3 Functions of More Than One Variable

Learning Outcome:

3. Understand single-valued functions of two or three variables and their derivatives, perform associated computations, and apply understanding and computations to solve problems.

3.1 Introduction During the ﬁrst and second terms of calculus you worked with functions of one variable, which were usually denoted y = f(x) (for example, y = x2). We could think of such a function as a “machine” named f, with the variable x being the input to the machine and the variable y being the output. We will now investigate functions of more than one input variable; the ones with which we will work will have the form z = f(x, y) or w = f(x, y, z). Examples would be things like f(x, y)= x2 + y2 and w = xyz +3y + z2. Consider the function w. There are three input variables x, y and z, and one output variable, w. We say that w is a function of three variables, the number of input variables. Similarly, f is a function of two variables. OK, so those two examples are bit abstract. Some more concrete (and practical!) examples would be the following: • The perceived temperature (“wind-chill”) depends on the actual air temperature and the speed of the wind. • The temperature in a well depends on the depth of the well and the time of year. • The pressure at a point in a room depends on the location (x, y, z) of the point. • The temperature on a plate of metal depends on the location (x, y) of the point.

In the first two semesters of calculus the two main themes were differentiation and integration of functions of one variable. We will now take up differentiation and, a bit later, integration of functions of more than one variable.

3.2 Representing Functions of More Than One Variable

Learning Outcome:

3. (a) Determine a function value of a function of more than one variable from an equation, a table of values, or a level curve plot.

In the past, when you were given a function of one variable, you often considered several representations of such a function: an equation of the function, a table of corresponding input and output values, a graph of the function. Each has their own advantages and disadvantages. Exact function values can be obtained from either the equation or the table, but the table has a limited number of values, whereas the equation can be used to get an unlimited number of input-output pairs. Exact input-output values are impossible to obtain from a graph, but the overall behavior of a function can be obtained quickly from looking at a graph. Functions of three or more variables are challenging to represent in any form other than an equation, so we will often focus on functions of two variables, which are more easily represented. There are four common representations of such functions; we will use three of the four regularly. • An equation can be given, from which function values can be obtained by simply inserting input values and obtaining corresponding output values. The two equations f(x, y) = x2 + y2 and w = xyz +3y + z2 are examples of this. An equation can easily be used to describe a function of any number of variables. The following three methods are only possible for functions of two variables.

• A table of values can be given. Such a table lists values of one input variable horizontally across the top or bottom of the table and values of the other input variable along one of the sides (usually the left side). The output value is for a given pair of input values is then located in the cell of the table where the row and column of the two input values intersect. An example is the table below, which gives (in the “body” of the table) the wind-chill temperature W (in degrees Fahrenheit) for various actual temperature/wind speed combinations (in that order). Wind speed is in miles per hour and actual temperature is given in degrees Fahrenheit.

Actual Temperature (◦F ) T v −10 0 10 20 30 40 5 −22 −11 1 13 25 36 10 −28 −16 −4 9 21 34 15 −32 −19 −7 6 19 32 −35 −22 −9 Wind Speed (mph) 20 4 17 30

• A level curve plot (contour plot) is essentially a “map” of the two input variables, with points having certain ﬁxed output values connected by curves. The one below represents temperatures T on a rectangular sheet of metal. The temperatures are in degrees Fahrenheit, and the distances are in feet (with x-values being on the horizontal axis and y-values on the vertical, as usual).

40 10 90 50 80 60 70 70

60 5 50 90 80

5 10 • A surface plot, which is a three-dimensional representation of the function, where the input values represent points in the xy-plane and the output values are z-values. When the z-values are plotted above (or below) their corresponding x and y values, a surface is obtained. Such plots are valuable for seeing the behavior of the function, but they are challenging to draw by hand, and almost impossible to get input-output values from. We will rarely (never?) use these. For examples of these, see page 792 1st Ed., page 779 2nd Ed.

