Let be function of one variable and suppose that . According to the second-derivative test, has
a local minimum at if
a local maximum at if
We have a similar test for functions of two variables. As one might expect, the test is somewhat more complicated to state and definitely more difficult to prove.
THEOREM THE SECOND-PARTIALS TEST
Suppose that
has continuous second-order partials in a neighborhood of
and that
.
Set
and form the discriminant
.
1. If , then is a saddle point.
2. If , then has
a local minimum at if
a local maximum at if
The test is geometrically evident for functions of the form
The graph of such a function is a paraboloid:
The gradient is
at the origin
.
Moreover
and
If
,
then
and
have opposite signs and the surface has a saddle point.
Suppose now that If , then and the surface has a minimum point; if then and the surface has a maximum point.
In the examples that follow we apply the second-partials test to a variety of functions.
Example 1. For the function
we have
Setting both partials equal to zero, we have
The only simultaneous solution to these equations is
.
The point
is thus the only stationary point.
The second partials are constant:
Thus
and
Since
it follows from the second-partials test that
is a local minimum.
Example 2. The function
has partial derivatives
Setting both of these partials equal to zero, we have
The only simultaneous solution to these equations is
The point
is thus the only stationary point.
The second partials are
Evaluating these partials at the point
,
we find that
,
and thus
By the second-partials test,
is a saddle point.
We could have determined that
is a saddle point by comparing the value of
at
with its value at nearby points
For small
,
this expression is positive if
and negative if
Example 3. The function
has partial derivatives
Setting both partials equal to zero, we have
The only simultaneous solutions are and . The points and are the only stationary points.
The second partials are
At
,
we have
,
and thus
Since
,
we know that
is a local minimum. At
,
we have
.
Thus
The origin is a saddle point.
Example 4. Here we test the function
on the open square
In the first place
Setting both of these partials equal to zero, we have
Together these equations imply that
.
Since both
and
lie between
and
,
we can conclude that
.
The condition
now gives
Into this last equation we substitute the identity
and thereby obtain the relation
This is a quadratic in
with solutions
We throw out the solution
because
has to lie strictly between
and
.
The other possibility,
,
gives
.
Since
,
we also have
.
This shows that
is the only stationary point.
The second partials are
At
we have
and thus
Since
,
we can conclude that
is a local maximum.
The second-derivative test applies to points
where
but
If
,
the second-derivative test provides no information. The second-partials test
suffers from a similar limitation. It applies to points
where
but
If
the second-partials test provides no information. Consider, for example, the
functions
Each of these functions has zero gradient at the origin, and in each case
Yet,
(1) for , gives a local minimum;
(2) for , gives a local maximum;
(3) for , is a saddle point.
Statements (1) and (2) are obvious. To confirm (3), note that , but in every neighborhood of the function takes on both positive and negative values:
for while for .
When we ask for the distance from a point
to a line
we are asking for the minimum value of
with
subject to the side condition
When we ask for the distance from a point
to a plane
we are asking for the minimum value of
with
subject to the side condition
Our interest here is to present techniques for handling problems of this sort in general. In the two-variable case, the problems will take the form of maximizing (or minimizing) some expression subject to a side condition . In the three-variable case, we will seek to maximize (or minimize) some expression subject to a side condition . We begin with two simple problems-
Problem 1. Maximize the product subject to the side condition
Solution. The condition gives . Our initial problem can therefore be solved simply by maximizing the product . The derivative is only at . Since , we know from the second-derivative test that is the desired maximum.
Problem 2. Find the maximum volume of a rectangular solid given that the sum of the lengths of its edges is
Solution. We denote the dimensions of the solid by
The volume is given by
The stipulation on the edges requires that
Solving this last equation for
,
we find that
Substituting this expression for
in the volume formula, we have
Since
and
must all remain positive, our problem is to find the maximum value of
on the interior of the triangle bounded by the lines
and
The first partials are
Setting both partials equal to zero, we have
Since
and
are assumed positive, we can divide by
and
and get
and .
Solving these equations simultaneously, we find that The point , which does lie within the triangle, is the only stationary point. The value of at that point is . The conditions of the problem make it clear that this is a maximum.
The last two problems were easy. They were easy in part because the side conditions were such that we could solve for one of the variables in terms of the other(s). In general this is not possible and a more sophisticated approach is required.
Throughout the discussion
will be a function of two or three variables continuously differentiable on
some open set
.
We take
as a curve that lies entirely in
and has at each point a nonzero tangent vector
.
The basic result is this:
if maximizes (or minimizes) on ,
then is perpendicular to at
Proof. Choose so that
The composition
has a maximum (or minimum) at
.
