Constrained Maxima and Minima EXAMPLE 1 Finding a Minimum with Constraint

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1 1038 Chapter 14: Partial Derivatives 14.8 Lagrange Multipliers HISTORICAL BIOGRAPHY Joseph Louis Lagrange ( ) Sometimes we need to find the etreme values of a function whose domain is constrained to lie within some particular subset of the plane a disk, for eample, a closed triangular region, or along a curve. In this section, we eplore a powerful method for finding etreme values of constrained functions: the method of Lagrange multipliers. Constrained Maima and Minima EXAMPLE 1 Finding a Minimum with Constraint Find the point P(,, z) closest to the origin on the plane + - z - 5 = 0. Solution The problem asks us to find the minimum value of the function ƒ OP 1 ƒ = s - 0d + s - 0d + sz - 0d = + + z

2 14.8 Lagrange Multipliers 1039 subject to the constraint that Since ƒ OP 1 ƒ has a minimum value wherever the function has a minimum value, we ma solve the problem b finding the minimum value of ƒ(,, z) subject to the constraint + - z - 5 = 0 (thus avoiding square roots). If we regard and as the independent variables in this equation and write z as our problem reduces to one of finding the points (, ) at which the function has its minimum value or values. Since the domain of h is the entire -plane, the First Derivative Test of Section 14.7 tells us that an minima that h might have must occur at points where This leads to and the solution + - z - 5 = 0. ƒs,, zd = + + z z = + - 5, hs, d = ƒs,, + - 5d = + + s + - 5d h = + s + - 5dsd = 0, h = + s + - 5d = = 0, = 10, = 5 3, = 5 6. We ma appl a geometric argument together with the Second Derivative Test to show that these values minimize h. The z-coordinate of the corresponding point on the plane z = is z z = a 5 3 b = Therefore, the point we seek is Closest point: P a 5 3, 5 6, b. The distance from P to the origin is 5> 6 L.04. (1, 0, 0) ( 1, 0, 0) Attempts to solve a constrained maimum or minimum problem b substitution, as we might call the method of Eample 1, do not alwas go smoothl. This is one of the reasons for learning the new method of this section. z 1 EXAMPLE Finding a Minimum with Constraint Find the points closest to the origin on the hperbolic clinder - z - 1 = 0. FIGURE The hperbolic clinder - z - 1 = 0 in Eample. Solution 1 The clinder is shown in Figure We seek the points on the clinder closest to the origin. These are the points whose coordinates minimize the value of the function ƒs,, zd = + + z Square of the distance

3 1040 Chapter 14: Partial Derivatives subject to the constraint that - z - 1 = 0. If we regard and as independent variables in the constraint equation, then z = - 1 and the values of ƒs,, zd = + + z on the clinder are given b the function hs, d = + + s - 1d = To find the points on the clinder whose coordinates minimize ƒ, we look for the points in the -plane whose coordinates minimize h. The onl etreme value of h occurs where h = 4 = 0 and h = = 0, The hperbolic clinder z 1 On this part, On this part, z z 1 z FIGURE The region in the plane from which the first two coordinates of the points (,, z) on the hperbolic clinder - z = 1 are selected ecludes the band in the -plane (Eample ). that is, at the point (0, 0). But there are no points on the clinder where both and are zero. What went wrong? What happened was that the First Derivative Test found (as it should have) the point in the domain of h where h has a minimum value. We, on the other hand, want the points on the clinder where h has a minimum value. Although the domain of h is the entire plane, the domain from which we can select the first two coordinates of the points (,, z) on the clinder is restricted to the shadow of the clinder on the -plane; it does not include the band between the lines = -1 and = 1 (Figure 14.51). We can avoid this problem if we treat and z as independent variables (instead of and ) and epress in terms of and z as With this substitution, ƒs,, zd = + + z becomes and we look for the points where k takes on its smallest value. The domain of k in the zplane now matches the domain from which we select the - and z-coordinates of the points (,, z) on the clinder. Hence, the points that minimize k in the plane will have corresponding points on the clinder. The smallest values of k occur where or where = z = 0. This leads to = z + 1. ks, zd = sz + 1d + + z = z k = = 0 and k z = 4z = 0, = z + 1 = 1, = ;1. The corresponding points on the clinder are s ;1, 0, 0d. We can see from the inequalit ks, zd = z Ú 1 that the points s ;1, 0, 0d give a minimum value for k. We can also see that the minimum distance from the origin to a point on the clinder is 1 unit. Solution Another wa to find the points on the clinder closest to the origin is to imagine a small sphere centered at the origin epanding like a soap bubble until it just touches the clinder (Figure 14.5). At each point of contact, the clinder and sphere have the same tangent plane and normal line. Therefore, if the sphere and clinder are represented as the level surfaces obtained b setting ƒs,, zd = + + z - a and gs,, zd = - z - 1

