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
[ f(x,y)  f(x0 , y0) / f(x,y)  f(x0 , y0) ]
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
[ f(x,y)  f(x0 , y0) / f(x,y)  f(x0 , y0) ]
for all points (x,y) in the domain .
A point at which either a minimum or a maximum occurs is called an
extremum.
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
(1) fx = fy = 0 (where the tangent plane is parallel to the xy plane), or
(2) f is not differentiable.
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
will be like an inflection point in single variable calculus
or
will be what is called a saddle point. (See the figure on page 210.)
Does the function
f(x,y) = 2(x2+y2)e
–(x2+y2)
have any extrema?
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.
Does the function
f(x,y) =
–(x2+y2)
2
2
2(x +y )e
have any extrema?
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 = 4x e
fy = 4y e
–(x2+y2)
–(x2+y2)
–
4x(x2+y2)
–(x2+y2)
–
4y(x2+y2)
e
e
–(x2+y2)
–(x2+y2)
The critical points are (0,0) and all points on the circle x2 + y2 = 1.
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.
Find all critical points of f(x,y) = x2y + xy2
fx = 2xy + y2
fy = 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
us
h1
f(x0 + h)  f(x0) + fx(x0) fy(x0)
h2
1
+ —
2
1
 f(x0 + h) – f(x0)  — h1 h2
2
1
= —
2
h1 h2
fxx(x0) fyx(x0)
h1
fxy(x0) fyy(x0)
h2
fxx(x0) fyx(x0)
h1
fxy(x0) fyy(x0)
h2
fxx(x0) h12 + 2fxy(x0) h1h2 + fyy(x0) h22
h22
h12
= —— fxx(x0) —2 + 2fxy(x0)
2
h2
A x2
B
h1
— + fyy(x0)
h2
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 .
A x2
B x
C
h2 2
h12
h1
f(x0 + h) – f(x0)  —— fxx(x0) —2 + 2fxy(x0) — + fyy(x0)
2
h2
h2
will either always be positive or always be negative when
4[fxx(x0)][fyy(x0)] – [2fxy(x0)]2 > 0 
[fxx(x0)][fyy(x0)] – [fxy(x0)]2 > 0 
det
fxx(x0) fyx(x0)
fxy(x0) fyy(x0)
> 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 local extremum , but if det(A) = fxx(x0) fyy(x0) – [fxy(x0)]2 < 0, then
x0 is a saddle point.
The determinant det(A) = fxx(x0) fyy(x0) – [fxy(x0)]2 is called the
discriminant 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 local minimum , but if fxx(x0) and fyy(x0) are both
negative, x0 is a local 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 = 2x + 3y
fy = 3x + 2y
The only critical point is (0,0).
fxx = 2
fyy = 2
D = (2)(2) – (3)2 = – 5
Since D < 0, then (0,0) is a saddle point.
fxy = 3
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 = 6x – 5y
fy = –5x + 6y
The only critical point is (0,0).
fxx = 6
fyy = 6
fxy = –5
D = (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 = sin y
fy = 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 = 0
fyy = – x sin y
fxy = cos y
D = – cos2y
Since D < 0 for all critical points, then all critical points are saddle
points.
(–x2–y2) / 2
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 = x[2 – x2 + y2] e
(–x2–y2) / 2
fy = y[– 2 – x2 + y2] e
(–x2–y2) / 2
The critical points are
(0,0), (2,0), (–2,0), (0,2), (0,–2)
2 + x2(x2 – y2) + y2] e
[2
–
5x
fxx =
(–x2–y2) / 2
fyy = [– 2 + 5y2 + y2(x2 – y2) – x2] e
2 – y2) e
xy(x
fxy =
(–x2–y2) / 2
(–x2–y2) / 2
fxx(0,0) = 2
fyy(0,0) = – 2
D= –4
fxy(0,0) = 0
(0,0) is a saddle point
fxx(2,0) = fxx(–2,0) = – 4e–1 D = 16e–2
fyy(2,0) = fyy(–2,0) = – 4e–1
fxy(2,0) = fxy(–2,0) = 0
fxx(0,2) = fxx(0,–2) = 4e–1
fyy(0,2) = fyy(0,–2) = 4e–1
fxy(0,2) = fxy(0,–2) = 0
(2,0) and (–2,0) are
each a local maximum
D = 16e–2
(0,2) and (0,–2) are
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
 (x–0)2 + (y–0)2 + (z–0)2 .
Minimizing this distance is the same as minimizing x2 + y2 + z2 .
If (x,y,z) is a point on the plane, then z = x + 3y – 6 .
Consequently, we want to minimize the function f(x,y) =
x2 + y2 + (x + 3y – 6)2 .
fx = 2x + 2(x + 3y – 6)
fy = 2y + 6(x + 3y – 6)
The only critical point is (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 (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 minimum distance from the origin to a point on the plane is the
distance from the origin to the point (6/11 , 18/11 , –6/11) .
This distance is 6/11 .
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
 (x–0)2 + (y–0)2 + (z–0)2 .
Minimizing this distance is the same as minimizing x2 + y2 + z2 .
If (x,y,z) is a point on the surface, then z = 1/ (xy) .
Consequently, we want to minimize the function f(x,y) =
x2 + y2 + 1/(x2y2) .
fx = 2x – 2/(x3y2)
fy = 2y – 2/(x2y3)
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.
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.
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 3 , which must be the minimum distance.