Differentiation of Functions of Several Variables

# 27 Maxima/Minima Problems

### Learning Objectives

- Use partial derivatives to locate critical points for a function of two variables.
- Apply a second derivative test to identify a critical point as a local maximum, local minimum, or saddle point for a function of two variables.
- Examine critical points and boundary points to find absolute maximum and minimum values for a function of two variables.

One of the most useful applications for derivatives of a function of one variable is the determination of maximum and/or minimum values. This application is also important for functions of two or more variables, but as we have seen in earlier sections of this chapter, the introduction of more independent variables leads to more possible outcomes for the calculations. The main ideas of finding critical points and using derivative tests are still valid, but new wrinkles appear when assessing the results.

### Critical Points

For functions of a single variable, we defined critical points as the values of the function when the derivative equals zero or does not exist. For functions of two or more variables, the concept is essentially the same, except for the fact that we are now working with partial derivatives.

Let be a function of two variables that is defined on an open set containing the point The point is called a critical point of a function of two variables if one of the two following conditions holds:

- Either does not exist.

Find the critical points of each of the following functions:

- First, we calculate

Next, we set each of these expressions equal to zero:

Then, multiply each equation by its common denominator:

Therefore, and so is a critical point of

We must also check for the possibility that the denominator of each partial derivative can equal zero, thus causing the partial derivative not to exist. Since the denominator is the same in each partial derivative, we need only do this once:

This equation represents a hyperbola. We should also note that the domain of consists of points satisfying the inequality

Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. To put the hyperbola in standard form, we use the method of completing the square:

Dividing both sides by puts the equation in standard form:

Notice that point is the center of the hyperbola. - First, we calculate

Next, we set each of these expressions equal to zero, which gives a system of equations in

Subtracting the second equation from the first gives Substituting this into the first equation gives so Therefore is a critical point of ((Figure)). There are no points in that make either partial derivative not exist.

Find the critical point of the function

Calculate and then set them equal to zero.

The main purpose for determining critical points is to locate relative maxima and minima, as in single-variable calculus. When working with a function of one variable, the definition of a local extremum involves finding an interval around the critical point such that the function value is either greater than or less than all the other function values in that interval. When working with a function of two or more variables, we work with an open disk around the point.

Let be a function of two variables that is defined and continuous on an open set containing the point Then *f* has a *local maximum* at if

for all points within some disk centered at The number is called a *local maximum value*. If the preceding inequality holds for every point in the domain of then has a *global maximum* (also called an *absolute maximum*) at

The function has a *local minimum* at if

for all points within some disk centered at The number is called a *local minimum value*. If the preceding inequality holds for every point in the domain of then has a *global minimum* (also called an *absolute minimum*) at

If is either a local maximum or local minimum value, then it is called a *local extremum* (see the following figure).

In Maxima and Minima, we showed that extrema of functions of one variable occur at critical points. The same is true for functions of more than one variable, as stated in the following theorem.

Let be a function of two variables that is defined and continuous on an open set containing the point Suppose and each exists at If has a local extremum at then is a critical point of

### Second Derivative Test

Consider the function This function has a critical point at since However, does not have an extreme value at Therefore, the existence of a critical value at does not guarantee a local extremum at The same is true for a function of two or more variables. One way this can happen is at a saddle point. An example of a saddle point appears in the following figure.

In this graph, the origin is a saddle point. This is because the first partial derivatives of are both equal to zero at this point, but it is neither a maximum nor a minimum for the function. Furthermore the vertical trace corresponding to is (a parabola opening upward), but the vertical trace corresponding to is (a parabola opening downward). Therefore, it is both a global maximum for one trace and a global minimum for another.

Given the function the point is a saddle point if both and but does not have a local extremum at

The second derivative test for a function of one variable provides a method for determining whether an extremum occurs at a critical point of a function. When extending this result to a function of two variables, an issue arises related to the fact that there are, in fact, four different second-order partial derivatives, although equality of mixed partials reduces this to three. The second derivative test for a function of two variables, stated in the following theorem, uses a discriminant that replaces in the second derivative test for a function of one variable.

