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

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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

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BUSINESS MATHEMATICS

Extreme values in two dimensions

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

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EXTREME VALUES IN TWO DIMENSIONS

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EXTREME VALUES IN ONE DIMENSION

Consider a smooth function 𝑔 𝑦

  • 1. Necessary first-order condition for extreme value

𝑒𝑔 𝑦 𝑒𝑦 = 0 β‡’ stationary point

  • 2. Sufficient second-order condition at stationary point

β–ͺ

𝑒2𝑔 𝑦 𝑒𝑦2

< 0 β‡’ maximum β–ͺ

𝑒2𝑔 𝑦 𝑒𝑦2

> 0 β‡’ minimum β–ͺ

𝑒2𝑔 𝑦 𝑒𝑦2

= 0 β‡’?

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EXTREME VALUES IN ONE DIMENSION

Consider a smooth function 𝑔 𝑦

  • 1. Necessary first-order condition for extreme value

𝑒𝑔 𝑦 𝑒𝑦 = 0 β‡’ stationary point

  • 2. Sufficient second-order condition at stationary point

β–ͺ

𝑒2𝑔 𝑦 𝑒𝑦2

< 0 β‡’ maximum β–ͺ

𝑒2𝑔 𝑦 𝑒𝑦2

> 0 β‡’ minimum β–ͺ

𝑒2𝑔 𝑦 𝑒𝑦2

= 0 β‡’?

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

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

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

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MODIFIED SECOND-ORDER CONDITIONS

Additional sufficient second-order condition for extreme point β–ͺ not

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦2

> 0 and

πœ–2𝑔 𝑦,𝑧 πœ–π‘§2

> 0 β–ͺ nor

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦2

< 0 and

πœ–2𝑔 𝑦,𝑧 πœ–π‘§2

< 0 But…..

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MODIFIED SECOND-ORDER CONDITIONS

  • 1. Testing for extreme point:

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦2 πœ–2𝑔 𝑦,𝑧 πœ–π‘§2

βˆ’

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦πœ–π‘§ 2

> 0 [𝑔

𝑦𝑦𝑔 𝑧𝑧 βˆ’ 𝑔 𝑦𝑧 2 > 0 or 𝑔 𝑦𝑦𝑔 𝑧𝑧 βˆ’ 𝑔 𝑦𝑧𝑔 𝑧𝑦 > 0]

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MODIFIED SECOND-ORDER CONDITIONS

  • 1. Testing for extreme point:

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦2 πœ–2𝑔 𝑦,𝑧 πœ–π‘§2

βˆ’

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦πœ–π‘§ 2

> 0 [𝑔

𝑦𝑦𝑔 𝑧𝑧 βˆ’ 𝑔 𝑦𝑧 2 > 0 or 𝑔 𝑦𝑦𝑔 𝑧𝑧 βˆ’ 𝑔 𝑦𝑧𝑔 𝑧𝑦 > 0] Recall that 𝑔

𝑦𝑧 = 𝑔 𝑧𝑦

so that 𝑔

𝑦𝑧 2 = 𝑔 𝑦𝑧𝑔 𝑧𝑦

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MODIFIED SECOND-ORDER CONDITIONS

  • 1. Testing for extreme point:

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦2 πœ–2𝑔 𝑦,𝑧 πœ–π‘§2

βˆ’

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦πœ–π‘§ 2

> 0 [𝑔

𝑦𝑦𝑔 𝑧𝑧 βˆ’ 𝑔 𝑦𝑧 2 > 0 or 𝑔 𝑦𝑦𝑔 𝑧𝑧 βˆ’ 𝑔 𝑦𝑧𝑔 𝑧𝑦 > 0]

  • 2. Then, maximum point:

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦2

< 0 or

πœ–2𝑔 𝑦,𝑧 πœ–π‘§2

< 0 [𝑔

𝑦𝑦 < 0 or 𝑔 𝑧𝑧 < 0]

minimum point:

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦2

> 0 or

πœ–2𝑔 𝑦,𝑧 πœ–π‘§2

> 0 [𝑔

𝑦𝑦 > 0 or 𝑔 𝑧𝑧 > 0]

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EXAMPLE

Consider 𝑔 𝑦, 𝑧 = 𝑦2 βˆ’ 𝑦𝑧 + 𝑧2 βˆ’ 4𝑧 + 5

  • 1. Finding stationary points:

β–ͺ

πœ–π‘” πœ–π‘¦ = 2𝑦 βˆ’ 𝑧 = 0

β–ͺ

πœ–π‘” πœ–π‘§ = βˆ’π‘¦ + 2𝑧 βˆ’ 4 = 0

Result: 𝑦, 𝑧 =

4 3, 8 3

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EXAMPLE CONTINUTED

  • 2. Second-order test for 𝑦, 𝑧 =

4 3, 8 3 :

πœ–2𝑔 πœ–π‘¦2 πœ–2𝑔 πœ–π‘§2 βˆ’ πœ–2𝑔 πœ–π‘¦πœ–π‘§

2

= 2 Γ— 2 βˆ’ βˆ’1 2 = 3 So

4 3, 8 3 is an extreme point.

