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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

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Topic 6: Optimization I

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  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

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