Let be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point Suppose and Define the quantity

- If and then has a local minimum at
- If and then has a local maximum at
- If then has a saddle point at
- If then the test is inconclusive.

See (Figure).

To apply the second derivative test, it is necessary that we first find the critical points of the function. There are several steps involved in the entire procedure, which are outlined in a problem-solving strategy.

Let be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point To apply the second derivative test to find local extrema, use the following steps:

- Determine the critical points of the function where Discard any points where at least one of the partial derivatives does not exist.
- Calculate the discriminant for each critical point of
- Apply (Figure) to determine whether each critical point is a local maximum, local minimum, or saddle point, or whether the theorem is inconclusive.

Find the critical points for each of the following functions, and use the second derivative test to find the local extrema:

- Step of the problem-solving strategy involves finding the critical points of To do this, we first calculate and then set each of them equal to zero:

Setting them equal to zero yields the system of equations

The solution to this system is and Therefore is a critical point of

Step 2 of the problem-solving strategy involves calculating To do this, we first calculate the second partial derivatives of

Therefore,

Step 3 states to check (Figure). Since and this corresponds to case 1. Therefore, has a local minimum at as shown in the following figure.

- For step 1, we first calculate and then set each of them equal to zero:

Setting them equal to zero yields the system of equations

To solve this system, first solve the second equation for*y.*This gives Substituting this into the first equation gives

Therefore, or Substituting these values into the equation yields the critical points and

Step 2 involves calculating the second partial derivatives of

Then, we find a general formula for

Next, we substitute each critical point into this formula:

In step 3, we note that, applying (Figure) to point leads to case which means that is a saddle point. Applying the theorem to point leads to case 1, which means that corresponds to a local minimum as shown in the following figure.

Use the second derivative to find the local extrema of the function

is a saddle point, is a local maximum.

Follow the problem-solving strategy for applying the second derivative test.

### Absolute Maxima and Minima

When finding global extrema of functions of one variable on a closed interval, we start by checking the critical values over that interval and then evaluate the function at the endpoints of the interval. When working with a function of two variables, the closed interval is replaced by a closed, bounded set. A set is *bounded* if all the points in that set can be contained within a ball (or disk) of finite radius. First, we need to find the critical points inside the set and calculate the corresponding critical values. Then, it is necessary to find the maximum and minimum value of the function on the boundary of the set. When we have all these values, the largest function value corresponds to the global maximum and the smallest function value corresponds to the absolute minimum. First, however, we need to be assured that such values exist. The following theorem does this.

A continuous function on a closed and bounded set in the plane attains an absolute maximum value at some point of and an absolute minimum value at some point of

Now that we know any continuous function defined on a closed, bounded set attains its extreme values, we need to know how to find them.

Assume is a differentiable function of two variables defined on a closed, bounded set Then will attain the absolute maximum value and the absolute minimum value, which are, respectively, the largest and smallest values found among the following:

- The values of at the critical points of in
- The values of on the boundary of

The proof of this theorem is a direct consequence of the extreme value theorem and Fermat’s theorem. In particular, if either extremum is not located on the boundary of then it is located at an interior point of But an interior point of that’s an absolute extremum is also a local extremum; hence, is a critical point of by Fermat’s theorem. Therefore the only possible values for the global extrema of on are the extreme values of on the interior or boundary of

Let be a continuous function of two variables defined on a closed, bounded set and assume is differentiable on To find the absolute maximum and minimum values of on do the following:

- Determine the critical points of in
- Calculate at each of these critical points.
- Determine the maximum and minimum values of on the boundary of its domain.
- The maximum and minimum values of will occur at one of the values obtained in steps

Finding the maximum and minimum values of on the boundary of can be challenging. If the boundary is a rectangle or set of straight lines, then it is possible to parameterize the line segments and determine the maxima on each of these segments, as seen in (Figure). The same approach can be used for other shapes such as circles and ellipses.

If the boundary of the set is a more complicated curve defined by a function for some constant and the first-order partial derivatives of exist, then the method of Lagrange multipliers can prove useful for determining the extrema of on the boundary. The method of Lagrange multipliers is introduced in Lagrange Multipliers.

