2. Definition: Let z = f(x, y), a point (a, b, z = f(a,b)) is a relative
maximum if f(a, b) > f(x, y) for all (x, y) in a circle with (a, b) as
the center.
Minimum, Maximum, and Saddle Points
3. Definition: Let z = f(x, y), a point (a, b, z = f(a,b)) is a relative
maximum if f(a, b) > f(x, y) for all (x, y) in a circle with (a, b) as
the center.
Minimum, Maximum, and Saddle Points
A point (a, b, z=f(a, b)) is a absolute maximum if f(a, b) > f(x, y)
for all (x, y) in the domain.
4. Definition: Let z = f(x, y), a point (a, b, z = f(a,b)) is a relative
maximum if f(a, b) > f(x, y) for all (x, y) in a circle with (a, b) as
the center.
Minimum, Maximum, and Saddle Points
A point (a, b, z=f(a, b)) is a absolute maximum if f(a, b) > f(x, y)
for all (x, y) in the domain.
We have similar definition for relative and absolute minimum
with the inequlities reversed.
5. Definition: Let z = f(x, y), a point (a, b, z = f(a,b)) is a relative
maximum if f(a, b) > f(x, y) for all (x, y) in a circle with (a, b) as
the center.
Minimum, Maximum, and Saddle Points
A point (a, b, z=f(a, b)) is a absolute maximum if f(a, b) > f(x, y)
for all (x, y) in the domain.
We have similar definition for relative and absolute minimum
with the inequlities reversed.
Maxima and minima are called extrema.
6. Definition: Let z = f(x, y), a point (a, b, z = f(a,b)) is a relative
maximum if f(a, b) > f(x, y) for all (x, y) in a circle with (a, b) as
the center.
Minimum, Maximum, and Saddle Points
A point (a, b, z=f(a, b)) is a absolute maximum if f(a, b) > f(x, y)
for all (x, y) in the domain.
Absolute max.
Relative max.
We have similar definition for relative and absolute minimum
with the inequalities reversed. Maxima and minima are called
extrema.
Relative max.
Absolute max.
Absolute min.
(a, b, z=f(a, b))
7. Definition: Let z = f(x, y), a point (a, b, z = f(a,b)) is a relative
maximum if f(a, b) > f(x, y) for all (x, y) in a circle with (a, b) as
the center.
Minimum, Maximum, and Saddle Points
A point (a, b, z=f(a, b)) is a absolute maximum if f(a, b) > f(x, y)
for all (x, y) in the domain.
Absolute max.
Relative max.
Absolute min.
We have similar definition for relative and absolute minimum
with the inequalities reversed. Maxima and minima are called
extrema. Some saddle points are shown also.
(a, b)
(a, b, z=f(a, b))
Relative max.
Absolute max.
Saddle point
Saddle point
8. Given z = f(x, y) over a domain, it may not have any
maximum or minimum.
Minimum, Maximum, and Saddle Points
9. Given z = f(x, y) over a domain, it may not have any
maximum or minimum.
Minimum, Maximum, and Saddle Points
Example A. The function
z = 1/(x2
– y2
) over the
domain x>0, y>0 does not
have a maximum nor a
minimum.
10. Given z = f(x, y) over a domain, it may not have any
maximum or minimum.
Minimum, Maximum, and Saddle Points
Example A. The function
z = 1/(x2
– y2
) over the
domain x>0, y>0 does not
have a maximum nor a
minimum.
x
y
11. Given z = f(x, y) over a domain, it may not have any
maximum or minimum.
Minimum, Maximum, and Saddle Points
Example A. The function
z = 1/(x2
– y2
) over the
domain x>0, y>0 does not
have a maximum nor a
minimum.
x
y
Theorem (Existence of Extrema): Let z = f(x, y) be
a continuous function over a closed and bounded
domain (i.e. a set that is contained in a circle and it
includes its boundary). Then f(x, y) has absolute
maximum and absolute minimum in D.
14. For example:
Minimum, Maximum, and Saddle Points
bounded domain
but not closeed
closeed domain
but not bounded
closed and
bounded domain
15. For example:
Minimum, Maximum, and Saddle Points
The absolute maximum and absolute minimum of z = f(x, y), a
continuous function over a closed and bounded domain, is
either in the interior of D, or on the boundary of D.
bounded domain
but not closeed
closeed domain
but not bounded
closed and
bounded domain
16. For example:
Minimum, Maximum, and Saddle Points
The absolute maximum and absolute minimum of z = f(x, y), a
continuous function over a closed and bounded domain, is
either in the interior of D, or on the boundary of D.
