is a maximizer. However this is not the case. We will now give a graphical argument explaining why argue a further condition must be satisfied.

Transcription

1 D. Maimization with two variables D. Sufficient conditions for a maimum Suppose that the second order conditions hold strictly. It is tempting to believe that this might be enough to ensure that is a maimizer. However this is not the case. We will now give a graphical argument eplaining why argue a further condition must be satisfied. Eample : f ( ) 6 The gradient vector is f (,6 ). Let m ( ) be the maimizing value of. This must satisfy the first order condition (, ). Therefore the vertical line in the figure below. Similarly, let m ( ) be the maimizing value of This must satisfy the first order condition (, ) 6 horizontal line. For m ( ). This is. Therefore m( ) 3. This is the to satisfy the first order conditions it must be the interaction point of both lines.. Therefore (,3) is the unique potential maimizer of ( ) f.

2 Since both partial derivatives are negative it is labelled as indicate the directions in which the function f( ) Z(, ) is increasing.. The directions of the arrows The other three zones are similarly labelled. Consider the point in the dotted zone Z(, ) to the North-East of. In this zone, the level set f ( ) f ( ) implicitly defines a function g( ), i.e. f g (, ( )) f ( ). Appealing to the Implicit Function Rule, the slope of this function is g ( ) f Since the partial derivatives are both strictly negative in zone Z(, ), the slope of the implicit function is negative. That is, the slope of the level set is negative. Arguing in the same way in each of the zones it follows that the slope of the level set is positive in Z(, ) and Z(,, ) but negative in Z(, ). Then the level set must be as depicted below. Alternatively, the derivative of the equation f (, g( )) f ( ) is g ( ). Since the partial derivatives are both negative, g ( ) is negative

3 Since the arrows show the directions in which f( ) is increasing, the points inside this ellipse are points in the superlevel set f ( ) f ( ). Therefore f ( ) f ( ). Since this argument holds it follows that is the unique maimizer. Eample : f ( ) 5 5 The gradient vector is (5,5 ). Let m ( ) be the maimizing value of. This must satisfy the first order condition (, ) 5. Therefore ( m ( ), ) 5 m ( ). Rearranging this equation, it follows that m ( ) (5 ). A symmetrical argument establishes that the maimizing value of is m ( ) (5 ) As you may readily confirm, (5,5) is the unique point satisfying both of these equations. The two maimizing functions are depicted below. Again we divide the depicted below. Arguing eactly as in Eample, the superlevel set through any point -plane into four zones. These are is an ellipse. 3

4 Since is in the superlevel set it follows that f ( ) f ( ). The 3D surface map for f( ) is depicted below. Eample 3: f ( )

5 The gradient vector is (3 4,3 4 ). As you may readily confirm, the maimizing functions are m ( ) (3 ) and 4 m ( ) (3 ) 4 (5,5) is the unique point satisfying both of these equations. The two cross sections through parallel to the aes are shown below. As you may check, the slope of each cross section is strictly decreasing. The full surface chart is shown below. 5

6 As a mountain climber, if you were standing at the point on the hill with coordinates (5,5) would see that while the slope to the North and East is zero, this is a saddle point, not a mountain top. Again we divide the -plane into four zones. These are depicted below. Consider zone Z(, ). Since one partial derivative is positive in this zone and the other is negative, the slope of the, you level set is positive. Moreover, from any in this zone, the points in the superlevel set f ( ) f ( ) all lie to the South-East. Thus f ( ) f ( ) and so is not a maimizer. In Eamples and 3, the cross partial derivative f. As a result, the slopes of the two boundary lines are negative. As Eample shows, is a maimizer of the line labelled () has a more negative slope than the line labelled (). The second line m ( ) has a slope of m ( ). The first line m ( ) has a slope of / m ( ). Thus is a maimizer if m ( ). m ( ) 6

7 Multiplying both sides of the inequality by a negative number reverses the direction of the inequality. Therefore m ( ) m ( ) The remaining possibility is that the cross partial derivative is strictly positive. Applying essentially the same argument we have the following result. Sufficient condition for to be a maimum for a quadratic function If the second order necessary conditions hold strictly and the maimizing functions m( ) and m ( ) implicitly defined by ( ) and ( ) satisfy m ( ) m ( ), then satisfying the first order conditions is the unique maimizer. 7