3.3 Average Rates of Change

Learning Outcome:

3. (b) Determine the average rate of change of a function of two or three variables from one point to another from either a level curve graph or the equation of the function. • Examples: 780: 8, 9 • Exercises: 786: 40, 41, 43, 45 Answer to 786: 40 is −1.67 × 10−4 ppt/◦C

Looking at either the table of values or the level curve plot from the previous page, it is clear that the values of the functions (wind chill temperatures for the table, temperatures on the metal plate for the level curves) vary as the input values change. The main focus of our next couple weeks will be the rates at which the function values change as the input values change. We will begin by calculating average rates of change as one or the other (or perhaps both) of the input values change. Suppose that for a function f(x, y) we go from inputs of (x1,y1) to (x2,y2). Then the corresponding outputs go from f(x1,y1) to f(x2,y2). The average rate of change of the function is then the change in output divided by the change in input, which we show symbolically like this: f(x ,y ) − f(x ,y ) Ave ROC = 2 2 1 1 “distance” from (x1,y1) to (x2,y2) x1, y1, x2 and y2 should all be the appropriate numbers, and instead of f you should use the letter for whatever function you are working with. The “distance” from (x1,y1) to (x2,y2) will always be either a subtraction or an application of the Pythagorean Theorem. I’ve put distance in quotes because in many cases it will not actually represent a geometric distance, but simply a distance between numbers. Note that we generally find the difference by taking the final values minus the initial values. Here are a few examples, demonstrating the way that I’d like you to show your calculations: ⋄ Example 3.3(a): For the temperature function with level curves shown on the previous page, find the average rate of change of the temperature from (7, 1) to (7, 5).

T (7, 5) − T (7, 1) 65 − 38 Ave ROC = = =6.75◦F/foot ♠ 5 − 1 4

⋄ Example 3.3(b): For the function given by the table on the previous page, ﬁnd the rate of change in wind-chill from a actual temperature/wind speed combination of 0◦F and −10 mph to 0◦F and −20 mph.

W (0, −20) − W (0, −10) −22 − (−16) Ave ROC = = = −0.6◦F/mph ♠ 60 − 30 10 Note that in this last example the “distance” is really the change in wind speed. In both of the above example we have held one of the input variables ﬁxed while the other changes. In the next example we let both of the input variables vary. This only makes sense to do when both input variables measure the same quantity and have the same units. (In this case the quantity is true distance, measured in feet.

⋄ Example 3.3(c): For the temperature function from Exercise 1 of Assignment 8, ﬁnd the average rate of change in temperature from (5, 6) to (9, 1). T (9, 1) − T (5, 6) 60 − 75 Ave ROC = = = −2.3◦F/foot ♠ (9 − 5)2 + (1 − 6)2 6.4 p The quantity (9 − 5)2 + (1 − 6)2, is just the distance from (5, 6) to (9, 1), calculated using the distance formulap for two dimensions.

⋄ Example 3.3(d): In some location the temperature T in degrees Fahrenheit of the ground at a depth of d feet and at time t days from some zero date is modeled by

T (d, t)=30e−0.2d cos(0.0172t − 0.2d)+60 . Find the average rate of change of temperature at time ﬁfty days, from a depth of 3 feet to a depth of 40 feet.

T (40, 50) − T (3, 50) 60.0066 − 75.9110 Ave ROC = = = −0.4282◦F/foot ♠ 40 − 3 37

3.4 Partial Derivatives

Learning Outcome:

3. (c) Find and interpret (give location, direction, change in dependent variable per unit of change in independent variable) partial derivatives of a function at a point. • Examples: 794: 1, 2, 3, 7 • Exercises: 802: 13 - 33 odd, 37, 41 43, 57, 59, 61

Consider Example 3.3(d). Suppose that we next computed the average rate of change from (3, 50) to (10, 50), then from (3, 50) to (5, 50), then from (3, 50) to (3.5, 50), from (3, 50) to (3.1, 50), and so on. Note that we are holding the day ﬁxed at 50 and the initial depth at 3 feet. The ﬁnal depth, however, is getting closer and closer to 3 feet. What we should see is the average rates of change approaching some value, which is the instantaneous rate of change of temperature with respect to depth. The process of computing all those average rates of change would get a bit tedious; fortunately it is not necessary. All we need to do is compute the partial derivative of temperature T with respect ∂T to depth d, denoted by either or T (d, t). The computation of the partial derivative I’ll leave ∂d d to class and your textbook, but it turns out to be