To look for max. and min. over D, we find the
1. max. or min. in the interior
bounded domain
but not closeed
closeed domain
but not bounded
closed and
bounded domain
17. For example:
Minimum, Maximum, and Saddle Points
The absolute maximum and absolute minimum of z = f(x, y), a
continuous function over a closed and bounded domain, is
either in the interior of D, or on the boundary of D.
To look for max. and min. over D, we find the
1. max. or min. in the interior
2. max. and min. on the boundary
bounded domain
but not closeed
closeed domain
but not bounded
closed and
bounded domain
18. For example:
Minimum, Maximum, and Saddle Points
The absolute maximum and absolute minimum of z = f(x, y), a
continuous function over a closed and bounded domain, is
either in the interior of D, or on the boundary of D.
To look for max. and min. over D, we find the
1. max. or min. in the interior
2. max. and min. on the boundary
Then get the extrema over D by comparing their answers.
bounded domain
but not closeed
closeed domain
but not bounded
closed and
bounded domain
19. Recall the following fact.
Zero Gradient Theorem: Let f(x, y) be differentiable at
P = (a, b, c) a. If ∇f(a, b) = 0, then all Du(a, b) = 0 for all
directional vectors u’s and that P could be a maximum,
a minimum, or a saddle point.
Directional Derivatives and the Gradient
Maximum
Minimum
Regular Saddle Point
Saddle Points
Let f(x,y) be differentiable, a point P = (a, b) is a critical point
if ∇f(a, b) = 0 or if any of the partial derivatives fails to exist at P.
20. Extrema Test: Suppose f(x, y) is differentiable in a circle
containing (a, b) and ∇f(a, b) = 0, i.e. fx(a, b) = fy(a, b) = 0.
Let D = fxxfyy – (fxy)2
|(a,b).
Minimum, Maximum, and Saddle Points
1. If D < 0, then (a, b) is a saddle point.
2. a. If D > 0 and fxx|(a,b) < 0, then (a, b) is a max.
b. If D > 0 and fxx|(a,b) > 0, then (a, b) is a min.
3. If D = 0 then the test fails.
Example A. Find and classify the extremum of z = x2
– y2
.
To find (a, b) where ∇f(a, b) = 0, we solve the system:
fx = 2x = 0, fy = –2y = 0 and get x = 0 and y = 0.
Calculate the 2nd
partials:
fxx = 2, fyy = –2, and fxy = 0,
which gives D = fxxfyy – (fxy)2
|(0,0)
= –4 < 0. So (0, 0) is a saddle point
and it’s the only extremum
on the surface.
x
y
z
z = x2
– y2
21. Minimum, Maximum, and Saddle Points
Example: Find and classify the extrema of z = f(x, y) =√x2
+ y2
.
22. Minimum, Maximum, and Saddle Points
Example: Find and classify the extrema of z = f(x, y) =√x2
+ y2
.
fx = x/√x2
+ y2
and fy = y/√x2
+ y2
.
23. Minimum, Maximum, and Saddle Points
Example: Find and classify the extrema of z = f(x, y) =√x2
+ y2
.
fx = x/√x2
+ y2
and fy = y/√x2
+ y2
. Solve the system
fx = 0
fy = 0
24. Minimum, Maximum, and Saddle Points
Example: Find and classify the extrema of z = f(x, y) =√x2
+ y2
.
fx = x/√x2
+ y2
and fy = y/√x2
+ y2
. Solve the system
There is no solution since fx and fy are
not defined at (0, 0).
fx = 0
fy = 0
25. Minimum, Maximum, and Saddle Points
Example: Find and classify the extrema of z = f(x, y) =√x2
+ y2
.
fx = x/√x2
+ y2
and fy = y/√x2
+ y2
. Solve the system
There is no solution since fx and fy are
not defined at (0, 0).
fx = 0
fy = 0
Therefore (0, 0) is the only critical point due to failure of the
partial derivatives and clearly it is the absolute minimum.
