SECTION 11.7 MAXIMUM AND MINIMUM VALUES 645 absolute minimum FIGURE 1 local maimum local minimum absolute maimum Look at the hills and valles in the graph of f shown in Figure 1. There are two points a, b where f has a local maimum, that is, where f a, b is larger than nearb values of f,. The larger of these two values is the absolute maimum. Likewise, f has two local minima, where f a, b is smaller than nearb values. The smaller of these two values is the absolute minimum. 1 DEFINITION A function of two variables has a local maimum at a, b if f, f a, b when, is near a, b. [This means that f, f a, b for all points, in some disk with center a, b.] The number f a, b is called a local maimum value. If f, f a, b when, is near a, b, then f a, b is a local minimum value. f If the inequalities in Definition 1 hold for all points, in the domain of f, then has an absolute maimum (or absolute minimum) at a, b. Notice that the conclusion of Theorem can be stated in the notation of gradient vectors as f a, b. THEOREM If f has a local maimum or minimum at a, b and the firstorder partial derivatives of f eist there, then f a, b and f a, b. PROOF Let t f, b. If f has a local maimum (or minimum) at a, b, then t has a local maimum (or minimum) at a, so t a b Fermat s Theorem (see Theorem 4.1.4). But t a f a, b (see Equation 11.3.1) and so f a, b. Similarl, b appling Fermat s Theorem to the function G f a,, we obtain f a, b. If we put f a, b and f a, b in the equation of a tangent plane (Equation 11.4.), we get. Thus the geometric interpretation of Theorem is that if the graph of f has a tangent plane at a local maimum or minimum, then the tangent plane must be horiontal. A point a, b is called a critical point (or stationar point) of f if f a, b and f a, b, or if one of these partial derivatives does not eist. Theorem sas that if f has a local maimum or minimum at a, b, then a, b is a critical point of f. However, as in single-variable calculus, not all critical points give rise to maima or minima. At a critical point, a function could have a local maimum or a local minimum or neither. EXAMPLE 1 Let f, 6 14. Then f, f, 6 (1, 3, 4) These partial derivatives are equal to when 1 and 3, so the onl critical point is 1, 3. B completing the square, we find that f, 4 1 3 FIGURE = + --6+14 Since 1 and 3, we have f, 4 for all values of and. Therefore, f 1, 3 4 is a local minimum, and in fact it is the absolute minimum of f. This can be confirmed geometricall from the graph of f, which is the elliptic paraboloid with verte 1, 3, 4 shown in Figure.
646 CHAPTER 11 PARTIAL DERIVATIVES EXAMPLE Find the etreme values of f,. SOLUTION Since f and f, the onl critical point is,. Notice that for points on the -ais we have, so f, (if ). However, for points on the -ais we have, so f, (if ). Thus ever disk with center, contains points where f takes positive values as well as points where f takes negative values. Therefore, f, can t be an etreme value for f, so f has no etreme value. FIGURE 3 = - Eample illustrates the fact that a function need not have a maimum or minimum value at a critical point. Figure 3 shows how this is possible. The graph of f is the hperbolic paraboloid, which has a horiontal tangent plane ( ) at the origin. You can see that f, is a maimum in the direction of the -ais but a minimum in the direction of the -ais. Near the origin the graph has the shape of a saddle and so, is called a saddle point of f. We need to be able to determine whether or not a function has an etreme value at a critical point. The following test, which is proved in Appendi B, is analogous to the Second Derivative Test for functions of one variable. 3 SECOND DERIVATIVES TEST Suppose the second partial derivatives of f are continuous on a disk with center a, b, and suppose that f a, b and f a, b [that is, a, b is a critical point of f ]. Let D D a, b f a, b f a, b f a, b (a) If D and f a, b, then f a, b is a local minimum. (b) If D and f a, b, then f a, b is a local maimum. (c) If D, then f a, b is not a local maimum or minimum. NOTE 1 In case (c) the point a, b is called a saddle point of f and the graph of f crosses its tangent plane at a, b. NOTE If D, the test gives no information: f could have a local maimum or local minimum at a, b, or a, b could be a saddle point of f. NOTE 3 To remember the formula for D it s helpful to write it as a determinant: D f f f f f f f V EXAMPLE 3 Find the local maimum and minimum values and saddle points of f, 4 4 4 1. SOLUTION We first locate the critical points: f 4 3 4 Setting these partial derivatives equal to, we obtain the equations 3 To solve these equations we substitute one. This gives f 4 3 4 and 3 3 from the first equation into the second 9 8 1 4 1 4 1 1 1 4 1