Use the problem-solving strategy for finding absolute extrema of a function to determine the absolute extrema of each of the following functions:

- on the domain defined by and
- on the domain defined by

- Using the problem-solving strategy, step involves finding the critical points of on its domain. Therefore, we first calculate and then set them each equal to zero:

Setting them equal to zero yields the system of equations

The solution to this system is and Therefore is a critical point of Calculating gives

The next step involves finding the extrema of on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph:

is the line segment connecting and and it can be parameterized by the equations for Define This gives Differentiating*g*leads to Therefore, has a critical value at which corresponds to the point Calculating gives the*z-*value

is the line segment connecting and and it can be parameterized by the equations for Again, define This gives Then, has a critical value at which corresponds to the point Calculating gives the*z-*value

is the line segment connecting and and it can be parameterized by the equations for Again, define This gives The critical value corresponds to the point So, calculating gives the*z-*value

is the line segment connecting and and it can be parameterized by the equations for This time, and the critical value correspond to the point Calculating gives the*z-*value

We also need to find the values of at the corners of its domain. These corners are located at

*** QuickLaTeX cannot compile formula: \begin{array}{}\\ \hfill f\left(0,0\right)& =\hfill & {\left(0\right)}^{2}-2\left(0\right)\left(0\right)+4{\left(0\right)}^{2}-4\left(0\right)-2\left(0\right)+24\hfill & =\hfill & 24\hfill \\ \hfill f\left(4,0\right)& =\hfill & {\left(4\right)}^{2}-2\left(4\right)\left(0\right)+4{\left(0\right)}^{2}-4\left(4\right)-2\left(0\right)+24\hfill & =\hfill & 24\hfill \\ \hfill f\left(4,2\right)& =\hfill & {\left(4\right)}^{2}-2\left(4\right)\left(2\right)+4{\left(2\right)}^{2}-4\left(4\right)-2\left(2\right)+24\hfill & =\hfill & 20\hfill \\ \hfill f\left(0,2\right)& =\hfill & {\left(0\right)}^{2}-2\left(0\right)\left(2\right)+4{\left(2\right)}^{2}-4\left(0\right)-2\left(2\right)+24\hfill & =\hfill & 36.\hfill \end{array} *** Error message: Missing # inserted in alignment preamble. leading text: $\begin{array}{} Missing $ inserted. leading text: $\begin{array}{}\\ \hfill f\left Missing $ inserted. leading text: $\begin{array}{}\\ \hfill f\left(0,0\right)& Extra alignment tab has been changed to \cr. leading text: $\begin{array}{}\\ \hfill f\left(0,0\right)& Extra alignment tab has been changed to \cr. leading text: ...ay}{}\\ \hfill f\left(0,0\right)& =\hfill & Missing $ inserted. leading text: ... \hfill f\left(0,0\right)& =\hfill & {\left Extra }, or forgotten $. leading text: ...left(0,0\right)& =\hfill & {\left(0\right)} Missing } inserted. leading text: ...-4\left(0\right)-2\left(0\right)+24\hfill & Extra }, or forgotten $. leading text: ...-4\left(0\right)-2\left(0\right)+24\hfill &

The absolute maximum value is which occurs at and the global minimum value is which occurs at both and as shown in the following figure.

- Using the problem-solving strategy, step involves finding the critical points of on its domain. Therefore, we first calculate and then set them each equal to zero:

Setting them equal to zero yields the system of equations

The solution to this system is and Therefore, is a critical point of Calculating we get

The next step involves finding the extrema of*g*on the boundary of its domain. The boundary of its domain consists of a circle of radius centered at the origin as shown in the following graph.

The boundary of the domain of can be parameterized using the functions for Define

Setting leads to

This equation has two solutions over the interval One is and the other is For the first angle,

Therefore, and so is a critical point on the boundary and

For the second angle,

Therefore, and so is a critical point on the boundary and

The absolute minimum of*g*is which is attained at the point which is an interior point of*D*. The absolute maximum of*g*is approximately equal to 44.844, which is attained at the boundary point These are the absolute extrema of*g*on*D*as shown in the following figure.