26. Minimum, Maximum, and Saddle Points
Example: Find and classify the extrema of z = f(x, y) =√x2
+ y2
.
fx = x/√x2
+ y2
and fy = y/√x2
+ y2
. Solve the system
There is no solution since fx and fy are
not defined at (0, 0).
fx = 0
fy = 0
Therefore (0, 0) is the only critical point due to failure of the
partial derivatives and clearly it is the absolute minimum.
z = f(x, y) =√x2
+ y2
.
27. Minimum, Maximum, and Saddle Points
Example: Find and classify the extrema of z = f(x, y) =√x2
+ y2
.
fx = x/√x2
+ y2
and fy = y/√x2
+ y2
. Solve the system
There is no solution since fx and fy are
not defined at (0, 0).
fx = 0
fy = 0
Therefore (0, 0) is the only critical point due to failure of the
partial derivatives and clearly it is the absolute minimum.
Example: A. Find and classify
the extrema of f(x, y) = y4
+ x4
– 4xy.
z = f(x, y) =√x2
+ y2
.
28. Minimum, Maximum, and Saddle Points
Example: Find and classify the extrema of z = f(x, y) =√x2
+ y2
.
fx = x/√x2
+ y2
and fy = y/√x2
+ y2
. Solve the system
There is no solution since fx and fy are
not defined at (0, 0).
fx = 0
fy = 0
Therefore (0, 0) is the only critical point due to failure of the
partial derivatives and clearly it is the absolute minimum.
Example: A. Find and classify
the extrema of f(x, y) = y4
+ x4
– 4xy.
z = f(x, y) =√x2
+ y2
.
fx = 4x3
– 4y = 0
fy = 4y3
– 4x = 0
Solve the system:
Eqn 1
Eqn 2
29. Minimum, Maximum, and Saddle Points
Example: Find and classify the extrema of z = f(x, y) =√x2
+ y2
.
fx = x/√x2
+ y2
and fy = y/√x2
+ y2
. Solve the system
There is no solution since fx and fy are
not defined at (0, 0).
fx = 0
fy = 0
Therefore (0, 0) is the only critical point due to failure of the
partial derivatives and clearly it is the absolute minimum.
Example: A. Find and classify
the extrema of f(x, y) = y4
+ x4
– 4xy.
z = f(x, y) =√x2
+ y2
.
fx = 4x3
– 4y = 0
fy = 4y3
– 4x = 0
Solve the system:
Eqn 1
Eqn 2
From equation 1, we get y =x3
.
30. Minimum, Maximum, and Saddle Points
Example: Find and classify the extrema of z = f(x, y) =√x2
+ y2
.
fx = x/√x2
+ y2
and fy = y/√x2
+ y2
. Solve the system
There is no solution since fx and fy are
not defined at (0, 0).
fx = 0
fy = 0
Therefore (0, 0) is the only critical point due to failure of the
partial derivatives and clearly it is the absolute minimum.
Example: A. Find and classify
the extrema of f(x, y) = y4
+ x4
– 4xy.
z = f(x, y) =√x2
+ y2
.
fx = 4x3
– 4y = 0
fy = 4y3
– 4x = 0
Solve the system:
Eqn 1
Eqn 2
From equation 1, we get y =x3
.
Substitue this into equation 2.
33. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0
34. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
35. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
36. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
37. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
fxx = 12x2
,
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
38. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
fxx = 12x2
, fyy = 12y2
,
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
39. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
fxx = 12x2
, fyy = 12y2
, fxy = -4
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
40. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
fxx = 12x2
, fyy = 12y2
, fxy = -4
D = fxxfyy – (fxy)2
|(0,0)
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
41. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
fxx = 12x2
, fyy = 12y2
, fxy = -4
D = fxxfyy – (fxy)2
|(0,0)
= 0 – (-4)2
= -16.
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
42. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
fxx = 12x2
, fyy = 12y2
, fxy = -4
D = fxxfyy – (fxy)2
|(0,0)
= 0 – (-4)2
= -16.
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
(0, 0) is a saddle point.
43. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
fxx = 12x2
, fyy = 12y2
, fxy = -4
D = fxxfyy – (fxy)2
|(0,0)
= 0 – (-4)2
= -16.
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
(0, 0) is a saddle point.
