_lecture_06 F_minima_maxima_classification.pdf

Math 2321 (Multivariable Calculus)
Lecture #10 of 38 ∼ February 10, 2021
Critical Points, Minima, and Maxima
Critical Points
Minima, Maxima, Saddle Points
Classification of Critical Points
This material represents §2.5.1 from the course notes.

Motivation
Now that we have developed the basic ideas of derivatives for
functions of several variables, we will tackle one of the primary
motivating questions for the development of the derivative: finding
minima and maxima of functions of several variables.
We will primarily discuss functions of two variables, because
there is a nice criterion for deciding whether a critical point is
a minimum or a maximum in that situation.
Classifying critical points for functions of more than two
variables requires some results from linear algebra, so we will
not treat functions of more than two variables.
We will then discuss various different flavors of optimization
problems.

Critical Points, I
We would first like to determine where a function f can have a
minimum or maximum value.
We know that the gradient vector ∇f , evaluated at a point P,
gives a vector pointing in the direction along which f
increases most rapidly at P.
If ∇f (P) 6= 0, then f cannot have a local extreme point at P,
since f increases along ∇f (P) and decreases along −∇f (P).
Thus, extreme points can only occur at points where the
gradient ∇f is the zero vector, or if it is undefined.

Critical Points, II
Definition
A critical point of a function f (x, y) is a point where ∇f is zero or
undefined. Equivalently, (x0, y0) is a critical point if
fx (x0, y0) = fy (x0, y0) = 0, or either fx (x0, y0) or fy (x0, y0) is
undefined.
We have a similar definition for functions of 3+ variables.
By the remarks on the previous slide, a local minimum or
maximum of a function of several variables can only occur at a
critical point. (Note that this is the same state of affairs as with
functions of one variable.)
Most critical points will arise from places where all the partial
derivatives are zero, since we will rarely encounter functions (e.g.,
absolute values) that are defined but whose derivatives are not.

Critical Points, III
Solving systems of equations in several variables (to find the
critical points) can sometimes be quite tricky. Some tips:
1. Identify an equation that you can solve for one variable in
terms of the other (or others). Use it to eliminate that
variable from the other equations, and repeat until you get an
equation in one variable.
2. Factor an equation and break into cases. If you have an
expression like [something] · [another thing] = 0 then you
know one of the two terms must be zero. This will give you
two separate cases to analyze.
3. Manipulate the equations algebraically: add, subtract,
multiply, or divide them.
In some cases, one of the equations may be much easier to work
with than the other(s). Start with whichever one seems simplest.

Critical Points, IV
Example: Find all critical points for each given function:
1. f (x, y) = x2 + y2.
2. g(x, y) = x2 + 2x − y2 − 6y + 4.
3. p(x, y) = x3 + y3 − 3xy.
4. q(x, y) = x2 + 4xy + 2y2 + 6x − 4y + 3.
5. j(x, y) = x2y2 − x2 − y2.
6. t(x, y) = y cos(x).
In each case, the partial derivatives are both defined
everywhere, so we only need to find where they are both zero.

Critical Points, V
1. f (x, y) = x2 + y2.

Critical Points, V
1. f (x, y) = x2 + y2.
We have fx = 2x and fy = 2y.
Clearly fx = 0 requires x = 0 while fy = 0 requires y = 0.
Therefore, we get one critical point: (x, y) = (0, 0).
2. g(x, y) = x2 + 2x − y2 − 6y + 4.

Critical Points, V
1. f (x, y) = x2 + y2.
We have fx = 2x and fy = 2y.
Clearly fx = 0 requires x = 0 while fy = 0 requires y = 0.
Therefore, we get one critical point: (x, y) = (0, 0).
2. g(x, y) = x2 + 2x − y2 − 6y + 4.
We have gx = 2x + 2 and gy = −2y − 6.
Then gx = 0 requires x = −1 while gy = 0 requires 2y = −6
so y = −3.
Therefore, we get one critical point: (x, y) = (−1, −3).

Critical Points, VI
3. p(x, y) = x3 + y3 − 3xy.

