A local minimum or local maximum for a function of two variables f(x,y) are defined analogously to the way they are defined for a function of one variable:A point (x0 , y0) is called a [local minimum / local maximum] , if

for all points (x,y) in a disk of positive radius about the point (x0 , y0) .A point (x0 , y0) is called a [global minimum / global maximum] , if

for all points (x,y) in the domain .A point at which either a minimum or a maximum occurs is called an

In single variable calculus, we begin the search for extrema by setting the first derivative equal to zero to obtain critical points. The critical points of a function f(x,y) are all points at which either

[ f(x,y) f(x0 , y0) / f(x,y) f(x0 , y0) ]

[ f(x,y) f(x0 , y0) / f(x,y) f(x0 , y0) ]

extremum.

(1) fx = fy = 0 (where the tangent plane is parallel to the xy plane), or (2) f is not differentiable.

Does the function f(x,y) = 2(x2+y2)e have any extrema?–(x2+y2)

If (x0 , y0) is a local extremum of f(x,y), and the first partial derivatives exist at (x0 , y0), then (x0 , y0) must be a critical point of f(x,y).

Theorem 4 on page 208 states this for any function f from Rn to R .It is possible that (x0 , y0) could be a critical point of f(x,y) and not be a local extremum of f(x,y). Such a point

By inspection, we find that

Find the critical points of the function.

since f(0,0) = 0 and f(x,y) is never negative, the point (0 , 0) is a global minimum.

will be like an inflection point in single variable calculusorwill be what is called a saddle point. (See the figure on page 210.)

Does the function f(x,y) = 2(x2+y2)e have any extrema?–(x2+y2)

By inspection, we find that since f(0,0) = 0 and f(x,y) is never negative, the point (0 , 0) is a global minimum.

Find the critical points of the function.

f(x,y) = 2(x2+y2) e

fx =

fy =

–(x2+y2)

4x e – 4x(x2+y2) e–(x2+y2) –(x2+y2)

4y e – 4y(x2+y2) e–(x2+y2) –(x2+y2)

The critical points are

Since the value of the function can never be negative, (0,0) must be a global minimum point. From the graph on page 210, we see that each point on the circle x2 + y2 = 1 is a global maximum point.

(0,0) and all points on the circle x2 + y2 = 1.

Find all critical points of f(x,y) = x2y + xy2

fx = fy =2xy + y2 x2 + 2xy

2xy + y2 = 0 when either y = 0 or y = –2x

x2 + 2xy = 0 when either x = 0 or y = –x/2

The only critical point is (0,0). Notice that if x = y = t, then f(x,y) = 2t3 which can be either positive or negative no matter how close (t,t) is to (0,0). This implies that (0,0) is not a local extremum.

Suppose f(x,y) is differentiable at a critical point x0 = (x0 , y0). Then, for points x0 + h which are “close” to x0 = (x0 , y0) the Taylor expansion tells usf(x0 + h) f(x0) +

fx(x0) fy(x0)h1

h2

fxx(x0) fyx(x0)

fxy(x0) fyy(x0)h1 h2

h1

h2

1— 2

f(x0 + h) – f(x0)

fxx(x0) fyx(x0)

fxy(x0) fyy(x0) + h1 h2

h1

h2

1— 2

= fxx(x0) h12 + 2fxy(x0) h1h2 + fyy(x0) h2

2 1— 2

= fxx(x0) + 2fxy(x0) + fyy(x0) h2

2

—— 2

h12

—h2

2

h1—h2

A x2 B x C

Recall from Precalculus that when the roots of Ax2 + Bx + C are complex numbers, then this function will either always be positive or always be negative (since the graph of the function can never cross the x axis forcing the graph to be either always above the x axis or always below the x axis.)Consequently, Ax2 + Bx + C will either always be positive or always be negative if B2 – 4AC < 0 which is the same as 4AC – B2 > 0 .

will either always be positive or always be negative when

f(x0 + h) – f(x0) fxx(x0) + 2fxy(x0) + fyy(x0) h2

2

—— 2

h12

—h2

2

h1—h2

A x2 B x C

4[fxx(x0)][fyy(x0)] – [2fxy(x0)]2 > 0

[fxx(x0)][fyy(x0)] – [fxy(x0)]2 > 0

fxx(x0) fyx(x0)

fxy(x0) fyy(x0)det > 0

We shall label this Hessian matrix as A.

This suggests that if det(A) = fxx(x0) fyy(x0) – [fxy(x0)]2 > 0, then x0 is a , but if det(A) = fxx(x0) fyy(x0) – [fxy(x0)]2 < 0, then

x0 is alocal extremum

saddle point.

The determinant det(A) = fxx(x0) fyy(x0) – [fxy(x0)]2 is called thediscriminant of the Hessian and is often represented by D.Suppose det(A) = fxx(x0) fyy(x0) – [fxy(x0)]2 > 0. Then if fxx(x0) and fyy(x0) are both positive, x0 is a , but if fxx(x0) and fyy(x0) are both negative, x0 is a

local minimumlocal maximum.

