MATHEMATICS 1 CONTENTS Extreme values in one dimension Extreme - PowerPoint PPT Presentation

Extreme values in two dimensions BUSINESS MATHEMATICS 1

CONTENTS Extreme values in one dimension Extreme values in two dimensions Modified second-order conditions Example Old exam question Further study 2

EXTREME VALUES IN TWO DIMENSIONS 3

EXTREME VALUES IN ONE DIMENSION Consider a smooth function 𝑔 𝑦 1. Necessary first-order condition for extreme value 𝑒𝑔 𝑦 = 0 ⇒ stationary point 𝑒𝑦 2. Sufficient second-order condition at stationary point 𝑒 2 𝑔 𝑦 < 0 ⇒ maximum ▪ 𝑒𝑦 2 𝑒 2 𝑔 𝑦 > 0 ⇒ minimum ▪ 𝑒𝑦 2 𝑒 2 𝑔 𝑦 ▪ = 0 ⇒? 𝑒𝑦 2 4

EXTREME VALUES IN ONE DIMENSION Consider a smooth function 𝑔 𝑦 1. Necessary first-order condition for extreme value 𝑒𝑔 𝑦 = 0 ⇒ stationary point 𝑒𝑦 2. Sufficient second-order condition at stationary point 𝑒 2 𝑔 𝑦 < 0 ⇒ maximum ▪ 𝑒𝑦 2 𝑒 2 𝑔 𝑦 > 0 ⇒ minimum ▪ 𝑒𝑦 2 𝑒 2 𝑔 𝑦 ▪ = 0 ⇒? 𝑒𝑦 2 5

EXERCISE 1 Given is 𝑔 𝑦, 𝑧 = 2𝑦𝑧 − 2x + y − 2 Find the stationary points. 6

EXTREME VALUES IN TWO DIMENSIONS In some cases: true (maximum) 8

EXTREME VALUES IN TWO DIMENSIONS In other cases: false (saddle point) 9

MODIFIED SECOND-ORDER CONDITIONS Additional sufficient second-order condition for extreme point 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 ▪ not > 0 and > 0 𝜖𝑦 2 𝜖𝑧 2 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 ▪ nor < 0 and < 0 𝜖𝑦 2 𝜖𝑧 2 But….. 10

MODIFIED SECOND-ORDER CONDITIONS 1. Testing for extreme point: 2 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 − > 0 𝜖𝑦 2 𝜖𝑧 2 𝜖𝑦𝜖𝑧 2 > 0 or 𝑔 [ 𝑔 𝑧𝑦 > 0 ] 𝑦𝑦 𝑔 𝑧𝑧 − 𝑔 𝑦𝑦 𝑔 𝑧𝑧 − 𝑔 𝑦𝑧 𝑔 𝑦𝑧 11

MODIFIED SECOND-ORDER CONDITIONS 1. Testing for extreme point: Recall that 𝑔 𝑦𝑧 = 𝑔 𝑧𝑦 2 2 = 𝑔 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 so that 𝑔 𝑦𝑧 𝑔 𝑦𝑧 𝑧𝑦 − > 0 𝜖𝑦 2 𝜖𝑧 2 𝜖𝑦𝜖𝑧 2 > 0 or 𝑔 [ 𝑔 𝑧𝑦 > 0 ] 𝑦𝑦 𝑔 𝑧𝑧 − 𝑔 𝑦𝑦 𝑔 𝑧𝑧 − 𝑔 𝑦𝑧 𝑔 𝑦𝑧 12

MODIFIED SECOND-ORDER CONDITIONS 1. Testing for extreme point: 2 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 − > 0 𝜖𝑦 2 𝜖𝑧 2 𝜖𝑦𝜖𝑧 2 > 0 or 𝑔 [ 𝑔 𝑧𝑦 > 0 ] 𝑦𝑦 𝑔 𝑧𝑧 − 𝑔 𝑦𝑦 𝑔 𝑧𝑧 − 𝑔 𝑦𝑧 𝑔 𝑦𝑧 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 2. Then, maximum point: < 0 or < 0 𝜖𝑦 2 𝜖𝑧 2 [ 𝑔 𝑦𝑦 < 0 or 𝑔 𝑧𝑧 < 0 ] 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 minimum point: > 0 or > 0 𝜖𝑦 2 𝜖𝑧 2 [ 𝑔 𝑦𝑦 > 0 or 𝑔 𝑧𝑧 > 0 ] 13

EXAMPLE Consider 𝑔 𝑦, 𝑧 = 𝑦 2 − 𝑦𝑧 + 𝑧 2 − 4𝑧 + 5 1. Finding stationary points: 𝜖𝑔 ▪ 𝜖𝑦 = 2𝑦 − 𝑧 = 0 𝜖𝑔 ▪ 𝜖𝑧 = −𝑦 + 2𝑧 − 4 = 0 Result: 𝑦, 𝑧 = 4 3 , 8 3 14

