Second Order Derivatives. Background to Topic 6 Maximisation and Minimisation

Transcription

1 Second Order Derivatives Course Manual Background to Topic 6 Maximisation and Minimisation Jacques (4 th Edition): Chapter 4.6 & 4.7

2 Y Y=a+bX a X 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

3 Y Y=X α α>1 X Y Y=X α α>0 and <1 X

4 Y= f (X) = X α First Derivative 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 = (α-1)α(y/x 2 ) 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 Maximisation and Minimisation Stationary Points Second-order derivatives Applications

6 Y Max B Min A X*=1 X*=4 X Stationary points are the turning points or critical points of a function Slope of tangent to curve is zero at stationary points, f (X) = 0 Are these a Max or Min point of the function?

7 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 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

8 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 e.g. inflection point Y f =0 & f =0 X

9 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)

10 Find the Maxima and Minima of the following functions: 2 (i) y = x + 2x + 1 F.O.C. : slope=0 at stationary point dy dx = 2 x + 2 = 2x = 2 x = 1 0 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) Y X

11 Example: A firm faces the demand curve P=8-0.5Q and total cost function TC= 1 / 3 Q 3-3Q 2 +12Q. Find the level of Q that maximises total profit and verify that this value of Q is where MC=MR Solution: P = 8-0.5Q inverse demand function TR (Q) = P.Q = 8Q - ½Q 2 TC (Q) = 1 / 3 Q 3-3Q Q MAX Π(Q) = TR (Q) - TC (Q) Π (Q) = -4Q + 2 ½ Q 2 1 / 3 Q 3

12 MAX Π (Q) = -4Q + 2 ½ Q 2 1 / 3 Q 3 First Order Condition: dπ/dq = f (Q)= Q Q 2 = 0 ( ) b ± 2 b 4 ac (solve quadratic Q 2 + 5Q 4 by applying formula: Q = 2a ) 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) So profit is max at output Q = 4

13 TR (Q) = 8Q - ½Q 2 dtr MR = = 8 Q dq Q = 4 then MR = 4 TC (Q) = 1 / 3 Q 3 MC = dtc dq 2 = Q 6Q Q Q Q = 4 then MC = = +4 At Q = 4, MR = MC

14 Maximisation and Minimisation Tax Example Jacques (4 th Chapter 4.6 Edition): 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?

15 Set Supply equal to Demand In equilibrium, Q D = Q S Q t = 80 3Q Solve for Q Q e = 18 ¼ t Tax Rev. T = t.q e = t(18 ¼ t) MAX T(t) = 18t ¼ t 2 t * First Order Condition for max: dt/dt = 18 ½ t = 0 t * = 36 Second Order Condition for max: d 2 T/dt 2 = -½ < 0 (Max)

16 Results t * = 36 Q e = 18 ¼ t * = 9 T = t *.Q e = 18t * ¼ t *2 = 324 P e = Q e t * = 53 If t = 0, then Q e = 18 P e = Q e + 8 = 26 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 Yet the tax is t * = 36 but the price increase is only 27 (75% paid by consumer)

17 Another Example Manual, Section 5, Q 6 Cost Producing Q units output given 2 C = 8K + Q 2 capital K: K (a) if K=20 in Short Run, find the level of Q at which AC is minimised. Show that MC and AC are equal at this point. 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: AC is at min when dac/dq = 0 slope AC: dac/dq = - 160/Q = 0 Q 2 = 1600 Q = 40

18 Second Order Condition: d 2 AC/dQ 2 >0 min. d 2 AC/dQ 2 = + 320/Q 3 at Q = 40, d 2 AC/dQ 2 = 320 / 40 3 >0 min AC at Q = 40 AC at Q=40: 160/40 + (0.1*40) = 8 and MC at Q = 40: 0.2*40 = 8 MC = AC at min AC when Q=40

19 b) In Long Run, K changes. What level of K minimises costs when Q = 1000? 2 C = 8K + Q 2 K dc/dk = 8 (2Q 2 / K 2 ) = 0 F.O.C. 8K 2 = 2Q 2 K 2 = ¼ Q 2 optimal K = ¼Q 2 = ½ Q if Q = 1000, optimal K = 500 if Q = Q 0, optimal K = Q 0 /2 min cost producing Q 0 = + 8 Q Q Q 0 2 = 4Q 0 + 4Q 0 = 8Q 0

20 Questions Covered: Section 6: Maximisation and Minimisation Q1: Identifying the max and min of various functions Q2: Identifying the max and min of various functions sketch graphs Q3: Finding value of t that maximises tax revenues, given D and S functions (as in lecture example) Q4: Similar to Q3 above, but where Qd = a bp and Qs = c + dp and find impact of t on sales and value of t that maximises tax revenue

21 Q5: Showing that max of total revenue curve occurs where elasticity of demand is unity. Q6: Identifying AC min output level for a given K. And min Costs given optimal K. (lecture example). Q7: (a) Identifying all local max and min of various functions. (b) Identifying profit max output level. Q8: (a) Differentiate various functions. (b) more on finding t that maximises tax revenue