If det(A) = fxx(x0) fyy(x0) – [fxy(x0)]2 = 0, then we cannot tell whether x0 is an extremum or not.

Theorem 6 on page 216 is a statement of the Second Derivative test for functions of two variables.

Find all critical points of f(x,y) = x2 + 3xy + y2 , and determine whether each is a local minimum, a local maximum, or a saddle point.

fx = fy =2x + 3y 3x + 2y

The only critical point is (0,0).

fxx = fyy = fxy =

D =

2 32

(2)(2) – (3)2 = – 5

Since D < 0, then (0,0) is a saddle point.

Find all critical points of f(x,y) = 3x2 – 5xy + 3y2 , and determine whether each is a local minimum, a local maximum, or a saddle point.

fx = fy =6x – 5y –5x + 6y

The only critical point is (0,0).

fxx = fyy = fxy =

D =

6 –56

(6)(6) – (–5)2 = 11

Since D > 0, and fxx(0,0) > 0, then (0,0) is a local minimum.

Find all critical points of f(x,y) = x sin y , and determine whether each is a local minimum, a local maximum, or a saddle point.

fx = fy =sin y x cos y

sin y = 0 when y = n for any integer n

x cos y = 0 when either x = 0 or y = (n + 1/2) for any integer n

The critical points are (0 , n) for any integer n.

fxx = fyy = fxy =

D =

0 cos y– x sin y

– cos2y

Since D < 0 for all critical points, then all critical points are saddle points.

Find all critical points of z = (x2 – y2) e , and determine whether each is a local minimum, a local maximum, or a saddle point.

fx =

fy =

(–x2–y2) / 2

x[2 – x2 + y2] e (–x2–y2) / 2

y[– 2 – x2 + y2] e (–x2–y2) / 2

The critical points are

(0,0), (2,0), (–2,0), (0,2), (0,–2)

fxx =

fyy =

fxy =

[2 – 5x2 + x2(x2 – y2) + y2] e (–x2–y2) / 2

[– 2 + 5y2 + y2(x2 – y2) – x2] e (–x2–y2) / 2

xy(x2 – y2) e (–x2–y2) / 2

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

D = (0,0) is

fxx(2,0) = fxx(–2,0) = D =

fyy(2,0) = fyy(–2,0) = (2,0) and (–2,0) are

fxy(2,0) = fxy(–2,0) =

fxx(0,2) = fxx(0,–2) = D =

fyy(0,2) = fyy(0,–2) = (0,2) and (0,–2) are

fxy(0,2) = fxy(0,–2) =

2 – 2 0

– 4 a saddle point

– 4e–1

– 4e–1

0

16e–2

each a local maximum

4e–1

4e–1

0

16e–2

each a local minimum

Find the minimum distance from the origin to a point on the plane

x + 3y – z = 6 .

The distance from the origin to any point (x,y,z) is

Minimizing this distance is the same as minimizing

If (x,y,z) is a point on the plane, then

Consequently, we want to minimize the function f(x,y) =

fx = fy =

The only critical point is

2x + 2(x + 3y – 6) 2y + 6(x + 3y – 6)

(6/11 , 18/11) .

(x–0)2 + (y–0)2 + (z–0)2 .

x2 + y2 + z2 .

z = x + 3y – 6 .

x2 + y2 + (x + 3y – 6)2 .

Since f(x,y) only becomes large as x and y become large, then f(x,y) must attain a minimum somewhere and at (6/11 , 18/11) is the only possibility.

The minimum distance from the origin to a point on the plane is the distance from the origin to the point

This distance is

(6/11 , 18/11 , –6/11) .

6/11 .

(6/11 , 18/11) .

Since f(x,y) only becomes large as x and y become large, then f(x,y) must attain a minimum somewhere and at (6/11 , 18/11) is the only possibility.

The only critical point is

Find the minimum distance from the origin to a point on the surface

g(x,y) = 1/ (xy) .

The distance from the origin to any point (x,y,z) is

Minimizing this distance is the same as minimizing

If (x,y,z) is a point on the surface, then

Consequently, we want to minimize the function f(x,y) =

fx = fy =

The critical points are

2x – 2/(x3y2) 2y – 2/(x2y3)

(1,1), (–1,1), (1,–1), (–1,–1) .

(x–0)2 + (y–0)2 + (z–0)2 .

x2 + y2 + z2 .

z = 1/ (xy) .

x2 + y2 + 1/(x2y2) .

Since f(x,y) only becomes large as (x,y) moves away from any one of these four points in the same quadrant, at least one of these four points must be a minimum point.

The distance from the origin to each of the points

(1,1,1), (–1,1,–1), (1,–1,–1), (–1,–1,1)

on the surface is , which must be the minimum distance.3

The critical points are (1,1), (–1,1), (1,–1), (–1,–1) .Since f(x,y) only becomes large as (x,y) moves away from any one of these four points in the same quadrant, at least one of these four points must be a minimum point.