Consequently, its derivative
must be zero at
:
This shows that
Since
is tangent to
at
,
is perpendicular to
at
We are now ready for the side condition problems. Suppose that is a continuously differentiable function of two or three variables defined on a subset of the domain of . Lagrange made the following observation:
if
maximizes (or minimizes)
subject to the side condition
,
then
and
are parallel. Thus, if
,
then there exists a scalar
,
such that
Such a scalar has come to be called a Lagrange multiplier.
Proof. Let us suppose that maximizes (or minimizes) subject to the side condition . If , the result is trivially true: every vector is parallel to the zero vector.
In the two-variable case we suppose the side condition and
The side condition defines a curve that has a nonzero tangent vector at Since maximizes (or minimizes) on , we know that is perpendicular to at Since is also perpendicular to at , the two gradients are therefore parallel.
In the three-variable case we have the side condition and The side condition defines a surface that lies in the domain of . Now let be a curve that lies on and passes through with nonzero tangent vector. We know that maximizes (or minimizes) on . Consequently, is perpendicular to at . Since this is true for each such curve , must be perpendicular to itself. But is also perpendicular to at . It follows that and are parallel.
We come now to some problems that are susceptible to Lagrange's method. In
each case
is not
where
is
and therefore we can focus entirely on those points
that satisfy a Lagrange condition
Problem 3. Maximize and minimize
on the unit circle
Solution. Since
is continuous and the unit circle is closed and bounded, it is clear that both
a maximum and a minimum exist. To apply Lagrange's method we set
We want to maximize and minimize
subject to the side condition .
The gradients are
Setting
we obtain
Multiplying the first equation by
and the second equation by
,
we find that
and thus
The side condition
now implies that
and therefore that
. The only points that can give rise to an extreme value are
At the first and fourth points
takes on the value
At the second and third points
takes on the value
Clearly then
is the maximum value and
the minimum value.
Problem 4. Find the minimum value taken on by the function
on the hyperbola .
Solution. This minimum is simply the square of the distance from the
point
to the hyperbola and thus clearly exists. Now set
We want to minimize
subject to the side condition
Here
The Lagrange condition
gives
which we can simplify to
The side condition shows that cannot be zero. Dividing by , we get . This means that and therefore . With , the side condition gives The points to be checked are therefore and . At each of these points takes on the value . This is the desired minimum.
Remark. The last problem could have been solved more simply by rewriting the side condition as and eliminating from by substitution. It would then have been only a matter of minimizing
Problem 5. Maximize
subject to the side condition
with
Solution. The set of all points
that satisfy the side condition can be shown to be closed and bounded. Since
is continuous, we can be sure that the desired maximum exists. We begin by
setting
so that the side condition becomes
.
We seek those triples
that simultaneously satisfy a Lagrange condition
and the side condition .
The gradients are
The Lagrange condition
gives
Multiplying the first equation by
,
the second by
,
and the third by
,
we get
and consequently
We can
exclude
because, if
,
then
,
or
would have to be zero. That would force
to be
,
and
is obviously not a maximum. Having excluded
,
we can divide by
and get
and thus
.
The side condition
now gives
The desired maximum is
.
Problem 6. Show that, of all the triangles inscribed in a fixed circle, the equilateral triangle has the largest perimeter.
Solution. It is intuitively clear that this maximum exists and
geometrically clear that the triangle that offers this maximum contains the
center of the circle in its interior or on its boundary. We denote by
the central angles that subtend the three sides. It can verified by
trigonometry that the perimeter of the triangle is given by the function
As a side condition we have
To maximize the perimeter we form the gradients
The Lagrange condition
gives
and therefore
With
all in
,
we can conclude that
Since the central angles are equal, the sides are equal. The triangle is
therefore equilateral.
The Lagrange equation can be replaced by a cross-product equation: points that
satisfy
satisfy
In two variables this reduces to
Problem 7. Maximize and minimize on the unit circle
Solution. This problem was solved earlier by means of the Lagrange
equation. This time we will use the cross-product equation instead. As before
we set
so that the side condition takes the form
.
Since
we have
This gives
As before, the side condition
now implies that
and therefore that
.
The points under consideration are
At the first and fourth points
takes on the value
At the second and third points
takes on the value
.
The first is a maximum and the second a minimum.
Find the extreme value of subject to the condition (Maximum value is no minimum)
Find the maximum and minimum distances from the origin to the curve (Maximum is , minimum is )
Assume and are fixed positive numbers.
Find the extreme values of subject to the condition (Maximum is at minimum is at
Find the extreme values of subject to the condition (Minimum is at at no maximum)
Find the extreme values of subject to the side condition (Maximum is at the points where is any integer; minimum is at the points where is any integer)
Find the extreme values of the scalar field on the sphere (Maximum is at (minimum is at (-
Find the points of the surface nearest to the origin. ( and )
Find the shortest distance from the point to the parabola ()