# 14.7 - 14.7 MAXIMUM AND MINIMUM VALUES KIAM HEONG KWA 1...

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14.7: MAXIMUM AND MINIMUM VALUES KIAM HEONG KWA 1. Local Extreme Values and Critical Points A function f of two variables is said to have a local maximum at a point ( x M ,y M ) if f ( x,y ) f ( x M ,y M ) for all ( x,y ) suﬃciently close to ( x M ,y M ) . The number f ( x M ,y M ) is called a local maximum value of f . Similarly, f is said to have a local minimum at a point ( x m ,y m ) if f ( x,y ) f ( x m ,y m ) for all ( x,y ) suﬃciently near ( x m ,y m ) . The number f ( x m ,y m ) is called a local minimum value of f . Local maximum values and local minimum values are collectively known as local extreme values . On the other hand, a point ( x 0 ,y 0 ) in the domain of a function f ( x,y ) is called a critical point (or stationary point) of f if f x ( x 0 ,y 0 ) = f y ( x 0 ,y 0 ) = 0, i.e., f ( x 0 ,y 0 ) = 0 , or if one of these partial derivatives does not exist. Theorem 1. If a function f ( x,y ) has a local extreme value at a point ( x 0 ,y 0 ) , then the point ( x 0 ,y 0 ) must be a critical point of f . Proof. If f x ( x 0 ,y 0 ) does not exist, then there is nothing to prove. Sup- pose f x ( x 0 ,y 0 ) exists. Deﬁne g ( x ) = f ( x,y 0 ) for all x suﬃciently close to x 0 . Then g ( x 0 ) must be a local extreme value of g since f ( x 0 ,y 0 ) is a local extreme value of f , from which it follows that f x ( x 0 ,y 0 ) = g 0 ( x 0 ) = 0. Likewise, either f y ( x 0 ,y 0 ) does not exist or one can show that f y ( x 0 ,y 0 ) = 0. ± Remark 1. The idea used in the proof follows from the geometrical observation that if ( x 0 ,y 0 ,f ( x 0 ,y 0 )) is the highest point in one of its suﬃciently small neighborhood, then it is also the highest point along the curve r ( x ) = < x,y 0 ,f ( x,y 0 ) > in the neighborhood. Date : September 30, 2010. 1

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2 KIAM HEONG KWA Corollary 1. If f ( x,y ) is a diﬀerentiable function and has a local extreme value at a point ( x 0 ,y 0 ) , then all its directional derivatives vanish at ( x 0 ,y 0 ) . Proof. Since f is diﬀerentiable, f ( x 0 ,y 0 ) exist. Furthermore, for any unit vector u , D u f ( x 0 ,y 0 ) = f ( x 0 ,y 0 ) · u . The conclusion now follows from the previous theorem. ± Second Derivatives Test. Let f have continuous second partial deriva- tives near a critical point ( x 0 ,y 0 ). In particular, f x ( x 0 ,y 0 ) = f y ( x 0 ,y 0 ) = 0. The Hessian matrix of f at ( x 0 ,y 0 ) is the matrix ± f xx ( x 0 ,y 0 ) f xy ( x 0 ,y 0 ) f yx ( x 0 ,y 0 ) f yy ( x 0 ,y 0 ) ² . Note that f xy ( x 0 ,y 0 ) = f yx ( x 0 ,y 0 ) by the Clairaut’s theorem in section 14.3. The discriminant of f at ( x 0 ,y 0 ) is the determinant of the Hessian matrix, i.e., D ( x 0 ,y 0 ) = ³ ³ ³ ³ f xx ( x 0 ,y 0 ) f xy ( x 0 ,y 0 ) f yx ( x 0 ,y 0 ) f yy ( x 0 ,y 0 ) ³ ³ ³ ³ (1.1) = f xx ( x 0 ,y 0 ) f yy ( x 0 ,y 0 ) - [ f xy ( x 0 ,y 0 )] 2 . Suppose D ( x 0 ,y 0 ) > 0. If f xx ( x 0 ,y 0 ) < 0 or f yy ( x 0 ,y 0 ) < 0, then f ( x 0 ,y 0 ) is a local maximum. If
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14.7 - 14.7 MAXIMUM AND MINIMUM VALUES KIAM HEONG KWA 1...

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