0 = 8 Q 0 29

Topic 6: Maximisation and Minimisation o o o Second De. 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 o o Second-order derivatives Stationary Points Optimisation I Applications 31