SECTION 11.7 MAXIMUM AND MINIMUM VALUES 647 so there are three real roots:, 1, 1. The three critical points are,, 1, 1, and 1, 1. Net we calculate the second partial derivatives and D, : f f f 1 1 4 FIGURE 4 =$+$-4+1 D, f f f 144 16 Since D, 16, it follows from case (c) of the Second Derivatives Test that the origin is a saddle point; that is, f has no local maimum or minimum at,. Since D 1, 1 18 and f 1, 1 1, we see from case (a) of the test that f 1, 1 1 is a local minimum. Similarl, we have D 1, 1 18 and f 1, 1 1, so f 1, 1 1 is also a local minimum. The graph of f is shown in Figure 4. A contour map of the function f in Eample 3 is shown in Figure 5. The level curves near 1, 1 and 1, 1 are oval in shape and indicate that as we move awa from 1, 1 or 1, 1 in an direction the values of f are increasing. The level curves near,, on the other hand, resemble hperbolas. The reveal that as we move awa from the origin (where the value of f is 1), the values of f decrease in some directions but increase in other directions. Thus, the contour map suggests the presence of the minima and saddle point that we found in Eample 3. _.5.5.9 1 1.1 1.5 3 FIGURE 5 In Module 11.7 ou can use contour maps to estimate the locations of critical points. V EXAMPLE 4 Find the shortest distance from the point 1,, to the plane 4. SOLUTION The distance from an point,, to the point 1,, is d s 1 but if,, lies on the plane 4, then 4 and so we have d s 1 6. We can minimie d b minimiing the simpler epression B solving the equations d f, 1 6 f 1 6 4 4 14 Eample 4 could also be solved using vectors. Compare with the methods of Section 1.5. f 4 6 4 1 4 ( 11 6, 5 3) we find that the onl critical point is. Since f 4, f 4, and f 1, we have D, f f f 4 and f, so b the Second Derivatives
648 CHAPTER 11 PARTIAL DERIVATIVES ( 11 6, 5 3 ) Test f has a local minimum at. Intuitivel, we can see that this local minimum is actuall an absolute minimum because there must be a point on the given plane that is closest to 1,,. If 11 and 5 6 3, then d s 1 6 s( 5 6) ( 5 3) ( 5 6) 5s6 6 FIGURE 6 The shortest distance from 1,, to the plane 4 is 5s6 6. V EXAMPLE 5 A rectangular bo without a lid is to be made from 1 m of cardboard. Find the maimum volume of such a bo. SOLUTION Let the length, width, and height of the bo (in meters) be,, and, as shown in Figure 6. Then the volume of the bo is We can epress V as a function of just two variables and b using the fact that the area of the four sides and the bottom of the bo is Solving this equation for, we get 1, so the epression for V becomes V We compute the partial derivatives: V 1 1 1 V 1 V 1 If V is a maimum, then V V, but or gives V, so we must solve the equations 1 1 These impl that and so. (Note that and must both be positive in this problem.) If we put in either equation we get 1 3, which gives,, and 1 1. We could use the Second Derivatives Test to show that this gives a local maimum of V, or we could simpl argue from the phsical nature of this problem that there must be an absolute maimum volume, which has to occur at a critical point of V, so it must occur when,, 1. Then V 1 4, so the 3 maimum volume of the bo is 4 m. ABSOLUTE MAXIMUM AND MINIMUM VALUES For a function f of one variable the Etreme Value Theorem sas that if f is continuous on a closed interval a, b, then f has an absolute minimum value and an absolute maimum value. According to the Closed Interval Method in Section 4.1, we found these b evaluating f not onl at the critical numbers but also at the endpoints a and b.