D = fxxfyy – (fxy)2
|(1,1)
44. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
fxx = 12x2
, fyy = 12y2
, fxy = -4
D = fxxfyy – (fxy)2
|(0,0)
= 0 – (-4)2
= -16.
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
(0, 0) is a saddle point.
D = fxxfyy – (fxy)2
|(1,1)
= 144 – (-4)2
>0
45. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
fxx = 12x2
, fyy = 12y2
, fxy = -4
D = fxxfyy – (fxy)2
|(0,0)
= 0 – (-4)2
= -16.
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
(0, 0) is a saddle point.
D = fxxfyy – (fxy)2
|(1,1)
= 144 – (-4)2
>0
D = fxxfyy – (fxy)2
|(-1,-1)
46. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
fxx = 12x2
, fyy = 12y2
, fxy = -4
D = fxxfyy – (fxy)2
|(0,0)
= 0 – (-4)2
= -16.
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
(0, 0) is a saddle point.
D = fxxfyy – (fxy)2
|(1,1)
= 144 – (-4)2
>0
D = fxxfyy – (fxy)2
|(-1,-1)
= 144 – (-4)2
>0
47. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
fxx = 12x2
, fyy = 12y2
, fxy = -4
D = fxxfyy – (fxy)2
|(0,0)
= 0 – (-4)2
= -16.
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
(0, 0) is a saddle point.
D = fxxfyy – (fxy)2
|(1,1)
= 144 – (-4)2
>0
(1, 1) and (-1, -1) are rel. min.
D = fxxfyy – (fxy)2
|(-1,-1)
= 144 – (-4)2
>0
48. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
fxx = 12x2
, fyy = 12y2
, fxy = -4
D = fxxfyy – (fxy)2
|(0,0)
= 0 – (-4)2
= -16.
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
(0, 0) is a saddle point.
D = fxxfyy – (fxy)2
|(1,1)
= 144 – (-4)2
>0
(1, 1) and (-1, -1) are rel. min.
D = fxxfyy – (fxy)2
|(-1,-1)
= 144 – (-4)2
>0
49. Minimum, Maximum, and Saddle Points
We get 4(x3
)3
– 4x = 0
x9
– x = 0
x(x8
– 1) = 0 x = 0, x = 1, x = -1
fxx = 12x2
, fyy = 12y2
, fxy = -4
D = fxxfyy – (fxy)2
|(0,0)
= 0 – (-4)2
= -16.
Since y = x3
, the critical points are (0, 0), (1, 1), and (-1, -1).
Apply the test using the 2nd
partials:
(0, 0) is a saddle point.
D = fxxfyy – (fxy)2
|(1,1)
= 144 – (-4)2
>0
(1, 1) and (-1, -1) are rel. min.
D = fxxfyy – (fxy)2
|(-1,-1)
= 144 – (-4)2
>0
(0, 0, 0)
(1, 1, –2)(–1, –1, –2)
Two views of
z = y4
+ x4
– 4xy
50. Minimum, Maximum, and Saddle Points
B. Find the absolute extrema of f(x, y) = x4
+ y4
– 4xy
over the domain {(x, y)| -2 < x < 2, -2 < y < 2}
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
51. Minimum, Maximum, and Saddle Points
B. Find the absolute extrema of f(x, y) = x4
+ y4
– 4xy
over the domain {(x, y)| -2 < x < 2, -2 < y < 2}
We have to check the boundary for their max. and min.
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
52. Minimum, Maximum, and Saddle Points
B. Find the absolute extrema of f(x, y) = x4
+ y4
– 4xy
over the domain {(x, y)| -2 < x < 2, -2 < y < 2}
We have to check the boundary for their max. and min.
At the boundary x =2, z = y4
– 8y + 16
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
53. Minimum, Maximum, and Saddle Points
B. Find the absolute extrema of f(x, y) = x4
+ y4
– 4xy
over the domain {(x, y)| -2 < x < 2, -2 < y < 2}
We have to check the boundary for their max. and min.
At the boundary x =2, z = y4
– 8y + 16
At x = 2, z(y) = y4
– 8y + 16
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
54. Minimum, Maximum, and Saddle Points
B. Find the absolute extrema of f(x, y) = x4
+ y4
– 4xy
over the domain {(x, y)| -2 < x < 2, -2 < y < 2}
We have to check the boundary for their max. and min.