4 14.8 Lagrange Multipliers 1041 z 1 0 z z a 0 FIGURE 14.5 A sphere epanding like a soap bubble centered at the origin until it just touches the hperbolic clinder - z - 1 = 0 (Eample ). equal to 0, then the gradients ƒ and g will be parallel where the surfaces touch. At an point of contact, we should therefore be able to find a scalar l ( lambda ) such that or ƒ = l g, i + j + zk = lsi - zkd. Thus, the coordinates,, and z of an point of tangenc will have to satisf the three scalar equations = l, = 0, z = -lz. For what values of l will a point (,, z) whose coordinates satisf these scalar equations also lie on the surface - z - 1 = 0? To answer this question, we use our knowledge that no point on the surface has a zero -coordinate to conclude that Z 0. Hence, = l onl if = l, or l = 1. For l = 1, the equation z = -lz becomes z = -z. If this equation is to be satisfied as well, z must be zero. Since = 0 also (from the equation = 0), we conclude that the points we seek all have coordinates of the form s, 0, 0d. What points on the surface - z = 1 have coordinates of this form? The answer is the points (, 0, 0) for which - s0d = 1, = 1, or = ;1. The points on the clinder closest to the origin are the points s ;1, 0, 0d.

5 104 Chapter 14: Partial Derivatives The Method of Lagrange Multipliers In Solution of Eample, we used the method of Lagrange multipliers. The method sas that the etreme values of a function ƒ(,, z) whose variables are subject to a constraint gs,, zd = 0 are to be found on the surface g = 0 at the points where ƒ = l g for some scalar l (called a Lagrange multiplier). To eplore the method further and see wh it works, we first make the following observation, which we state as a theorem. THEOREM 1 The Orthogonal Gradient Theorem Suppose that ƒ(,, z) is differentiable in a region whose interior contains a smooth curve C: rstd = gstdi + hstdj + kstdk. If P 0 is a point on C where ƒ has a local maimum or minimum relative to its values on C, then ƒ is orthogonal to C at P 0. Proof We show that ƒ is orthogonal to the curve s velocit vector at P 0. The values of ƒ on C are given b the composite ƒ(g(t), h(t), k(t)), whose derivative with respect to t is dƒ dt = 0ƒ dg 0 dt + 0ƒ dh 0 dt At an point P 0 where ƒ has a local maimum or minimum relative to its values on the curve, dƒ>dt = 0, so ƒ # v = ƒ dk 0z dt = ƒ # v. B dropping the z-terms in Theorem 1, we obtain a similar result for functions of two variables. COROLLARY OF THEOREM 1 At the points on a smooth curve rstd = gstdi + hstdj where a differentiable function ƒ(, ) takes on its local maima and minima relative to its values on the curve, ƒ # v = 0, where v = dr>dt. Theorem 1 is the ke to the method of Lagrange multipliers. Suppose that ƒ(,, z) and g(,, z) are differentiable and that P 0 is a point on the surface gs,, zd = 0 where ƒ has a local maimum or minimum value relative to its other values on the surface. Then ƒ takes on a local maimum or minimum at P 0 relative to its values on ever differentiable curve through P 0 on the surface gs,, zd = 0. Therefore, ƒ is orthogonal to the velocit vector of ever such differentiable curve through P 0. So is g, moreover (because g is orthogonal to the level surface g = 0, as we saw in Section 14.5). Therefore, at P 0, ƒ is some scalar multiple l of g.