−0.2d −0.2d Td(d, t) = 6e sin(0.0172t − 0.2d) − 6e cos(0.0172t − 0.2d) = 6e−0.2d [sin(0.0172t − 0.2d) − cos(0.0172t − 0.2d)] The reason for two parts is that we need to apply the product rule in order to obtain the derivative. ◦ If we now evaluate Td(3, 50) we obtain Td(3, 50) = −2.34 F/ft. What does this tell us? Well,any interpretation needs to include the following things: (i) the physical quantity that we are investigating (output variable of the function) (ii) the rate at which the quantity is increasing or decreasing (iii) the input variable in which the change is occurring (iv) the “place” where that rate of change is occurring For our example we might say On the ﬁftieth day, the temperature at a depth of three feet is decreasing with depth at a rate of 2.34 degrees Fahrenheit per foot. The physical quantity being studied is temperature, and the rate of change and “location” should be clear. The variable in which the change is occurring is indicated by the words “with depth” (and, less obviously,by the units). By convention, the rate of change for the output variable is always for a positive change in the input variable.

3.5 Partial Diﬀerential Equations

Learning Outcome:

3. (d) Determine whether a given function is a solution to a given partial diﬀerential equation. • Examples: 800: 11 • Exercises: 804: 76, 77, 79, 80, 82

3.6 Partial Derivatives for a Discrete Function

Learning Outcome:

3. (e) Use a forward, backward or centered diﬀerence to approximate a partial derivative of a function of two variables at a point, when the function is given in table form.

Suppose that we desire a partial derivative of a function of two variables that is represented with a table. It is impossible to determine the value of such a partial derivative, but we can find an average rate of change that approximates the partial derivative. When approximating partial derivatives from tables, there are three approximations we can get, a forward difference approximation, a backward difference approximation, and a centered difference approximation:

• Forward Difference Approximation: Here we simply compute an average rate of change from the point where we want the partial derivative to the next data point in the direction of increase in the changing input variable. • Backward Difference Approximation: In this case we compute an average rate of change from the point “before” (in the sense of the changing input variable) where we want the partial derivative to the point where we want the partial derivative. • Centered Difference Approximation: This is just the average of the forward difference and backward difference. It is equivalent to the average rate of change from the “before” point to the ahead or “after” point. Let’s look at some examples:

⋄ Example 3.6(a): For the wind-chill function, find the forward difference approximation of WT (20, 10). W (30, 10) − W (20, 10) 21 − 9 W (20, 10) ≈ = =1.2◦F/◦F ♠ T 30 − 20 10 Note that 30 is “forward from 20, so the forward approximation is as T goes from 20 to 30. To obtain the backward difference, we need to compute the average rate of change as T goes from 10 to 20:

⋄ Example 3.6(b): For the wind-chill function, ﬁnd the backward diﬀerence approximation of WT (30, −15). W (20, 10) − W (10, 10) 9 − (−4) W (20, 10) ≈ = =1.3◦F/◦F ♠ T 20 − 10 10

⋄ Example 3.6(c): For the wind-chill function, find the centered difference approximation of WT (20, 10). The centered difference approximation is just the average of the forward and backward differences: 1.2+1.4 W (20, 10) ≈ =1.3◦F/◦F T 2 It can also be found by taking the average rate of change from (10, 10) to (30, 10): W (30, 10) − W (10, 10) 22 − (−4) W (20, 10) ≈ = =1.25◦F/◦F ♠ T 30 − 10 20

3.7 Linear Approximations of Functions of Several Variables

Learning Outcome:

3. (f) Find and use local linear approximations (“linearizations”) of functions of one, two and three variables. Find vectors and a plane tangent to a surface at a point. (g) Find the error or percent error for a linear approximation. • Examples: 807: 1, sort of • Exercises: 811: 17, 19, 21