EXAMPLE CONTINUTED 2. Second-order test for 𝑦, 𝑧 = 3 : 4 3 , 8 2 𝜖 2 𝑔 𝜖 2 𝑔 𝜖 2 𝑔 = 2 × 2 − −1 2 = 3 𝜖𝑧 2 − 𝜖𝑦 2 𝜖𝑦𝜖𝑧 So 3 , 8 4 3 is an extreme point. 15

EXAMPLE CONTINUTED 2. Second-order test for 𝑦, 𝑧 = 3 : 4 3 , 8 2 𝜖 2 𝑔 𝜖 2 𝑔 𝜖 2 𝑔 = 2 × 2 − −1 2 = 3 𝜖𝑧 2 − 𝜖𝑦 2 𝜖𝑦𝜖𝑧 So 4 3 , 8 3 is an extreme point. 𝜖 2 𝑔 Further 𝜖𝑦 2 = 2 > 0 , so 4 3 is a minimum point with minimum value 𝑔 3 , 8 4 3 , 8 3 = − 1 3 16

EXAMPLE CONTINUTED 2. Second-order test for 𝑦, 𝑧 = 3 : 4 3 , 8 2 𝜖 2 𝑔 𝜖 2 𝑔 𝜖 2 𝑔 = 2 × 2 − −1 2 = 3 𝜖𝑧 2 − 𝜖𝑦 2 𝜖𝑦𝜖𝑧 So 4 3 , 8 3 is an extreme point. 𝜖 2 𝑔 Further 𝜖𝑦 2 = 2 > 0 , so 4 3 is a minimum point with minimum value 𝑔 3 , 8 4 3 , 8 3 = − 1 3 17

MODIFIED SECOND-ORDER CONDITIONS Other cases 2 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 ▪ < 0 ⇒ saddle point − 𝜖𝑦 2 𝜖𝑧 2 𝜖𝑦𝜖𝑧 2 < 0 ] [ 𝑔 𝑦𝑦 𝑔 𝑧𝑧 − 𝑔 𝑦𝑧 2 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 𝜖 2 𝑔 𝑦,𝑧 = 0 ⇒? (no conclusion from this method) ▪ − 𝜖𝑦 2 𝜖𝑧 2 𝜖𝑦𝜖𝑧 2 = 0 ] [ 𝑔 𝑦𝑦 𝑔 𝑧𝑧 − 𝑔 𝑦𝑧 18

EXTREME VALUES IN TWO DIMENSIONS In other cases: false (saddle point) 19

SUMMARY OPTIMALITY CONDITIONS 1. First-order conditions for a stationary point: 𝑦 = 0 and 𝑔 𝑔 𝑧 = 0 2. Second-order conditions (for solutions to 1.): 2 > 0 ⇒ yes, this is an extreme point 𝑔 𝑦𝑦 𝑔 𝑧𝑧 − 𝑔 𝑦𝑧 2.a 𝑦𝑦 < 0 ⇒ maximum (or 𝑔 𝑧𝑧 < 0 ) 𝑔 2.b 𝑦𝑦 > 0 ⇒ minimum (or 𝑔 𝑧𝑧 > 0 ) 𝑔 2 < 0 ⇒ saddle point 𝑔 𝑦𝑦 𝑔 𝑧𝑧 − 𝑔 𝑦𝑧 2 = 0 ⇒ ?(no conclusion) 𝑔 𝑦𝑦 𝑔 𝑧𝑧 − 𝑔 𝑦𝑧 20

EXERCISE 2 Given is 𝑔 𝑦, 𝑧 = 2𝑦𝑧 2 − 4𝑦 2 𝑧 3 . Determine the formula for distinguishing extreme values from other stationary points. 21

EXAMPLE Consider the function 𝑔 𝑦, 𝑧 = 1 2 𝑦 2 𝑓 𝑧 − 1 3 𝑦 3 − 𝑧𝑓 3𝑧 3 and 𝑓 − 1 1 1 Are 0, − 6 extreme points of 𝑔 ? 6 , − What is the nature of the extreme points? 23

EXAMPLE CONTINUED First, use first-order conditions to verify stationary points 𝜖𝑔 𝜖𝑦 = 𝑦𝑓 𝑧 − 𝑦 2 ▪ 𝜖𝑔 2 𝑦 2 𝑓 𝑧 − 𝑓 3𝑧 − 3𝑧𝑓 3𝑧 1 ▪ 𝜖𝑧 = Then check the nature of these points (if any) with the second-order conditions 𝜖 2 𝑔 𝜖𝑦 2 = 𝑓 𝑧 − 2𝑦 ▪ 𝜖 2 𝑔 1 2 𝑦 2 𝑓 𝑧 − 6𝑓 3𝑧 − 9𝑧𝑓 3𝑧 ▪ 𝜖𝑧 2 = 𝜖 2 𝑔 𝜖𝑦𝜖𝑧 = 𝑦𝑓 𝑧 ▪ 24

OLD EXAM QUESTION 23 April 2015, Q2c 25

OLD EXAM QUESTION 10 December 2014, Q2b 26

FURTHER STUDY Sydsæter et al. 5/E 13.1-13.3 Tutorial exercises week 4 local extreme value 27