MAXIMA & MINIMAThe single-variable definitions and theorems relating to extermals can be extended to apply to multivariable calculus.

Defn: f x0 y0,( ) is a Relative Maximum if there is a disk centered at x0 y0,( ) such that f x0 y0,( ) f x y,( )≥ for all points x y,( ) in the disk. If the disk includes all points in the domain of " f ", then that Relative Maximum is the Absolute Maximum.

Defn : f x0 y0,( ) is a Relative Minimum if there is a disk centered at x0 y0,( ) such that f x0 y0,( ) f x y,( )≤ for all points x y,( ) in the disk. If the disk includes all points in the domain of " f ", then that Relative Maximum is the Absolute Minimum.

Defn : Interior Point xi yi,( ) of a domain " D " of points lies within " D ". This implies that some disk centered at xi yi,( ) can be drawn so as to contain only points within " D ".

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Defn : Boundary Point xb yb,( ) of a domain " D " of points lies on the boundary of " D ". This implies that every disk centered at xb yb,( ) contains points within " D " and

outside of " D ".

Theorem : If f x y,( ) is continuous on a closed and bounded set " R ", then " f " has both an Absolute Maximum & Absolute Minimum on" R ". ( Extreme -Value Theorem)

Theorem : If f x y,( ) has a relative extremum at x0 y0,( ) and if the first partial derivatives exist at that point, then both must be zero at that point: fx x0 y0,( ) fy x0 y0,( )= 0= .

Defn : xi yi,( ) is a Critical Point of " f " if both first partial derivatives are zero at that point or at least one does not exist .

Theorem : If f x y,( ) has an Absolute Extremum at an Interior Point of its domain, then this extremum occurs at a Critical Point.

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The second derivative test for extremals in the single-variable calculus is not so easily extended to the 2-variable calculus.

Theorem : Suppose f x y,( ) has continuous second partials in some disk centered at the Critical Point x0 y0,( ). The Sign of the Number: " D ", where,D fxx x0 y0,( )⋅ fyy⋅ x0 y0,( )⋅ fxy x0 y0,( )⋅⎡⎣ ⎤⎦

2−= can be used to determine the nature of that critical point.

(a). D 0> , f x0 y0,( ) is a relative minimum OR a relative maximum.

(b). D 0< , x0 y0,( ) is a Saddle Point.

(c). D 0= , test is useless.

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D fxx 2 1−,( )⋅ fyy⋅ 2 1−,( )⋅ fxy 2 1−,( )⋅⎡⎣ ⎤⎦2−= 3 0>=

fxy 2 1−,( ) 1=fyy 2 1−,( ) 2=fxx 2 1−,( ) 2=

fxy x y,( ) 1=fyy x y,( ) 2=fxx x y,( ) 2=

Critical Point: 2 1−, 3−,( )

x2

− 3 2 x⋅−=

y x2

−=y 3 2 x⋅−=

2 x⋅ y+ 3− 0= x 2 y⋅+ 0=

Critical Points are where the first partials vanish.

yf∂

∂x 2 y⋅+=

xf∂

∂2 x⋅ y+ 3−=

f x y,( ) x2 x y⋅+ y2+ 3 x⋅−=

Example # 1: Locate all relative maxima, minima, and saddle points.

Because D 0> , " f " is either a Relative Maximum or Relative Minimum. " f " is a Relative Minimum if the "unmixed" second partials are positive and a Relative Maximum if they are negative. Since fxx 2 1−,( ) 0> , the Critical Point is a Relative Minimum.

xy

z

Relative Minimum @ (2,- 1,-3)

Example # 2: Locate all relative maxima, minima, and saddle points.

f x y,( ) e x2 y2+ 2 x⋅+( )−= Page 5 of 13

xf∂

∂2 x⋅ 2+( )− f x y,( )⋅=

yf∂

∂2 y⋅( )− f x y,( )⋅=

Critical Points are where the first partials vanish. Since f(x,y) can never be zero,