SECTION 11.7 MAXIMUM AND MINIMUM VALUES 649 FIGURE 7 (a) Closed sets (b) Sets that are not closed There is a similar situation for functions of two variables. Just as a closed interval contains its endpoints, a closed set in is one that contains all its boundar points. [A boundar point of D is a point a, b such that ever disk with center a, b contains points in D and also points not in D.] For instance, the disk D, 1 which consists of all points on and inside the circle 1, is a closed set because it contains all of its boundar points (which are the points on the circle 1). But if even one point on the boundar curve were omitted, the set would not be closed. (See Figure 7.) A bounded set in is one that is contained within some disk. In other words, it is finite in etent. Then, in terms of closed and bounded sets, we can state the following counterpart of the Etreme Value Theorem in two dimensions. 4 EXTREME VALUE THEOREM FOR FUNCTIONS OF TWO VARIABLES If f is continuous on a closed, bounded set D in, then f attains an absolute maimum value f 1, 1 and an absolute minimum value f, at some points 1, 1 and, in D. To find the etreme values guaranteed b Theorem 4, we note that, b Theorem, if f has an etreme value at 1, 1, then 1, 1 is either a critical point of f or a boundar point of D. Thus we have the following etension of the Closed Interval Method. 5 To find the absolute maimum and minimum values of a continuous function f on a closed, bounded set D: 1. Find the values of f at the critical points of f in D.. Find the etreme values of f on the boundar of D. 3. The largest of the values from steps 1 and is the absolute maimum value; the smallest of these values is the absolute minimum value. EXAMPLE 6 Find the absolute maimum and minimum values of the function f, on the rectangle D, 3,. (, ) L (, ) (3, ) SOLUTION Since f is a polnomial, it is continuous on the closed, bounded rectangle D, so Theorem 4 tells us there is both an absolute maimum and an absolute minimum. According to step 1 in (5), we first find the critical points. These occur when f f L L (, ) L (3, ) so the onl critical point is 1, 1, and the value of f there is f 1, 1 1. In step we look at the values of f on the boundar of D, which consists of the four line segments,,, shown in Figure 8. On we have and L 1 L L 3 L 4 L 1 FIGURE 8 f, 3
65 CHAPTER 11 PARTIAL DERIVATIVES This is an increasing function of, so its minimum value is f, and its maimum value is f 3, 9. On we have 3 and L f 3, 9 4 f 3, 9 and its mini- This is a decreasing function of, so its maimum value is mum value is f 3, 1. On we have and L 3 9 f, 4 4 3 B the methods of Chapter 4, or simpl b observing that f,, we see that the minimum value of this function is f, and the maimum value is f, 4. Finall, on we have and L 4 L 3 D FIGURE 9 f(, )= -+ L f, with maimum value f, 4 and minimum value f,. Thus, on the boundar, the minimum value of f is and the maimum is 9. In step 3 we compare these values with the value f 1, 1 1 at the critical point and conclude that the absolute maimum value of f on D is f 3, 9 and the absolute minimum value is f, f,. Figure 9 shows the graph of f. 11.7 EXERCISES 1. Suppose (1, 1) is a critical point of a function f with continuous second derivatives. In each case, what can ou sa about f? (a) f 1, 1 4, f 1, 1 1, f 1, 1 (b) f 1, 1 4, f 1, 1 3, f 1, 1. Use the level curves in the figure to predict the location of the critical points of f, 3 3 4 and whether f has a saddle point or a local maimum or minimum at each of those points. Eplain our reasoning. Then use the Second Derivatives Test to confirm our predictions. _.9 _.7 _.5 _1 1.5 _1 1 _1 3 14 Find the local maimum and minimum values and saddle point(s) of the function. If ou have three-dimensional.5 1 1.5 1.5 1.7 1.9 1 graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. 3. f, 9 4 4 4. f, 3 1 8 5. f, 4 4 4 6. f, e 4 7. f, 1 8. f, 3 5 9. f, e cos 1. f, 1 11. f, sin 1. f, 13. f, e 14. f, e ; 15 18 Use a graph and/or level curves to estimate the local maimum and minimum values and saddle point(s) of the function. Then use calculus to find these values precisel. 15. f, 3 3 3 3 16. f, e
SECTION 11.7 MAXIMUM AND MINIMUM VALUES 651 17. f, sin sin sin,, 18. f, sin sin cos, 4, 4 ; 19 Use a graphing device (or Newton s method or a rootfinder) to find the critical points of f correct to three decimal places. Then classif the critical points and find the highest or lowest points on the graph. 19.. 1.. f, 4 5 3 f, 5 1 4 3 4 f, 4 4 4 f, e 4 3 4 cos 3 8 Find the absolute maimum and minimum values of f on the set D. 3. f, 1 4 5, D is the closed triangular region with vertices,,,, and, 3 4. f, 3, D is the closed triangular region with vertices 1,, 5,, and 1, 4 5. f, 4, 1 D, 1, 6. f, 4 6, D, 4, 5 7. f, 4, D, 4 4 3, 8. f,, D,,, 3 ; 9. For functions of one variable it is impossible for a continuous function to have two local maima and no local minimum. But for functions of two variables such functions eist. Show that the function f, 1 1 has onl two critical points, but has local maima at both of them. Then use a computer to produce a graph with a carefull chosen domain and viewpoint to see how this is possible. ; 3. If a function of one variable is continuous on an interval and has onl one critical number, then a local maimum has to be an absolute maimum. But this is not true for functions of two variables. Show that the function f, 3e 3 e 3 has eactl one critical point, and that f has a local maimum there that is not an absolute maimum. Then use a computer to produce a graph with a carefull chosen domain and viewpoint to see how this is possible. 