The basic idea here is this: Consider again T (x, y) as the function giving the temperature T at any point (x, y) on a sheet of metal, and suppose we know that ◦ ◦ ◦ T (4, 5) = 78.3 F, Tx(4, 5) = −0.3 F/ft, Ty(4, 5)=0.2 F/ft What if we wanted to know the temperature at (4.7, 5.4)? Since this point is close to (4, 5), the temperature should be somewhat close to 78.3◦F. The diagram to the right shows the two points under considera- tion, and it shows how we can move from (4, 5) to (4.7, 5.4) by going ∆x = 0.7 feet in the positive x-direction and then ∆y = 0.4 feet in the positive y-direction. Now we note that as we pass through y =6 (4, 5) in the positive x-direction the temperature is decreasing by ◦ −0.3 F/ft, so as we move by ∆x = 0.7 feet we would expect the (4.7, 5.4) ∆y =0.4 temperature to change by approximately ◦ ◦ (4, 5) ∆Tx = (0.7ft)(−0.3 F/ft) = −0.21 F y =5

Similarly the change in temperature in the y-direction should be x =4 x =5 ◦ ◦ ∆x =0.7 ∆Ty = (0.4ft)(0.2 F/ft) = 0.08 F We can then approximate the temperature at (4.7, 5.4) by adding these tow changes to the temperature at (4, 5):

◦ T (4.7, 5.4) ≈ T (4, 5)+∆Tx + ∆Ty = 78.3+(−0.21)+0.08 = 78.17 F The reason this is an approximation rather than an actual value is that the rates of change only hold right at (4, 5), but they probably don’t change much if we don’t get too far away from (4, 5). There is a way to get a quantitative idea of what the error might be, but it fairly technical. At this point, one should just try to understand what has happened here. What we have done is use something called the linear approximation of the function T at (or sometimes we say “near”) (4, 5). The formal way of deﬁning the linear approximation f a function f of two variables at a point (x0,y0) is

f(x, y) ≈ f(x0,y0)+ fx(x0,y0)∆x + fy(x0,y0)∆y for any point (x, y) near (x0,y0). Note what this says: To get a function value at a point near (x0,y0) we begin with the function value at that point, f(x0,y0). We then adjust for the point (x, y) by adding the changes in the x- and y-directions, which are given by the two terms fx(x0,y0)∆x and fy(x0,y0)∆y. We can actually create a new function L(x, y) that gives us our approximations near (x0,y0), using the facts that ∆x = x − x0 and ∆y = y − y0 This new function looks like

L(x, y)= f(x0,y0)+ fx(x0,y0)(x − x0)+ fy(x0,y0)(y − y0) For our example above we would have L(x, y)=78.3+(−0.3)(x − 4)+(0.2)(y − 5) = 78.5 − 0.3x +0.2y The last expression is obtained by distributing and combining like terms. The function L is called the linearization of f at the point (x0,y0): Linearization of a Function at a Point

The linearization of a function f(x, y) at a point (x0,y0) is the function

L(x, y)= f(x0,y0)+ fx(x0,y0)(x − x0)+ fy(x0,y0)(y − y0)

For points (x, y) “near” (x0,y0), L(x, y) ≈ f(x, y). A similar formula holds for functions of three or more variables.

⋄ Example 3.7(a): Give the linearization of f(x, y, z)= x2y−3xz+y2 at the point (2, 5, −1) and use it to get an approximate value of f(2.07, 4.95, −1.03). Because we have three variables, the linearization will look like

L(x, y, z)= f(2, 5, −1)+ fx(2, 5, −1)(x − 2) + fy(2, 5, −1)(y − 5)+ fz(2, 5, −1)(z − (−1)) (1) Doing some calculations gives us

2 fx(x, y, z)=2xy − 3z, fy(x, y, z)= x +2y, fz(x, y, z)= −3x and

f(2, 5, −1) = 51, fx(2, 5, −1) = 23, fy(2, 5, −1) = 14, fz(2, 5, −1) = −6. Substituting into (1), the linearization is then L(x, y, z) = 51+23(x − 2) + 14(y − 5) − 6(z + 1) = 51+23x − 46+14y − 70 − 6z − 6 = −71+23x + 14y − 6z