Critical Points, VI
3. p(x, y) = x3 + y3 − 3xy.
We have px = 3x2 − 3y and py = 3y2 − 3x.
So we get the equations 3x2 − 3y = 0 and 3y2 − 3x = 0.
Neither equation gives us a value for x or y directly.
But we can solve the first equation for y in terms of x:

Critical Points, VI
3. p(x, y) = x3 + y3 − 3xy.
We have px = 3x2 − 3y and py = 3y2 − 3x.
So we get the equations 3x2 − 3y = 0 and 3y2 − 3x = 0.
But we can solve the first equation for y in terms of x:this
gives y = x2.
Now plugging into the second equation yields
3(x2)2 − 3x = 0, so that 3x4 − 3x = 0.
Factoring gives 3x(x3 − 1) = 0, which has the solutions x = 0
and x = 1.
If x = 0, then y = x2 = 0, so we get the point (0, 0).
If x = 1, then y = x2 = 1, so we get the point (1, 1).
In total, there are two critical points: (0, 0) and (1, 1).

Critical Points, VII
4. q(x, y) = x2 + 4xy + 2y2 + 6x − 4y + 3.

4. q(x, y) = x2 + 4xy + 2y2 + 6x − 4y + 3.
We have qx = 2x + 4y + 6 and qy = 4x + 4y − 4.
So we get the equations 2x + 4y + 6 = 0 and 4x + 4y − 4 = 0.
But we can solve the second equation for y in terms of x:

4. q(x, y) = x2 + 4xy + 2y2 + 6x − 4y + 3.
We have qx = 2x + 4y + 6 and qy = 4x + 4y − 4.
So we get the equations 2x + 4y + 6 = 0 and 4x + 4y − 4 = 0.
But we can solve the second equation for y in terms of x:this
gives y = 1 − x.
Now plugging into the first equation yields
2x + 4(1 − x) + 6 = 0, so that 10 − 2x = 0 and thus x = 5.
Then y = 1 − x = −4.
Therefore, there is one critical point: (x, y) = (5, −4).
There are many other ways to solve this system: for example, we
could have solved the first equation for y in terms of x, or for x in
terms of y, or we could have subtracted the two equations.

Critical Points, VIII
5. j(x, y) = x2y2 − x2 − y2.

5. j(x, y) = x2y2 − x2 − y2.
We have jx = 2xy2 − 2x and jy = 2x2y − 2y.
So we get the equations 2xy2 − 2x = 0 and 2x2y − 2y = 0.
We can factor the first equation:

5. j(x, y) = x2y2 − x2 − y2.
We have jx = 2xy2 − 2x and jy = 2x2y − 2y.
So we get the equations 2xy2 − 2x = 0 and 2x2y − 2y = 0.
We can factor the first equation: 2x(y2 − 1) = 0.
This means either 2x = 0 (so that x = 0) or y2 − 1 = 0 (so
that y = 1 or y = −1).
If x = 0, then the second equation gives −2y = 0 so that
y = 0. We get the point (0, 0).
If y = 1, then the second equation gives 2x2 − 2 = 0 so that
x = 1 or x = −1. We get points (1, 1) and (−1, 1).
If y = −1, then the second equation gives −2x2 + 2 = 0 so
that x = 1 or x = −1. We get points (1, −1) and (−1, −1).
In total we get five critical points: (0, 0), (1, 1), (−1, 1),
(1, −1), (−1, −1).

Critical Points, IX
6. t(x, y) = y cos(x).

Critical Points, IX
6. t(x, y) = y cos(x).
We have tx = −y sin(x) and ty = cos(x).
So we get the equations −y sin(x) = 0 and cos(x) = 0.
The second equation implies x = π/2 + πk for some integer k.
Then since cos(π/2 + πk) = (−1)k, the first equation requires
y = 0.
Therefore, the critical points are (x, y) = (π/2 + πk, 0) for
any integer k. (Note that there are infinitely many of them!)

Minima, Maxima, and Saddles, I
We have various different types of critical points:
Definition
A local minimum is a point where f is nearby always bigger.
A local maximum is a point where f is nearby always smaller.
A saddle point is a critical point where f nearby is bigger in some
directions and smaller in others.
From our earlier discussion, local minima and local maxima always
occur at critical points.