At the boundary x =2, z = y4
– 8y + 16
At x = 2, z(y) = y4
– 8y + 16
Need the extrema of this function
for -2 < y < 2.
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
55. Minimum, Maximum, and Saddle Points
B. Find the absolute extrema of f(x, y) = x4
+ y4
– 4xy
over the domain {(x, y)| -2 < x < 2, -2 < y < 2}
We have to check the boundary for their max. and min.
At the boundary x =2, z = y4
– 8y + 16
At x = 2, z(y) = y4
– 8y + 16
Need the extrema of this function
for -2 < y < 2.
Set dz/dy = 4y3
– 8 = 0
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
56. Minimum, Maximum, and Saddle Points
B. Find the absolute extrema of f(x, y) = x4
+ y4
– 4xy
over the domain {(x, y)| -2 < x < 2, -2 < y < 2}
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
We have to check the boundary for their max. and min.
At the boundary x =2, z = y4
– 8y + 16
At x = 2, z(y) = y4
– 8y + 16
Need the extrema of this function
for -2 < y < 2.
Set dz/dy = 4y3
– 8 = 0
y3
– 2 = 0
57. Minimum, Maximum, and Saddle Points
B. Find the absolute extrema of f(x, y) = x4
+ y4
– 4xy
over the domain {(x, y)| -2 < x < 2, -2 < y < 2}
We have to check the boundary for their max. and min.
At the boundary x =2, z = y4
– 8y + 16
At x = 2, z(y) = y4
– 8y + 16
Need the extrema of this function
for -2 < y < 2.
Set dz/dy = 4y3
– 8 = 0
y3
– 2 = 0 y = √23
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
58. Minimum, Maximum, and Saddle Points
B. Find the absolute extrema of f(x, y) = x4
+ y4
– 4xy
over the domain {(x, y)| -2 < x < 2, -2 < y < 2}
We have to check the boundary for their max. and min.
At the boundary x =2, z = y4
– 8y + 16
At x = 2, z(y) = y4
– 8y + 16
Need the extrema of this function
for -2 < y < 2.
Set dz/dy = 4y3
– 8 = 0
y3
– 2 = 0 y = √23
The extrema is obtained by
comparing the output with the
end points y = 2 and -2
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
59. Minimum, Maximum, and Saddle Points
B. Find the absolute extrema of f(x, y) = x4
+ y4
– 4xy
over the domain {(x, y)| -2 < x < 2, -2 < y < 2}
We have to check the boundary for their max. and min.
At the boundary x =2, z = y4
– 8y + 16
At x = 2, z(y) = y4
– 8y + 16
Need the extrema of this function
for -2 < y < 2.
Set dz/dy = 4y3
– 8 = 0
y3
– 2 = 0 y = √23
The extrema is obtained by
comparing the output with the
end points y = 2 and -2
z( ) = 4
– 8 + 16 ≈ 8.4√2
3
√2
3
√2
3
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
60. Minimum, Maximum, and Saddle Points
B. Find the absolute extrema of f(x, y) = x4
+ y4
– 4xy
over the domain {(x, y)| -2 < x < 2, -2 < y < 2}
We have to check the boundary for their max. and min.
At the boundary x =2, z = y4
– 8y + 16
At x = 2, z(y) = y4
– 8y + 16
Need the extrema of this function
for -2 < y < 2.
Set dz/dy = 4y3
– 8 = 0
y3
– 2 = 0 y = √23
The extrema is obtained by
comparing the output with the
end points y = 2 and -2
z( ) = 4
– 8 + 16 ≈ 8.4√2
3
√2
3
√2
3
z(-2) = (-2)4
– 8*(-2) + 16 = 48
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
61. Minimum, Maximum, and Saddle Points
B. Find the absolute extrema of f(x, y) = x4
+ y4
– 4xy
over the domain {(x, y)| -2 < x < 2, -2 < y < 2}
We have to check the boundary for their max. and min.
At the boundary x =2, z = y4
– 8y + 16
At x = 2, z(y) = y4
– 8y + 16
Need the extrema of this function
for -2 < y < 2.