The approximate value of f(2.07, 4.95, −1.03) obtained from the linearization is then f(2.07, 4.95, −1.03) ≈ L(2.07, 4.95, −1.03) = −71 + 23(2.07) + 14(4.95) − 6(−1.03) = 52.09 ♠

Let’s make it clear that the linearization does not give us exact function values except at the point where the linearization was done. However, the values it gives us should be close to the function values as long as we are at points “near” the point where the linearization was done. Exactly how near we need to be to be how close is a subject for another class! The absolute value of the diﬀerence between the actual function value at a point and the value given by the linearization at the same point is usually called the error. What is more useful than the error, however, is the percent error, which we’ll deﬁne by |actual value − approximate value| percent error = × 100 |actual value| ⋄ Example 3.7(b): Find the error and percent error when using the linearization of the function f(x, y, z) = x2y − 3xz + y2 from Example 3.7(a) at (2, 5, −1) to approximate the value of f(2.07, 4.95, −1.03). The actual value of f(2.07, 4.95, −1.03) is f(2.07, 4.95, −1.03) = (2.07)2(4.95) − 3(2.07)(−1.03)+ (4.95)2 = 52.109055, so the error is

|f(2.07, 4.95, −1.03) − L(2.07, 4.95, −1.03)| = |52.109055 − 52.09| =0.019055 The percent error is |f(2.07, 4.95, −1.03) − L(2.07, 4.95, −1.03)| × 100 = 0.037% ♠ |52.109055|

⋄ Example 3.7(c): The temperature T (in degrees Fahrenheit) at each point (x, y) (both in feet) on a plate of metal is a function of where on the plate one is; that is, T = T (x, y). Suppose we know that ◦ ◦ ◦ T (4, 2) = 93 F, Tx(4, 2) = −3 F/ft, Ty(4, 2)=2 F/ft Use this information to approximate the temperature at (4.1, 1.7). There are two approaches we can use here, one utilizing the linearization of the function and the other a bit less formal. Using the second approach, the temperature at (4.1, 1.7) should be approximately the temperature at (4, 2), but with an adjustment based on each of the partial derivatives. The adjustment in the x-direction will be −3◦F/ft × 0.1ft = −0.3◦F and the adjustment in the y- direction will be 2◦F/ft×−0.3ft = −0.6◦F. The temperature at (4.1, 1.7) will then be approximately 93 − 0.3 − 0.6=92.1◦F. The more formal approach is to ﬁnd the linearization L(x, y)=93 − 3(x − 4)+2(y − 2) = 101 − 3x +2y and, using it, obtain

f(4.1, 1.7) ≈ L(4.1, 1.7) = 101 − 3(4.1)+ 2(1.7) = 92.1◦F ♠ 3.8 Directional Derivatives

Learning Outcome:

3. (h) Find and interpret (location, direction, change in dependent variable per unit of change in independent variables) directional derivatives of a function at a point. (i) Determine the direction in which a function has the greatest rate of increase or decrease at a point, and give that rate. Determine the directions from a point in which a function remains constant. • Examples: 818: 6(b), 7 • Exercises: 823: 21 - 33 odd

Recently we have investigated partial derivatives, which represent the rates of change of a function as we travel in the positive x or y direction. You may have wondered why we don’t determine the rate of change of a function in some other direction. Well, now we are going to do that; such a rate of change is called a directional derivative. The authors of the two books we are referring to define this in slightly different ways than each other; we’ll go with the definition that I’m going to give here. Recall the definition of a derivative of a function f of one variable at a point x0:

′ f(x0 + h) − f(x0) f (x0) = lim . h→0 h f(x + h) − f(x ) Note that the quantity 0 0 is the average rate of change of the function from a point h x0 to the point x0 + h. If we then look at such rates of change for h getting closer and closer to zero, the values we get approach some number that we call the derivative of f at x0, denoted by ′ f (x0). Now suppose that we have a function f of two variables. The partial derivative of f with respect to x at a point (x0,y0) is then deﬁned by

f(x0 + h, y0) − f(x0,y0) fx(x0,y0) = lim . h→0 h f(x + h, y ) − f(x ,y ) Here the quantity 0 0 0 0 is the average rate of change of f from (x ,y ) to h 0 0 (x0 + h, y0). Note that the y value of the function is fixed at y0, and only the x value changes as we move from the first point to the second. The limit as the distance moved is reduced to zero is then the derivative of f with respect to x at (x0,y0). Suppose now that we want to find the derivative of a function f of two variables in the direction of a vector v = ai + bj. The idea is this: If we place the tail of some scalar multiple hv of the vector v at the point (x0,y0), its tip will be at the point (x0 + ha, y0 + hb). We then compute the average rate of change of the function from (x0,y0) to (x0 + ha, y0 + hb) and take its limit as h goes to zero. The notation we use for this derivative is Dvf(x0,y0), and from this discussion we get that it is

f(x0 + ha, y0 + hb) − f(x0,y0) Dvf(x0,y0) = lim h→0 h The value obtained is independent of the magnitude of the vector v, depending only on its direction. The diagram to the right shows the point (x0,y0) and the points (x0 + a,y0 + b) (x0 + ah, y0 + hb) for several values of h. Recognize that the notation h =1 Dv(x0,y) contains four pieces of information: • The letter D tells us it is a derivative. 1 • f is the function that it is a derivative of. h = 2

1 • (x0,y0) is the point where the derivative is being taken. h = 4

• v gives the direction in which the derivative is taken. (x0,y0)

OK, so that’s all ﬁne and dandy, but how do we go about actually computing directional derivatives? The key to this is a vector function, called the gradient associated with the function f. The notation for the gradient function is ∇f(x, y). It is a function of the same number of variables as f, but its “output” is a vector with the same number of components as the variables of f. We will use it to ﬁnd directional derivatives, but it has some special characteristics also.

The Gradient of a Function

The gradient of a function f(x, y) at a point (x0,y0) is the vector

∇f(x0,y0)= fx(x0,y0)i + fy(x0,y0)j The gradient vector at a point has the following characteristics: • It is perpendicular to the level curve (or, for a function of three variables, the level surface) through the point (x0,y0). • It points in the direction of greatest increase of the function at the point (x0,y0). • Its magnitude is the greatest rate of increase that can be experienced when passing through the point (x0,y0).

Functions of three (or more) variables have gradients that are deﬁned in the same way, but with the additional components corresponding to the additional variables. Here’s how the gradient is used to calculate directional derivatives:

Computing a Directional Derivative

The directional derivative of f(x, y) at (x0,y0) and in the direction of a unit vector u is

Duf(x0,y0)= ∇f(x0,y0) u

Note that one must use a unit vector when computing a directional derivative this way. ⋄ Example 3.7(b): Find the directional derivative of f(x, y, z) = x2y − 3xz + y2 at the point (2, 5, −1) in the direction of the vector v = 2, −2, 1. Doing some calculations gives us 2 fx(x, y, z)=2xy − 3z, fy(x, y, z)= x +2y, fz(x, y, z)= −3x and fx(2, 5, −1) = 23, fy(2, 5, −1) = 14, fz(2, 5, −1) = −6, so ∇f(2, 5, −1) = 23, 14, −6. We now need the unit vector u in the direction of v: v 2, −2, 1 2 2 1 u = = = , − , v 22 +(−2)2 +12 3 3 3 The directional derivative is then p 2 2 1 12 Du(2, 5, −1) = ∇f(2, 5, −1) u = 23, 14, −6 , − , = =4 ♠ 3 3 3 3

3.9 Relative Minima and Maxima of Functions of Several Variables

Learning Outcome:

3. (j) Find relative minima and maxima of a function of one or two variables. (This includes both the function values and where they occur.) • Examples: 836: 3, 4 • Exercises: 843: 7 - 21 odd