6 14.8 Lagrange Multipliers 1043 The Method of Lagrange Multipliers Suppose that ƒ(,, z) and g(,, z) are differentiable. To find the local maimum and minimum values of ƒ subject to the constraint gs,, zd = 0, find the values of,, z, and l that simultaneousl satisf the equations ƒ = l g and gs,, zd = 0. For functions of two independent variables, the condition is similar, but without the variable z. (1) FIGURE Eample 3 shows how to find the largest and smallest values of the product on this ellipse. EXAMPLE 3 Using the Method of Lagrange Multipliers Find the greatest and smallest values that the function takes on the ellipse (Figure 14.53) ƒs, d = 8 + = 1. Solution We want the etreme values of ƒs, d = subject to the constraint gs, d = = 0. To do so, we first find the values of,, and l for which The gradient equation in Equations (1) gives from which we find ƒ = l g and gs, d = 0. i + j = l i + lj, 4 = l 4, = l, and = l 4 sld = l 4, so that = 0 or l = ;. We now consider these two cases. Case 1: If = 0, then = = 0. But (0, 0) is not on the ellipse. Hence, Z 0. Case : If Z 0, then l = ; and = ;. Substituting this in the equation gs, d = 0 gives s ;d 8 + = 1, = 8 and = ;1. The function ƒs, d = therefore takes on its etreme values on the ellipse at the four points s ;, 1d, s ;, -1d. The etreme values are = and = -. The Geometr of the Solution The level curves of the function ƒs, d = are the hperbolas = c (Figure 14.54). The farther the hperbolas lie from the origin, the larger the absolute value of ƒ. We want

7 1044 Chapter 14: Partial Derivatives f i j g 1 i j FIGURE When subjected to the constraint gs, d = >8 + > - 1 = 0, the function ƒs, d = takes on etreme values at the four points s ;, ;1d. These are the points on the ellipse when ƒ (red) is a scalar multiple of g (blue) (Eample 3). to find the etreme values of ƒ(, ), given that the point (, ) also lies on the ellipse + 4 = 8. Which hperbolas intersecting the ellipse lie farthest from the origin? The hperbolas that just graze the ellipse, the ones that are tangent to it, are farthest. At these points, an vector normal to the hperbola is normal to the ellipse, so ƒ = i + j is a multiple sl = ;d of g = s>4di + j. At the point (, 1), for eample, ƒ = i + j, g = 1 i + j, and ƒ = g. At the point s -, 1d, ƒ = i - j, g =- 1 i + j, and ƒ = - g. EXAMPLE 4 Finding Etreme Function Values on a Circle Find the maimum and minimum values of the function ƒs, d = on the circle + = 1. Solution We model this as a Lagrange multiplier problem with ƒs, d = 3 + 4, gs, d = and look for the values of,, and l that satisf the equations ƒ = l g: 3i + 4j = li + lj gs, d = 0: = 0. The gradient equation in Equations (1) implies that l Z 0 and gives = 3 l, = l.