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EXAMPLE CONTINUTED

  • 2. Second-order test for 𝑦, 𝑧 =

4 3, 8 3 :

πœ–2𝑔 πœ–π‘¦2 πœ–2𝑔 πœ–π‘§2 βˆ’ πœ–2𝑔 πœ–π‘¦πœ–π‘§

2

= 2 Γ— 2 βˆ’ βˆ’1 2 = 3 So

4 3, 8 3 is an extreme point.

Further

πœ–2𝑔 πœ–π‘¦2 = 2 > 0, so 4

3, 8 3 is a minimum point with minimum value 𝑔 4 3, 8 3 = βˆ’1 3

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EXAMPLE CONTINUTED

  • 2. Second-order test for 𝑦, 𝑧 =

4 3, 8 3 :

πœ–2𝑔 πœ–π‘¦2 πœ–2𝑔 πœ–π‘§2 βˆ’ πœ–2𝑔 πœ–π‘¦πœ–π‘§

2

= 2 Γ— 2 βˆ’ βˆ’1 2 = 3 So

4 3, 8 3 is an extreme point.

Further

πœ–2𝑔 πœ–π‘¦2 = 2 > 0, so 4

3, 8 3 is a minimum point with minimum value 𝑔 4 3, 8 3 = βˆ’1 3

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MODIFIED SECOND-ORDER CONDITIONS

Other cases β–ͺ

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦2 πœ–2𝑔 𝑦,𝑧 πœ–π‘§2

βˆ’

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦πœ–π‘§ 2

< 0 β‡’ saddle point [𝑔

𝑦𝑦𝑔 𝑧𝑧 βˆ’ 𝑔 𝑦𝑧 2 < 0]

β–ͺ

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦2 πœ–2𝑔 𝑦,𝑧 πœ–π‘§2

βˆ’

πœ–2𝑔 𝑦,𝑧 πœ–π‘¦πœ–π‘§ 2

= 0 β‡’? (no conclusion from this method) [𝑔

𝑦𝑦𝑔 𝑧𝑧 βˆ’ 𝑔 𝑦𝑧 2 = 0]

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

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SUMMARY OPTIMALITY CONDITIONS

  • 1. First-order conditions for a stationary point:

𝑔

𝑦 = 0 and 𝑔 𝑧 = 0

  • 2. Second-order conditions (for solutions to 1.):

𝑔

𝑦𝑦𝑔 𝑧𝑧 βˆ’ 𝑔 𝑦𝑧 2 > 0 β‡’ yes, this is an extreme point

2.a 𝑔

𝑦𝑦 < 0 β‡’ maximum (or 𝑔 𝑧𝑧 < 0)

2.b 𝑔

𝑦𝑦 > 0 β‡’ minimum (or 𝑔 𝑧𝑧 > 0)

𝑔

𝑦𝑦𝑔 𝑧𝑧 βˆ’ 𝑔 𝑦𝑧 2 < 0 β‡’ saddle point

𝑔

𝑦𝑦𝑔 𝑧𝑧 βˆ’ 𝑔 𝑦𝑧 2 = 0 β‡’ ?(no conclusion)

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EXERCISE 2

Given is 𝑔 𝑦, 𝑧 = 2𝑦𝑧2 βˆ’ 4𝑦2𝑧3. Determine the formula for distinguishing extreme values from other stationary points.

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EXAMPLE

Consider the function 𝑔 𝑦, 𝑧 = 1 2 𝑦2𝑓𝑧 βˆ’ 1 3 𝑦3 βˆ’ 𝑧𝑓3𝑧 Are 0, βˆ’

1 3 and π‘“βˆ’1

6, βˆ’

1 6 extreme points of 𝑔?

What is the nature of the extreme points?

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EXAMPLE CONTINUED

First, use first-order conditions to verify stationary points β–ͺ

πœ–π‘” πœ–π‘¦ = 𝑦𝑓𝑧 βˆ’ 𝑦2

β–ͺ

πœ–π‘” πœ–π‘§ = 1 2 𝑦2𝑓𝑧 βˆ’ 𝑓3𝑧 βˆ’ 3𝑧𝑓3𝑧

Then check the nature of these points (if any) with the second-order conditions β–ͺ

πœ–2𝑔 πœ–π‘¦2 = 𝑓𝑧 βˆ’ 2𝑦

β–ͺ

πœ–2𝑔 πœ–π‘§2 = 1 2 𝑦2𝑓𝑧 βˆ’ 6𝑓3𝑧 βˆ’ 9𝑧𝑓3𝑧

β–ͺ

πœ–2𝑔 πœ–π‘¦πœ–π‘§ = 𝑦𝑓𝑧

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OLD EXAM QUESTION 23 April 2015, Q2c

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OLD EXAM QUESTION 10 December 2014, Q2b

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