Notes for Max/Min Problems
Section 12.8 – Math 2421 – Calculus 3
Review of Single Variable Min/Max Problems:
Take five minutes to fill in the blanks with the help of your neighbors.
Assume f = f (x) is a function of one variable.
1. If f has a local maximum at x = a then f (x)
f (a) for all x in some open interval.
2. If f has a local minimum at x = a then f (x)
f (a) for all x in some open interval.
3. If f has a local extremum at x = a then f ′ (a) =
4. If f has a critical point at x = a then either f ′ (a) =
or f ′ (a)
5. If f ′′ (a) < 0, the f is concave
at the point x = a.
6. If f ′′ (a) > 0, the f is concave
at the point x = a.
.
7. If f has a local maximum at x = a then f ′′ (a)
8. If f has a local minimum at x = a then f ′′ (a)
9. If f ′′ (a) = 0 then
10. Describe the procedure for finding the absolute maximum/minimum of f (x) on a closed interval x ∈ [a, b].
Multivariable Min/Max Problems:
Assume f = f (x, y) is a function of two variables.
• Definition: Local Maximum:
A function f has a local maximum at (a, b) if f (x, y) ≤ f (a, b) for all (x, y) in the domain of f and in some
open disk centered at (a, b).
• Definition: Local Minimum:
A function f has a local minimum at (a, b) if f (x, y) ≥ f (a, b) for all (x, y) in the domain of f and in some
open disk centered at (a, b).
NOTE: local maxima and minima are defined on open sets only
• Theorem: Gradient and Extreme Values:
If f has a local extremum at (a, b) and fx and fy exist at (a, b) then ∇f (a, b) = 0 = h0, 0i
• Definition: Critical Points:
A point (a, b) in the interior of the domain of f is called a critical point of f if either
(a) ∇f (a, b) = 0, or
(b) one, or both, of fx or fy does not exist at (a, b).
1
Write a few sentences describing how to find the critical point(s) of a multivariable function.
Example 1: Find the critical points for f (x, y) = xy(x − 2)(y + 3).
Hint: The first partial derivatives are: fx = 2y(x − 1)(y + 3) and fy = x(x − 2)(2y + 3).
• Definition: Saddle Point:
A function f has a saddle point at a critical point (a, b) if, in every open disk centered at (a, b), there are points
(x, y) for which f (x, y) > f (a, b) and points for which f (x, y) < f (a, b).
The prototypical example is a hyperbolic paraboloid: z = x2 − y 2 .
• Definition: The Hessian Matrix:
Let f be twice differentiable. The Hessian matrix is a 2 × 2 matrix of the second partial derivatives
H(x, y) =
"
fxx (x, y)
fyx (x, y)
fxy (x, y)
fyy (x, y)
#
• Definition: Discriminant:
The discriminant of a multivariable function, f , is the determinant of the Hessian matrix
f (x, y)
xx
D(x, y) = fyx (x, y)
fxy (x, y)
2
= fxx fyy − fxy
fyy (x, y)
• Theorem: The Second Derivative Test:
Suppose that the second partial derivatives of f are continuous throughout an open disk centered at critical point
(a, b).
1. If D(a, b) > 0 and fxx (a, b) < 0, then f has a local maximum value at (a, b).
2. If D(a, b) > 0 and fxx (a, b) > 0, then f has a local minimum value at (a, b).
3. If D(a, b) < 0 there f has a saddle point at (a, b).
4. If D(a, b) = 0 then the test is inconclusive.
Note: D(a, b) > 0 means that the surface has the same general behavior in all directions near (a, b); either the
surface rises in all directions or it falls in all directions.
2
Example 2:
Consider fx = 2y(x − 1)(y + 3) with the following first derivatives, second derivatives, and critical points:
fx = 2y(x − 1)(y + 3),
fxx = 2y(y + 3),
fy = x(x − 2)(2y + 3)
fxy = fyx = 2(2y + 3)(x − 1),
crit. points: (0, 0),
(2, 0),
(1, −3/2),
fyy = 2x(x − 2)
(0, −3),
(2, −3)
Fill in the following table to state whether f has a max, min, or saddle at each of the critical points.