Minima, Maxima, and Saddles, II
Example: f (x, y) = x2 + y2 has a local minimum at (0, 0):

Minima, Maxima, and Saddles, III
Example: f (x, y) = −x2 − y2 has a local maximum at (0, 0):

Minima, Maxima, and Saddles, IV
Example: f (x, y) = x2 − y2 has a saddle point at (0, 0):

Minima, Maxima, and Saddles, V
Local minima and local maxima are (presumably) familiar from the
one-variable setting. Saddle points, however, are a new kind of
critical point.
A saddle point will look like a local minimum along some
directions and a local maximum along other directions.
For example, f (x, y) = x2 − y2 looks like a minimum in the
x-direction (as x varies and y is held fixed at 0, the function is
f (x, 0) = x2) but a maximum in the y-direction (as y varies
and x is held fixed at 0, the function is f (0, y) = −y2).
In contrast, a local minimum looks like a minimum in every
direction, while a local maximum looks like a maximum in
every direction.

Classification of Critical Points, I
Once we can identify a function’s critical points, we would like to
know whether those points actually are minima or maxima of f .
We can determine this using a quantity called the discriminant:
Definition
The discriminant (also called the Hessian) at a critical point is the
value D = fxx · fyy − (fxy )2, where each of the second-order partials
is evaluated at the critical point.
One way to remember the definition of the discriminant is as the
determinant of the matrix of the four second-order partials:
D =
fxx fxy
fyx fyy
. (We are implicitly using the fact that fxy = fyx .)

Classification of Critical Points, II
Example: Find D at the critical points (0, 0) and (1, 1) for
f (x, y) = x3 + y3 − 3xy.
Remember that D = fxx · fyy − (fxy )2 =
fxx fxy
fyx fyy
.

Example: Find D at the critical points (0, 0) and (1, 1) for
f (x, y) = x3 + y3 − 3xy.
Remember that D = fxx · fyy − (fxy )2 =
fxx fxy
fyx fyy
.
We have fx = 3x2 − 3y, fy = 3y2 − 3x, so fxx = 6x,
fxy = −3, and fyy = 6y.
Therefore, D = (6x)(6y) − (−3)2.
Thus, at (x, y) = (0, 0) we get D = (0)(0) − 9 = −9.
Also, at (x, y) = (1, 1) we get D = (6)(6) − 9 = 27.

Our main result is that we can use D to classify critical points:
Theorem (Second Derivatives Test)
Suppose P is a critical point of f (x, y), and let D be the value of
the discriminant fxx fyy − f 2
xy at P.
If D > 0 and fxx > 0, then the critical point is a local minimum.
If D > 0 and fxx < 0, then the critical point is a local maximum.
If D < 0, then the critical point is a saddle point.
If D = 0, then the test is inconclusive.
Note that if D > 0 then fxx cannot be zero, because D = −f 2
xy in
that case.

Classification of Critical Points, III
Proof (outline):
Assume for simplicity that P is at the origin.
Then by our results on Taylor series, the function
f (x, y) − f (P) will be closely approximated by the polynomial
ax2 + bxy + cy2, where a = 1
2fxx , b = fxy , and c = 1
2fyy .
If D 6= 0, then the behavior of f (x, y) near the critical point P
will be the same as that quadratic polynomial.
Completing the square and examining whether the resulting
quadratic polynomial has any real roots and whether it opens
or downwards yields the test.
The behavior of the roots of the quadratic polynomial
ax2 + bxy + cy2 is determined by its discriminant b2 − 4ac.
Here, b2 − 4ac = f 2
xy − fxx fyy = −D. (That’s why D is called
the discriminant. Though yes, there is the minus sign....)

Classification of Critical Points, IV
Example: Classify the type of critical point that f (x, y) = x2 + y2
has at the origin (0, 0).

Classification of Critical Points, IV
Example: Classify the type of critical point that f (x, y) = x2 + y2
has at the origin (0, 0).
We saw earlier that (0, 0) is a critical point of this function.
To classify it, we compute fxx = 2, fxy = 0, and fyy = 2.
Then D = fxx fyy − (fxy )2 = 2 · 2 − 02 = 4. (Here, D is
constant, but normally we would need to evaluate it at our
point.)
So, by the second derivatives test, since D > 0 and fxx > 0 at
(0, 0), we see that (0, 0) is a local minimum.

Classification of Critical Points, V
Example: Classify the two critical points (0, 0) and (1, 1) for
p(x, y) = x3 + y3 − 3xy.
We saw earlier that (0, 0) and (1, 1) are the critical points of
this function and that D = (6x)(6y) − (−3)2.