2 x⋅ 2+ 0= 2− y⋅ 0=

x 1−= y 0=

Critical Point: 1− 0, e,( )

fxx x y,( ) 2−( ) f x y,( )⋅ 2 x⋅ 2+( )2 f x y,( )⋅+=

fyy x y,( ) 2−( ) f x y,( )⋅ 2 y⋅( )2 f x y,( )⋅+=

fxy x y,( )⋅ 2 x⋅ 2+( )− 2 y⋅( )⋅ f x y,( )⋅=

fxx 1− 0,( ) 2− e⋅= fyy 1− 0,( ) 2− e⋅=

fxy 1− 0,( ) 0=

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D fxx 1− 0,( )⋅ fyy⋅ 1− 0,( )⋅ fxy 1− 0,( )⋅⎡⎣ ⎤⎦2−= 4 e2⋅=

Because D 0> , " f " is either a Relative Maximum or Relative Minimum. " f " is a Relative Minimum if the "unmixed" second partials are positive and a Relative Maximum if they are negative. Since fyy 1− 0,( ) 0< , the Critical Point is a Relative Maximum.

z

y x

Relative Maximum @ (-1,0,e)

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In this case, D 0< , so this critical point is aSaddle Point.

fxy 3 1,( ) 1=fyy 3 1,( ) 0=fxx 3 1,( ) 0=

fxy x y,( ) 1=fyy x y,( ) 0=fxx x y,( ) 0=

Critical Point: 3 1, 3−,( )

y 1− 0= x 3− 0=

Critical Points are where the first partials vanish.

yf∂

∂x 3−=

xf∂

∂y 1−=

R : Triangular Region with these vertices:0 0,( ), 0 4,( ), 5 0,( ).

f x y,( ) x y⋅ x− 3 y⋅−=

Example # 3: Find the absolute extrema of the given function on the indicated closed and bounded set " R ".

Since " R " is "closed", there must exist both an Absolute Minimum & an Absolute Maximimum at Interior Points or on Boundary Points.

Interior Points : There is only one critical point: 3 1, 3−,( ) and it lies in the Interior.

Boundary Points : There are three sides to the triangular region, " R ". Look for extremals on all three sides.

0 1 2 3 4 5

1

2

3

4R:Triangular Region

x

y

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Side # 1 : x 0= .

f 0 y,( ) 3− y⋅= No Critical Points.

Side # 2 : y 0= .

f x 0,( ) x−= No Critical Points.

Side # 3 : y 4 45

x⋅−= .

f x 4 45

x⋅−⎛⎜⎝

⎞⎟⎠

,⎡⎢⎣

⎤⎥⎦

x 4 45

x⋅−⎛⎜⎝

⎞⎟⎠

⋅ x− 3 4 45

x⋅−⎛⎜⎝

⎞⎟⎠

⋅−=

f x 4 45

x⋅−⎛⎜⎝

⎞⎟⎠

,⎡⎢⎣

⎤⎥⎦

x 3−( ) 4 45

x⋅−⎛⎜⎝

⎞⎟⎠

⋅ x−=

xf x 4 4

5x⋅−⎛

⎜⎝

⎞⎟⎠

,⎡⎢⎣

⎤⎥⎦

dd

4 45

x⋅−⎛⎜⎝

⎞⎟⎠

45

x 3−( )⋅− 1−=

xf x 4 4

5x⋅−⎛

⎜⎝

⎞⎟⎠

,⎡⎢⎣

⎤⎥⎦

dd

5.4 85

x⋅−= 0=

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x 5 5.4( )⋅8

= 3.375=

y 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 Boundary 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 4,( ) 12−= f 5 0,( ) 5−=

Evidently , the two vertices at 0 0,( ) & 0 4,( ) yield, respectively, an Absolute Maximum & Absolute Minimum for " f x y,( ) " on " R ".

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y

z

x

Key Points in "R"

At the Saddle Point: 3 1, 3−,( ), the contours in the "x" and "y" directionsare horizontal while the contoursalong y x= and y x−= increase and decrease respectively.

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Saddle Point@ (3,1)

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