2
D(a, b) = fxx fyy − fxy
crit. pt. (a, b)
fxx (a, b)
Conclusion (min/max/saddle)
(0, 0)
(2, 0)
(1, −3/2)
(0, −3)
(2, −3)
Surface Plot: f(x,y) = xy(x−2)(y+3)
Contour Plot: f(x,y) = xy(x−2)(y+3)
0.5
0
3
−0.5
2
−1
0
y
z
1
−1.5
−1
−2
−2
−3
1
−2.5
0
2.5
2
−1
−3
1.5
−2
1
0.5
−3
y
0
−4
−0.5
−3.5
−0.5
x
0
0.5
1
x
3
1.5
2
2.5
Example 3:
Let f (x, y) = 2 − xy 2 . Find the critical points and determine the behavior of the surface at those critical points.
Critical Points:
Hessian:



H(x, y) = 
Second Derivative Test:


2
D(a, b) = fxx fyy − fxy
crit. pt. (a, b)
fxx (a, b)
Conclusion (min/max/saddle)
Surface Plot: f(x,y) = 2−xy2
Contour Plot: f(x,y) = 2−xy2
1
0.8
3
0.6
2.8
2.6
0.4
2.4
0.2
2
y
z
2.2
1.8
0
1.6
−0.2
1.4
1.2
−0.4
1
1
−0.6
1
0.5
0.5
0
−0.8
0
−0.5
y
−0.5
−1
−1
−1
−1
x
−0.8
−0.6
−0.4
−0.2
0
x
4
0.2
0.4
0.6
0.8
1
• Definition: Absolute Maximum:
If f (x, y) ≤ f (a, b) for ALL (x, y) in the domain of f , then f has an absolute maximum at (a, b).
• Definition: Absolute Minimum:
If f (x, y) ≥ f (a, b) for ALL (x, y) in the domain of f , then f has an absolute minimum at (a, b).
• Procedure: Finding absolute max/min on a closed, bounded set:
Let f be continuous on a closed, bounded, set R in R2 . To find the absolute max and min values of f on R:
1. Determine the values of f at all critical points in R.
2. Find the max and min values of f on the boundary of R.
3. The greatest function value found in steps (1) and (2) is the abs. max., and the least function value found
in steps (1) and (2) is the abs. min. of f on R.
5
Example 4:
2
2
Surface Plot: f(x,y) = x +y −2y+1 with R in the xy−plane
2
2
Contour Plot: f(x,y) = x + y − 2y + 1 with R
Find the absolute max and min of
2.5
2
20
1.5
18
f (x, y) = x2 + y 2 − 2y + 1
16
1
14
on the region
0.5
10
y
z
12
8
6
0
−0.5
4
2
2
R = {(x, y) : x + y ≤ 4}
2
−1
0
3
−1.5
2
3
1
−2
2
0
1
0
−1
.
−2
−2
−1.5
−1
−2
−3
y
−2.5
−2.5
−1
x
• Critical Points:
fx (x, y) =
fy (x, y) =
Crit.P t :
(a, b) = (
,
)
f (a, b) =
• Parameterize the Boundary:
Write a parametric description of the boundary of R.
Hint: The boundary of R is a circle. Let 0 ≤ t < 2π be the parameter.
x(t) =
y(t) =
• Define a new function, g, that takes the value of f on the boundary of R.
g(t) =
Now simplify to get
g(t) =
• Find the maximum/minimum value(s) of g for 0 ≤ t < 2π (this is a calc 1 problem)
g ′ (t) =
Crit.P ts :
t1 =
t2 =
F unctionV alues :
g(t1 ) =
g(t2 ) =
• Compare to get the absolute max and min
,y=
Absolute Max: x =
Absolute Min: x =
,y=
,z=
.
,z=
.
• Challenge: What if R were changed to R = {(x, y) : x2 + y 2 < 4}?
6
−0.5
0
x
−3
0.5
1
1.5
2
2.5