# SECOND-PARTIALS TEST

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 .

# MAXIMA AND MINIMA WITH SIDE CONDITIONS

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.

# The Method of Lagrange

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 .

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 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.

# Exercise

1. Find the extreme value of subject to the condition (Maximum value is no minimum)

2. Find the maximum and minimum distances from the origin to the curve (Maximum is , minimum is )

3. Assume and are fixed positive numbers.

1. Find the extreme values of subject to the condition (Maximum is at minimum is at

2. Find the extreme values of subject to the condition (Minimum is at at no maximum)

4. 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)

5. Find the extreme values of the scalar field on the sphere (Maximum is at (minimum is at (-

6. Find the points of the surface nearest to the origin. ( and )

7. Find the shortest distance from the point to the parabola ()

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