MAXIMA & MINIMA The single-variable definitions and theorems relating to etermals can be etended to appl to multivariable calculus. ( ) is a Relative Maimum if there ( ) such that ( ) f(, for all points (, in the Defn: f 0, 0 is a disk centered at 0, 0 f 0, 0 disk. If the disk includes all points in the domain of " f ", then that Relative Maimum is the Absolute Maimum. ( ) is a Relative Minimum if there ( ) such that ( ) f(, for all points (, in the Defn : f 0, 0 is a disk centered at 0, 0 f 0, 0 disk. If the disk includes all points in the domain of " f ", then that Relative Maimum is the Absolute Minimum. Defn : Interior Point ( i, i ) of a domain " D " of points lies within " D ". This implies that some disk centered at i, i so as to contain onl points within " D ". ( ) can be drawn Page 1 of 13
Defn : Boundar Point ( b, b ) of a domain " D " of points lies on the boundar of " D ". This implies that ever disk centered at b, b outside of " D ". ( ) contains points within " D " and Theorem : If f(, is continuous on a closed and bounded set " R ", then " f " has both an Absolute Maimum & Absolute Minimum on " R ". ( Etreme -Value Theorem) Theorem : If f(, has a relative etremum at ( 0, 0 ) and if the first partial derivatives eist at that point, then both must be zero at that point: f ( 0, 0 ) f ( 0, 0 ) 0. Defn : ( i, i ) is a Critical Point of " f " if both first partial derivatives are zero at that point or at least one does not eist. Theorem : If f(, has an Absolute Etremum at an Interior Point of its domain, then this etremum occurs at a Critical Point. Page 2 of 13
The second derivative test for etremals in the single-variable calculus is not so easil etended to the 2-variable calculus. Theorem : Suppose f(, has continuous second partials in some disk centered at the Critical Point ( 0, 0 ). The Sign of the Number: " D ", where, ( ) ( ) ( ) D f 0, 0 f 0, 0 f 0, 0 can be used to determine the nature of that critical point. (a). D > 0, f 0, 0 OR a relative maimum. (b). D < 0, 0, 0 2 ( ) is a relative minimum ( ) is a Saddle Point. (c). D 0, test is useless. Page 3 of 13
Eample # 1: Locate all relative maima, minima, and saddle points. f 2 f(, 2 + + 2 3 2 Critical Points are where the first partials vanish. + 3 0 + 2 0 3 2 + 3 f 3 2 2 Critical Point: ( 2, 1, 3) + 2 2 f (, 2 f ( 2, 1) 2 f (, 2 f ( 2, 1) 2 f (, 1 f ( 2, 1) 1 D f ( 2, 1) f ( 2, 1) f ( 2, 1) 3 > 0 2 Page of 13
Because D > 0, " f " is either a Relative Maimum or Relative Minimum. " f " is a Relative Minimum if the "unmied" second partials are positive and a Relative Maimum if the are negative. Since f ( 2, 1) > 0, the Critical Point is a Relative Minimum. Relative Minimum @ (2,- 1,-3) z Eample # 2: Locate all relative maima, minima, and saddle points. 2 + 2 + 2 f(, e Page 5 of 13 ( )
f ( 2 + 2) f(, f ( 2 ) f(, Critical Points are where the first partials vanish. Since f(, can never be zero, 2 + 2 0 2 0 1 0 Critical Point: ( 1, 0, e) f (, ( 2) f(, + ( 2 + 2) 2 f(, f (, ( 2) f(, + ( 2 ) 2 f(, f (, ( 2 + 2) ( 2 ) f(, f ( 1, 0) 2 e f ( 1, 0) 2 e f ( 1, 0) 0 Page 6 of 13
D f ( 1, 0) f ( 1, 0) f ( 1, 0) 2 e 2 Because D > 0, " f " is either a Relative Maimum or Relative Minimum. " f " is a Relative Minimum if the "unmied" second partials are positive and a Relative Maimum if the are negative. Since f ( 1, 0) < 0, the Critical Point is a Relative Maimum. Relative Maimum @ (-1,0,e) z Page 7 of 13
Eample # 3: Find the absolute etrema of the given function on the indicated closed and bounded set " R ". f(, 3 R : Triangular Region with these vertices: ( 0, 0), ( 0, ), ( 5, 0). f 1 f 3 Critical Points are where the first partials vanish. 1 0 3 0 Critical Point: ( 3, 1, 3) f (, 0 f ( 3, 1) 0 f (, 0 f ( 3, 1) 0 f (, 1 f ( 3, 1) 1 In this case, D < 0, so this critical point is a Saddle Point. Page 8 of 13
Since " R " is "closed", there must eist both an Absolute Minimum & an Absolute Maimimum at Interior Points or on Boundar Points. Interior Points : There is onl one critical point: ( 3, 1, 3) and it lies in the Interior. Boundar Points : There are three sides to the triangular region, " R ". Look for etremals on all three sides. R:Triangular Region 3 2 1 0 1 2 3 5 Page 9 of 13
Side # 1 : 0. f( 0, 3 No Critical Points. Side # 2 : 0. f(, 0) No Critical Points. Side # 3 : 5. f, 5 5 3 5 f, 5 ( 3) 5 d f, d 5 5 5 ( 3) 1 d f, d 5 8 5. 5 0 Page 10 of 13
5 ( 5.) 3.375 8 ( 3.375) 1.3 Critical Point : ( 3.375, 1.3, 2.89) Since the Interior Critical Point has a "z-value" less than the the Boundar Critical Point, the Interior Critical Point is candidate for the Absolute Minimum on " R ". Now we must evaluate " f " at the three vertices. f( 0, 0) 0 f( 0, ) 12 f( 5, 0) 5 Evidentl, the two vertices at ( 0, 0) & ( 0, ) ield, respectivel, an Absolute Maimum & Absolute Minimum for " f(, " on " R ". Page 11 of 13
Ke Points in "R" z At the Saddle Point: ( 3, 1, 3), the contours in the "" and "" directions are horizontal while the contours along and increase and decrease respectivel. Page 12 of 13
Saddle Point@ (3,1) Page 13 of 13