Set dz/dy = 4y3
– 8 = 0
y3
– 2 = 0 y = √23
The extrema is obtained by
comparing the output with the
end points y = 2 and -2
z( ) = 4
– 8 + 16 ≈ 8.4√2
3
√2
3
√2
3
z(-2) = (-2)4
– 8*(-2) + 16 = 48
z(2) = 24
– 8*2 + 16 = 16
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
62. Minimum, Maximum, and Saddle Points
B. Find the absolute extrema of f(x, y) = x4
+ y4
– 4xy
over the domain {(x, y)| -2 < x < 2, -2 < y < 2}
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
We have to check the boundary for their max. and min.
At the boundary x =2, z = y4
– 8y + 16
At x = 2, z(y) = y4
– 8y + 16
Need the extrema of this function
for -2 < y < 2.
Set dz/dy = 4y3
– 8 = 0
y3
– 2 = 0 y = √23
The extrema is obtained by
comparing the output with the
end points y = 2 and -2
z( ) = 4
– 8 + 16 ≈ 8.4√2
3
√2
3
√2
3
z(-2) = (-2)4
– 8*(-2) + 16 = 48
z(2) = 24
– 8*2 + 16 = 16
Hence the abs. max. is 48 and abs. min. is 8.4 when x = √2
3
63. Minimum, Maximum, and Saddle Points
At x = -2, z(y) = y4
+ 8y + 16
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
64. Minimum, Maximum, and Saddle Points
At x = -2, z(y) = y4
+ 8y + 16
Need the extrema of this function for -2 < y < 2.
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
65. Minimum, Maximum, and Saddle Points
At x = -2, z(y) = y4
+ 8y + 16
Need the extrema of this function for -2 < y < 2.
Set dz/dy = 4y3
+ 8 = 0 y3
+ 2 = 0 y = -√23
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
66. Minimum, Maximum, and Saddle Points
At x = -2, z(y) = y4
+ 8y + 16
Need the extrema of this function for -2 < y < 2.
Set dz/dy = 4y3
+ 8 = 0 y3
+ 2 = 0 y = -√23
The extrema is obtained by comparing
the output with the end points y = 2, -2
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
67. Minimum, Maximum, and Saddle Points
At x = -2, z(y) = y4
+ 8y + 16
Need the extrema of this function for -2 < y < 2.
Set dz/dy = 4y3
+ 8 = 0 y3
+ 2 = 0 y = -√23
The extrema is obtained by comparing
the output with the end points y = 2, -2
z(- ) =(- )4
+ 8(- ) + 16 ≈ 8.4√2
3
√2
3
√2
3
z(-2) = (-2)4
+ 8*(-2) + 16 = 16
z(2) = 24
+ 8*2 + 16 = 48
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
68. Minimum, Maximum, and Saddle Points
At x = -2, z(y) = y4
+ 8y + 16
Need the extrema of this function for -2 < y < 2.
Set dz/dy = 4y3
+ 8 = 0 y3
+ 2 = 0 y = -√23
The extrema is obtained by comparing
the output with the end points y = 2, -2
z(- ) =(- )4
+ 8(- ) + 16 ≈ 8.4√2
3
√2
3
√2
3
z(-2) = (-2)4
+ 8*(-2) + 16 = 16
z(2) = 24
+ 8*2 + 16 = 48
Hence the abs. max. is 48
and abs. min. is 8.4 when x = - √2
3
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
69. Minimum, Maximum, and Saddle Points
At x = -2, z(y) = y4
+ 8y + 16
Need the extrema of this function for -2 < y < 2.
Set dz/dy = 4y3
+ 8 = 0 y3
+ 2 = 0 y = -√23
The extrema is obtained by comparing
the output with the end points y = 2, -2
z(- ) =(- )4
+ 8(- ) + 16 ≈ 8.4√2
3
√2
3
√2
3
z(-2) = (-2)4
+ 8*(-2) + 16 = 16
z(2) = 24
+ 8*2 + 16 = 48
Hence the abs. max. is 48
and abs. min. is 8.4 when x = -
HW: Check the extrema for the
boundary y = ±2 are:
abs. max at x = ±2 and z = 48
Abs. min at x = ± and z = 8.4√2
3
So the abs. max over D are (2, –2, 48) and (–2, 2, 48),
the abs. min. over D are (1, 1, –2) and (–1, –1, –2).
√2
3
(0, 0, 0)
(1, 1, -2)
(-1, -1, -2)
