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Chapter 16 Partial Differentiation

Chapter 16 Partial Differentiation

Chapter 16 Partial Differentiation

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<strong>16</strong>.7 Maxima and minima 385D(x 0 ,y 0 ) = f xx (x 0 ,y 0 )f yy (x 0 ,y 0 ) − f xy (x 0 ,y 0 ) 2 . If D > 0 and f xx (x 0 ,y 0 ) < 0 there isa local maximum at (x 0 ,y 0 ); if D > 0 and f xx (x 0 ,y 0 ) > 0 there is a local minimum at(x 0 ,y 0 ); if D < 0 there is neither a maximum nor a minimum at (x 0 ,y 0 ); if D = 0, thetest fails.EXAMPLE <strong>16</strong>.7.2 Verify that f(x,y) = x 2 +y 2 has a minimum at (0,0).First, we compute all the needed derivatives:f x = 2x f y = 2y f xx = 2 f yy = 2 f xy = 0.The derivatives f x and f y are zero only at (0,0). Applying the second derivative test there:D(0,0) = f xx (0,0)f yy (0,0)−f xy (0,0) 2 = 2·2−0 = 4 > 0,so there is a local minimum at (0,0), and there are no other possibilities.EXAMPLE <strong>16</strong>.7.3 Find all local maxima and minima for f(x,y) = x 2 −y 2 .The derivatives:f x = 2x f y = −2y f xx = 2 f yy = −2 f xy = 0.Again there is a single critical point, at (0,0), andD(0,0) = f xx (0,0)f yy (0,0)−f xy (0,0) 2 = 2·−2−0 = −4 < 0,so there is neither a maximum nor minimum there, and so there are no local maxima orminima. The surface is shown in figure <strong>16</strong>.7.1.EXAMPLE <strong>16</strong>.7.4 Find all local maxima and minima for f(x,y) = x 4 +y 4 .The derivatives:f x = 4x 3 f y = 4y 3 f xx = 12x 2 f yy = 12y 2 f xy = 0.Again there is a single critical point, at (0,0), andD(0,0) = f xx (0,0)f yy (0,0)−f xy (0,0) 2 = 0·0−0 = 0,so we get no information. However, in this case it is easy to see that there is a minimumat (0,0), because f(0,0) = 0 and at all other points f(x,y) > 0.EXAMPLE <strong>16</strong>.7.5 Find all local maxima and minima for f(x,y) = x 3 +y 3 .

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