Suppose that we wished to find where the function f(x) = x2 +6x − 3 had a maximum or a minimum. First we would take the derivative and set it equal to zero - this would tell us where the graph of the function had a horizontal tangent, which is then a place where the function might have a maximum or a minimum: f(x)= x2 +6x − 3 =⇒ f ′(x)=2x +6 =⇒ f ′(x) = 0 when x = −3 So we have identified that the function might have a maximum or minimum at the critical point x = −3. How do we determine whether it has one of those and, if it does, which one? In first-term calculus you learned two methods for doing this: • The First Derivative Test: For this test we determine the sign of the derivative on either side of the critical point and use that information to determine what is happening at the critical point. In the case of our example, we can see that the derivative will be negative for values of x less than −3, and it will be positive for values of x greater than −3. This means the graph of f(x) is sloping downward to the left of x = −3 and upward to the right of −3, so the function has a minimum at x = −3. • The Second Derivative Test: For this test we determine the sign of the second derivative at the critical point; if the sign of the second derivative there is negative, the graph is concave down and there is a maximum at the critical point, and if the sign of the second derivative at the critical point is negative, then the graph is concave up and there is a minimum at the critical point. For the above example, f ′′(x)=2 > 0 for all values of x, so the function is concave up “everywhere” (for all values of x). Therefore f has a minimum at x = −3. The minimum at x = −3 is actually an absolute minimum, meaning that the function has a smaller value there than anywhere else. All that the first and second derivative tests really establish, however, is that there is a relative minimum at x = −3, meaning that the function value there is smaller than at nearby values of x. The situation for a function of two variables is more complicated than it is for a function of just one variable, because such a function has two first derivatives (considering only partial derivatives) and three second derivatives. For such a function we first define

Critical Points of a Function of Two Variables

Let f be a function of two variables. Then any point (x0,y0) at which BOTH fx(x0,y0) = 0 and fy(x0,y0)=0 is called a critical point of the function.

Determining critical points is essentially ﬁnding where the derivative is zero. After locating all such points, we must then determine whether there is a minimum or maximum value of the function at each such point, or whether there is a saddle point there. (A saddle point is roughly analogous to an inﬂection point of a function of one variable.) To do this we use the following second derivative test.

Second Partials Test Let f be a function of two variables with continuous second-order partial derivatives in some circle centered at a critical point (x0,y0), and let

2 D = fxx(x0,y0)fyy(x0,y0) − [fxy(x0,y0)] .

(a) If D > 0 and fxx(x0,y0) < 0, then f has a relative maximum at (x0,y0).

(b) If D > 0 and fxx(x0,y0) > 0, then f has a relative minimum at (x0,y0).

When considering maxima and minima, it is important to distinguish where the maxima or minima occur and what the maxima or minima are. The maxima and minima are the values of the function obtained by evaluating the function at all points (x0,y0) where maxima or minima occur. The maxima and minima that we ﬁnd using this procedure are what we call relative maxima and relative minima. Sometimes you will see the word “local” used instead of “relative.” These occur at points where the function value is higher or lower than it is at all “nearby” points. 3.10 Minima and Maxima of Functions on Closed and Bounded Regions

Learning Outcome:

3. (k) Given a level curve plot of a function of two variables, determine the locations and approximate values of absolute maxima and minima on a closed region. (l) Use calculus to ﬁnd absolute maxima and minima of a function of two variables on a closed region. • Examples: 838: 5, 6 • Exercises: 845: 37, 39, 41

In the last section we discussed determining where a function has a relative maximum or minimum. Some functions don’t have any relative maxima or minima (plural for maximum and minimum), however. The simplest example of such a function is a sloping plane. But if we restrict ourselves to what is called a closed and bounded region in R2, a function f(x, y) will have an absolute maximum and minimum for that region. What do we mean by closed and bounded? Bounded means essentially that our region will ﬁt inside of some circle, and closed means the region included its boundary. Let’s take a look at these ideas using the temperature function with level curves shown below:

40 10 90 50 80 60 70 70

60 5 50 90 80

5 10

First of all we note that the function deﬁnitely has a relative maximum of about 95◦F, located some- where around (11, 5.5); there is no relative minimum that we can see in the region for which we have information. Now consider the region enclosed by the dashed lines (including the lines themselves), which is described by 1 ≤ x ≤ 8, 0 ≤ y ≤ 4. There is no relative maximum or minimum of the function inside the rectangle. However, you should be able to see that the minimum temperature is perhaps 25◦F and occurs along the bottom edge of the rectangle at about (5, 0). The maximum temperature is about 65◦F at the point (8, 4). Note that the maximum value occurs at a corner of the region, and the minimum value occurs at a low point along one of the edges. Now look at the region described by 10 ≤ x ≤ 13, 3 ≤ y ≤ 10, which is the region enclosed by the rectangle with dotted sides. The maximum temperature in the region is about 95◦F at about (10.9, 5.4), which is a relative maximum of the function as well. The minimum value is about 44◦F at (13, 10); there is not a relative minimum at that point. Everything that can possibly happen when ﬁnding the minimum and maximum of a function on a closed and bounded region has occurred in the two situations that we just looked at. To summarize, the minimum and maximum values of the function will occur at one of three locations:

• an interior point where there is a relative minimum or maximum • a low or high point along an edge • a corner of the region Now suppose that we are given the equation of a function of two variables, and a rectangular region to which the two variables are restricted. Here is the process for determining the minimum and maximum values of the function in the region: 1) Determine all points within the region where both ﬁrst partial derivatives are zero. We will not need to conduct the second derivative test! 2) a. Create a function of one variable by ﬁxing one of the variables with the value it has along one edge of the rectangle. b. Take the derivative of the function with respect to the remaining variable and determine whether it is zero at any point(s) along that edge of the rectangle. If it is, make note of the point(s). c. Repeat for the other three edges of the rectangle. 3) Evaluate the function at each of the points you found in parts (1) and (2), and at each of the four corners. The largest value you obtain is the maximum value of the function in the region, and the minimum value you obtain is the minimum value of the function in the region.

⋄ Example 3.10(a): Find the absolute minimum and maximum and their locations for T (x, y)= x2 + xy + y2 − 6x +2 on the region bounded by 0 ≤ x ≤ 5 and −3 ≤ y ≤ 0. First we ﬁnd critical points on the interior of the region. The two partial derivatives are

Tx(x, y)=2x + y − 6 and Ty(x, y)= x +2y. We set each of these equal to zero and solve the resulting system of equations: 2x + y − 6=0 =⇒ 2x + y =6 2x +(−2) = 6 x +2y =0 =⇒ −2x − 4y =20 2x = 8 times − 2 −3y =6 2x = 8 y = −2 x = 4

The solution to the system of equations is (4, −2), so that is the interior critical point that must be checked. Now we need to ﬁnd critical points along each edge of the rectangle. When x =0 we have T (0,y)= y2 +2 ⇒ T ′(0,y)=2y

T ′(0,y)=0 when y =0, so the point along the edge where x =0 that we must check is (0, 0). (This is a corner point, so we would have checked it anyway.) We do the same thing for the other three edges: x = 5 : T (5,y)= y2 +5y − 3 ⇒ T ′(5,y)=2y +5=0 when y = −2.5, so check (5, −2.5) y = −3 : T (x, −3) = x2 − 9x +11 ⇒ T ′(x, −3)=2x − 9=0 when x =4.5, so check (4.5, −3) y = 0 : T (x, 0) = ⇒ T ′(x, 0)=2x − 6=0 when x =3, so check (3, 0) Those are all of the points we have to check along each edge. We must also check the four corner points, (0, 0), (5, 0), (0, −3) and (5, −3). So now we have eight points where (x, y) T (x, y) we must check the function value; this is generally best , organized in a table, as shown to the right. We conclude (4 −2) −10 that the minimum value of the function (0, 0) 2 (5, −2.5) −9.25 f(x, y)= x2 + xy + y2 − 6x +2 (4.5, −3) −9.25 on the rectangle bounded by (3, 0) −7 (5, 0) −3 0 ≤ x ≤ 5 and − 3 ≤ y ≤ 0 (0, −3) 11 is −10 at (4, −2). The maximum value is 11 at (0, −3). (5, −3) −9 ♠