31. Find the shortest distance from the point, 1, 1 to the plane 1. 3. Find the point on the plane 4 that is closest to the point 1,, 3. 33. Find the points on the cone that are closest to the point 4,,. 34. Find the points on the surface 9 that are closest to the origin. 35. Find three positive numbers whose sum is 1 and whose product is a maimum. 36. Find three positive numbers,, and whose sum is 1 such that a b c is a maimum. 37. Find the volume of the largest rectangular bo with edges parallel to the aes that can be inscribed in the ellipsoid 9 36 4 36 38. Solve the problem in Eercise 37 for a general ellipsoid a b c 1 39. Find the volume of the largest rectangular bo in the first octant with three faces in the coordinate planes and one verte in the plane 3 6. 4. Find the dimensions of the rectangular bo with largest volume if the total surface area is given as 64 cm. 41. Find the dimensions of a rectangular bo of maimum volume such that the sum of the lengths of its 1 edges is a constant c. 4. The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five times as much (per unit area) as glass, find the dimensions of the aquarium that minimie the cost of the materials. 43. A cardboard bo without a lid is to have a volume of 3, cm 3. Find the dimensions that minimie the amount of cardboard used. 44. A rectangular building is being designed to minimie heat loss. The east and west walls lose heat at a rate of 1 units m per da, the north and south walls at a rate of 8 units m per da, the floor at a rate of 1 unit m per da, and the roof at a rate of 5 units m per da. Each wall must be at least 3 m long, the height must be at least 4 m, and the volume must be eactl 4 m 3. (a) Find and sketch the domain of the heat loss as a function of the lengths of the sides. (b) Find the dimensions that minimie heat loss. (Check both the critical points and the points on the boundar of the domain.)
65 CHAPTER 11 PARTIAL DERIVATIVES (c) Could ou design a building with even less heat loss if the restrictions on the lengths of the walls were removed? 45. If the length of the diagonal of a rectangular bo must be L, what is the largest possible volume? 46. Three alleles (alternative versions of a gene) A, B, and O determine the four blood tpes A (AA or AO), B (BB or BO), O (OO), and AB. The Hard-Weinberg Law states that the proportion of individuals in a population who carr two different alleles is where p, q, and r are the proportions of A, B, and O in the population. Use the fact that p q r 1 to show that P is at most. 3 P pq pr rq 47. Suppose that a scientist has reason to believe that two quantities and are related linearl, that is, m b, at least approimatel, for some values of m and b. The scientist performs an eperiment and collects data in the form of points 1, 1,,,..., n, n, and then plots these points. The points don t lie eactl on a straight line, so the scientist wants to find constants m and b so that the line m b fits the points as well as possible. (See the figure.) (, ) Let d i i m i b be the vertical deviation of the point i, i from the line. The method of least squares determines and so as to minimie n m b d i, the sum of the squares of these deviations. Show that, according to this method, the line of best fit is obtained when m n m n m i +b i b n i bn n ( i, i ) i n i Thus the line is found b solving these two equations in the two unknowns m and b. 48. Find an equation of the plane that passes through the point 1,, 3 and cuts off the smallest volume in the first octant. d i i i 11.8 LAGRANGE MULTIPLIERS g(, )=k FIGURE 1 f(, )=11 f(, )=1 f(, )=9 f(, )=8 f(, )=7 Visual 11.8 animates Figure 1 for both level curves and level surfaces. In Eample 5 in Section 11.7 we maimied a volume function V subject to the constraint 1, which epressed the side condition that the surface area was 1 m. In this section we present Lagrange s method for maimiing or minimiing a general function f,, subject to a constraint (or side condition) of the form t,, k. It s easier to eplain the geometric basis of Lagrange s method for functions of two variables. So we start b tring to find the etreme values of f, subject to a constraint of the form t, k. In other words, we seek the etreme values of f, when the point, is restricted to lie on the level curve t, k. Figure 1 shows this curve together with several level curves of f. These have the equations f, c, where c 7, 8, 9, 1, 11. To maimie f, subject to t, k is to find the largest value of c such that the level curve f, c intersects t, k. It appears from Figure 1 that this happens when these curves just touch each other, that is, when the have a common tangent line. (Otherwise, the value of c could be increased further.) This means that the normal lines at the point, where the touch are identical. So the gradient vectors are parallel; that is, f, for some scalar. This kind of argument also applies to the problem of finding the etreme values of f,, subject to the constraint t,, k. Thus the point,, is restricted to lie on the level surface S with equation t,, k. Instead of the level curves in Figure 1, we consider the level surfaces f,, c and argue that if the maimum value of f is f,, c, then the level surface f,, c is tangent to the level surface t,, k and so the corresponding gradient vectors are parallel. t,