Classification of Critical Points, V
Example: Classify the two critical points (0, 0) and (1, 1) for
p(x, y) = x3 + y3 − 3xy.
We saw earlier that (0, 0) and (1, 1) are the critical points of
this function and that D = (6x)(6y) − (−3)2.
At (0, 0), we have D = −9, so (0, 0) is a saddle point.
At (1, 1), we have D = (6)(6) − (−3)2 = 27, and also
fxx = 6x = 6 > 0, so (1, 1) is a local minimum.

Classification of Critical Points, VI
Example: Classify the critical point (5, −4) for
q(x, y) = x2 + 4xy + 2y2 + 6x − 4y + 3.

Classification of Critical Points, VI
Example: Classify the critical point (5, −4) for
q(x, y) = x2 + 4xy + 2y2 + 6x − 4y + 3.
We have qx = 2x + 4y + 6 and qy = 4x + 4y − 4 so qxx = 2,
qxy = 4, and qyy = 4.
Therefore, D = (2)(4) − 42 = −8. Here, D is constant.
Then at the critical point, we have D = −8, so by the second
derivatives test, (5, −4) is a saddle point.

Classification of Critical Points, VII
Example: For f (x, y) = 3x2 + 2y3 − 6xy, find the critical points of
f and classify them as minima, maxima, or saddle points.

First, fx = 6x − 6y and fy = 6y2 − 6x. They are both defined
everywhere so we need only find where they are both zero.
Next, we can see that fx is zero only when y = x.
Then the equation fy = 0 becomes 6x2 − 6x = 0, which by
factoring we can see has solutions x = 0 or x = 1.
Since y = x, we see (0, 0) and (1, 1) are the critical points.

First, fx = 6x − 6y and fy = 6y2 − 6x. They are both defined
everywhere so we need only find where they are both zero.
Next, we can see that fx is zero only when y = x.
Then the equation fy = 0 becomes 6x2 − 6x = 0, which by
factoring we can see has solutions x = 0 or x = 1.
Since y = x, we see (0, 0) and (1, 1) are the critical points.
To classify them, we compute fxx = 6, fxy = −6, and
fyy = 12y. Then D(0, 0) = 6 · 0 − (−6)2 < 0 and
D(1, 1) = 6 · 12 − (−6)2 > 0. Also, fxx > 0 at (1, 1).
So, by the second derivatives test, (0, 0) is a saddle point and
(1, 1) is a local minimum.

Classification of Critical Points, VIII
Example: Let g(x, y) = x3y − 3xy3 + 8y.
1. Find the critical points of g.
2. Classify each critical point as a local minimum, local
maximum, or saddle point.

Since gx and gy are defined everywhere, the critical points are
obtained by solving gx = gy = 0.
Then we use the second derivatives test to classify the types.

Classification of Critical Points, IX

Classification of Critical Points, IX
We have gx = 3x2y − 3y3 and gy = x3 − 9xy2 + 8.
Since gx = 3y(x2 − y2) = 3y(x + y)(x − y), we see that
gx = 0 precisely when y = 0 or y = x or y = −x.
If y = 0, then gy = 0 implies x3 + 8 = 0, so that x = −2.
This yields the point (x, y) = (−2, 0).
If y = x, then gy = 0 implies −8x3 + 8 = 0, so that x = 1.
This yields the point (x, y) = (1, 1).
If y = −x, then gy = 0 implies −8x3 + 8 = 0, so that x = 1.
This yields the point (x, y) = (1, −1).
Thus, (−2, 0), (1, 1), and (1, −1) are the critical points.

(−2, 0), (1, 1), and (1, −1) are the critical points.

(−2, 0), (1, 1), and (1, −1) are the critical points.
To classify them, we compute gxx = 6xy, gxy = 3x2 − 9y2,
and gyy = −18xy.
Therefore, D = (6xy)(−18xy) − (3x2 − 9y2)2.
Then D(−2, 0) = 0 · 0 − (12)2 < 0,
D(1, 1) = 6 · (−18) − (−6)2 < 0, and
D(1, −1) = (−6) · (18) − (−6)2 < 0.
So, by the second derivatives test, (−2, 0), (1, 1), and (1, −1)
are all saddle points.

Summary
We introduced critical points and discussed how to find them.
We discussed how to classify critical points as local minima, local
maxima, or saddle points.
Next lecture: Applied optimization, optimization of a function on a
region.

_lecture_06 F_minima_maxima_classification.pdf

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