8 14.8 Lagrange Multipliers f 3i 4j g g 6 i 8 j 5 5 3, These equations tell us, among other things, that and have the same sign. With these values for and, the equation gs, d = 0 gives so a 3 l b + a l b - 1 = 0, 9 4l + 4 l = 1, = 4l, 4l = 5, and l = ; Thus, = 3 l = ;3 5, = l = ;4 5, FIGURE The function ƒs, d = takes on its largest value on the unit circle gs, d = = 0 at the point (3> 5, 4> 5) and its smallest value at the point s -3>5, -4>5d (Eample 4). At each of these points, ƒ is a scalar multiple of g. The figure shows the gradients at the first point but not the second. C g 1 f g g 0 g 1 0 FIGURE The vectors g 1 and g lie in a plane perpendicular to the curve C because g 1 is normal to the surface g 1 = 0 and g is normal to the surface g = 0. and ƒs, d = has etreme values at s, d = ;s3>5, 4>5d. B calculating the value of at the points ;s3>5, 4>5d, we see that its maimum and minimum values on the circle + = 1 are 3 a 3 5 b + 4 a4 5 b = 5 5 = 5 and 3 a- 3 5 b + 4 a- 4 5 b =-5 5 = -5. The Geometr of the Solution The level curves of ƒs, d = are the lines = c (Figure 14.55). The farther the lines lie from the origin, the larger the absolute value of ƒ. We want to find the etreme values of ƒ(, ) given that the point (, ) also lies on the circle + = 1. Which lines intersecting the circle lie farthest from the origin? The lines tangent to the circle are farthest. At the points of tangenc, an vector normal to the line is normal to the circle, so the gradient ƒ = 3i + 4j is a multiple sl = ;5>d of the gradient g = i + j. At the point (3> 5, 4> 5), for eample, ƒ = 3i + 4j, g = 6 5 i j, and ƒ = 5 g. Lagrange Multipliers with Two Constraints Man problems require us to find the etreme values of a differentiable function ƒ(,, z) whose variables are subject to two constraints. If the constraints are g 1 s,, zd = 0 and g s,, zd = 0 and g 1 and g are differentiable, with g 1 not parallel to g, we find the constrained local maima and minima of ƒ b introducing two Lagrange multipliers l and m (mu, pronounced mew ). That is, we locate the points P(,, z) where ƒ takes on its constrained etreme values b finding the values of,, z, l, and m that simultaneousl satisf the equations ƒ = l g 1 + m g, g 1 s,, zd = 0, g s,, zd = 0 Equations () have a nice geometric interpretation. The surfaces g 1 = 0 and g = 0 (usuall) intersect in a smooth curve, sa C (Figure 14.56). Along this curve we seek the points where ƒ has local maimum and minimum values relative to its other values on the curve. ()

9 1046 Chapter 14: Partial Derivatives These are the points where ƒ is normal to C, as we saw in Theorem 1. But g 1 and g are also normal to C at these points because C lies in the surfaces g 1 = 0 and g = 0. Therefore, ƒ lies in the plane determined b g 1 and g, which means that ƒ = l g 1 + m g for some l and m. Since the points we seek also lie in both surfaces, their coordinates must satisf the equations g 1 s,, zd = 0 and g s,, zd = 0, which are the remaining requirements in Equations (). z Clinder 1 EXAMPLE 5 Finding Etremes of Distance on an Ellipse The plane + + z = 1 cuts the clinder + = 1 in an ellipse (Figure 14.57). Find the points on the ellipse that lie closest to and farthest from the origin. P Solution We find the etreme values of ƒs,, zd = + + z (0, 1, 0) (1, 0, 0) P 1 Plane z 1 FIGURE On the ellipse where the plane and clinder meet, what are the points closest to and farthest from the origin? (Eample 5) (the square of the distance from (,, z) to the origin) subject to the constraints g 1 s,, zd = = 0 g s,, zd = + + z - 1 = 0. The gradient equation in Equations () then gives ƒ = l g 1 + m g i + j + zk = lsi + jd + msi + j + kd i + j + zk = sl + mdi + sl + mdj + mk or = l + m, = l + m, z = m. The scalar equations in Equations (5) ield = l + z Q s1 - ld = z, = l + z Q s1 - ld = z. (3) (4) (5) (6) Equations (6) are satisfied simultaneousl if either l = 1 and z = 0 or l Z 1 and = = z>s1 - ld. If z = 0, then solving Equations (3) and (4) simultaneousl to find the corresponding points on the ellipse gives the two points (1, 0, 0) and (0, 1, 0). This makes sense when ou look at Figure If =, then Equations (3) and (4) give = z - 1 = 0 = 1 The corresponding points on the ellipse are z = 1 - = ; z = 1 <. P 1 = a,, 1 - b and P = a-, -, 1 + b.

10 14.8 Lagrange Multipliers 1047 Here we need to be careful, however. Although and P both give local maima of ƒ on the ellipse, P is farther from the origin than P 1. The points on the ellipse closest to the origin are (1, 0, 0) and (0, 1, 0). The point on the ellipse farthest from the origin is P. P 1