Use the problem-solving strategy for finding absolute extrema of a function to find the absolute extrema of the function

on the domain defined by and

The absolute minimum occurs at

The absolute maximum occurs at

Calculate and and set them equal to zero. Then, calculate for each critical point and find the extrema of on the boundary of

Pro- company has developed a profit model that depends on the number *x* of golf balls sold per month (measured in thousands), and the number of hours per month of advertising *y*, according to the function

where is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold is and the maximum number of hours of advertising that can be purchased is Find the values of and that maximize profit, and find the maximum profit.

Using the problem-solving strategy, step involves finding the critical points of on its domain. Therefore, we first calculate and then set them each equal to zero:

Setting them equal to zero yields the system of equations

The solution to this system is and Therefore is a critical point of Calculating gives

The domain of this function is and as shown in the following graph.

is the line segment connecting and and it can be parameterized by the equations for We then define

Setting yields the critical point which corresponds to the point in the domain of Calculating gives

is the line segment connecting and and it can be parameterized by the equations for Once again, we define

This function has a critical point at which corresponds to the point This point is not in the domain of

is the line segment connecting and it can be parameterized by the equations for We define

This function has a critical point at which corresponds to the point which is not in the domain.

is the line segment connecting and it can be parameterized by the equations for We define

This function has a critical point at which corresponds to the point which is on the boundary of the domain. Calculating gives

We also need to find the values of at the corners of its domain. These corners are located at

The maximum critical value is which occurs at Therefore, a maximum profit of is realized when golf balls are sold and hours of advertising are purchased per month as shown in the following figure.

### Key Concepts

- A critical point of the function is any point where either or at least one of and do not exist.
- A saddle point is a point where but is neither a maximum nor a minimum at that point.
- To find extrema of functions of two variables, first find the critical points, then calculate the discriminant and apply the second derivative test.

### Key Equations

**Discriminant**

For the following exercises, find all critical points.

For the following exercises, find the critical points of the function by using algebraic techniques (completing the square) or by examining the form of the equation. Verify your results using the partial derivatives test.

Maximum at

Relative minimum at

For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these.

The second derivative test fails. Since for all *x* and *y* different from zero, and when either *x* or *y* equals zero (or both), then the absolute minimum occurs at

is a saddle point.

is a saddle point.

is a local maximum.

Relative minimum located at

is a saddle point.

and are saddle points; is a relative minimum.

is a relative maximum.

is a saddle point.

The relative maximum is at

is a saddle point and is the relative minimum.

For the following exercises, determine the extreme values and the saddle points. Use a CAS to graph the function.

**[T]**

A saddle point is located at

**[T]**

**[T]**

There is a saddle point at local maxima at and local minima at

Find the absolute extrema of the given function on the indicated closed and bounded set

is the triangular region with vertices

Find the absolute maximum and minimum values of on the region

is the absolute minimum and is the absolute maximum.

on

on

There is an absolute minimum at and an absolute maximum at

Find three positive numbers the sum of which is such that the sum of their squares is as small as possible.

Find the points on the surface that are closest to the origin.

Find the maximum volume of a rectangular box with three faces in the coordinate planes and a vertex in the first octant on the plane

The sum of the length and the girth (perimeter of a cross-section) of a package carried by a delivery service cannot exceed in. Find the dimensions of the rectangular package of largest volume that can be sent.

A cardboard box without a lid is to be made with a volume of ft^{3}. Find the dimensions of the box that requires the least amount of cardboard.

Find the point on the surface nearest the plane Identify the point on the plane.

Find the point in the plane that is closest to the origin.

A company manufactures two types of athletic shoes: jogging shoes and cross-trainers. The total revenue from units of jogging shoes and units of cross-trainers is given by where and are in thousands of units. Find the values of *x* and *y* to maximize the total revenue.

and

A shipping company handles rectangular boxes provided the sum of the length, width, and height of the box does not exceed in. Find the dimensions of the box that meets this condition and has the largest volume.

Find the maximum volume of a cylindrical soda can such that the sum of its height and circumference is cm.

cm^{3}

### Glossary

- critical point of a function of two variables
- the point is called a critical point of if one of the two following conditions holds:

- At least one of and do not exist

- discriminant
- the discriminant of the function is given by the formula

- saddle point
- given the function the point is a saddle point if both and but does not have a local extremum at