Homework 8 Solutions, November 007. (1 We calculate some derivatives: f x = f y = x (x + y + 1 y (x + y + 1 x = (x + y + 1 4x (x + y + 1 4 y = (x + y + 1 4y (x + y + 1 4 x y = 4xy (x + y + 1 4 Substituting in the point a = (1, 1, we get p 1 (x, y = f(1, 1 + f f (1, 1(x 1 + (1, 1(y + 1 x y = 1 3 + 9 (x 1 (y + 1 9 = 1 9 + 9 x 9 y p (x, y = p 1 (x, y + 1 ( f x (1, 1(x 1 + f y (1, 1(y + 1 + f (1, 1(x 1(y + 1 x y = 1 9 + 9 x 9 y + 7 (x 1 81 + 7 81 (y + 1 + 4 (x 1(y + 1 81 = 37 81 + 8 81 x 8 81 y + 7 81 x + 7 81 y + 4 81 xy ( The Hessian matrix is given by: Hf = = ( x x y ( 0 0 1 x y y
(3 We use the Hessian matrix calculated in the previous problem p (x, y = 1 + ( 0 (4 The derivative matrix is 0 ( x y + ( x y ( 0 0 = 1 + x + y ( x y ( 3x + xy x z yz + 6z The Hessian matrix is 6x + y x 0 x 0 z 0 z y + 1z Plugging in the point (1, 0, 1, the Taylor polynomial is given by: p (x, y, z = 3 + ( 3 0 6 x y z + ( x y z 6 0 0 0 1 x y z = 3 + 3x + 6z + 6x + 4xy 4yz + 1z (5 First, we compute some derivatives: f x = e x+y+3z f y = e x+y+3z f z = 3e x+y+3z f xx = e x+y+3z f xy = e x+y+3z f xz = 3e x+y+3z f yy = 4e x+y+3z f yz = 6e x+y+3z f zz = 9e x+y+3z
3 f xxx = e x+y+3z f xxy = e x+y+3z f xxz = 3e x+y+3z f xyy = 4e x+y+3z f xyz = 6e x+y+3z f xzz = 9e x+y+3z f yyy = 8e x+y+3z f yyz = 1e x+y+3z f yzz = 18e x+y+3z f zzz = 7e x+y+3z Evaluating these derivatives at (0, 0, 0 yields the coefficients in front of the exponential. The Taylor polynomial is given by a weighted sum, where derivatives with doubled letters are weighted by 1/, and derivatives with tripled letters are weighted by 1/6. We get: 1 + x + y + 3z + x + xy + 3xz + y + 6yz + 9z + x3 6 + x y + 3x z + xy + 6xyz + 9xz (6 (a We calculate some derivatives: f x = sin x sin y f y = cos x cos y f xx = cos x sin y f xy = sin x cos y f yy = cos x sin y + 4y3 3 + 6y z + 9yz + 9z3 Evaluating at (0, pi/, we find that the Taylor polynomial is: 1 1 0x + 0y 0x 0xy 1 1y = 1 y (b We calculate some more derivatives: f xxx = sin x sin y f xxy = cos x cos y f xyy = sin x sin y f yyy = cos x cos y
4 If we bound the values of these by 1, Lagrange s form of the remainder yields: R (x, (0, π/ = 1 6 f xi x j x k (zh i h j h k i,j,k=1 1 6 8 1 0.33 0.036 (7 (a We calculate some derivatives: f x = e x+y f y = e x+y f xx = e x+y f xy = e x+y f yy = 4e x+y Evaluating at (0, 0, we find that the Taylor polynomial is: 1 + x + y + x (b We calculate more derivatives: f xxx = e x+y f xxy = e x+y f xyy = 4e x+y f yyy = 8e x+y + xy + y We can bound the values of f by e 0.3, so Lagrange s form yields: R (x, (0, 0 1 6 e0.3 (0.1 3 + 1 e0.3 (0.1 3 + 1 4e0.3 (0.1 3 + 1 6 8e0.3 (0.1 3 = 9 000 e0.3 0.0061 (8 (a f x = 4 x and f y = 6 y and these functions vanish at the point (, 3. (b c f xx =, f yy =, and f xy = 0, so the Hessian matrix is negative definite. This means f has a local maximum at (, 3, and it is global, since there are no other critical points. (9 f x = y + 3x 1 and f y = 4y 3 4xy. The equation f y = 0 has solutions y = 0 and x = y. We plug these solutions into the equation f x = 0 and get (±1/ 3, 0 for the first type of solution,
and (1, ±1 for the second. The second derivatives are: f xx = 6x, f xy = 4y, and f yy = 1y 4x. The Hessians are: Hf(1/ ( 3 0 3, 0 = 0 4/ 3 Hf( 1/ ( 3 0 3, 0 = 0 4/ 3 6 4 Hf(1, 1 = 4 8 6 4 Hf(1, 1 = 4 8 The first two Hessians describe saddle points. The last two are positive definite matrices, since they both have positive determinant and positive f xx. Thus, they describe local minima. (10 f x = y 8/x and f y = x 1/y. The equation f y = 0 has solutions x = 1/y, and we plug this equation into f x = 0 to get y = 8y 4, with solutions y = 0 and y = 1/. We reject y = 0, since f is not defined on the x axis. Solving for x yields the point (4, 1/. The second derivatives are: f xx = 16x 3, f xy = 1, and f yy = y 3, so 1/4 1 Hf(4, 1/ = 1 16 This has determinant 3 and f xx is positive, so the critical point is a local minimum. (11 We calculate some derivatives: f x = kx y f y = x + ky f xx = k f xy = f yy = k The point (0,0 is always critical, and the Hessian matrix is: k The determinant is 4k 4, so the critical point is degenerate for k = ±1 and a saddle for k < 1. f xx is positive if and only if k > 0, so the critical point is a local minimum for k > 1 and a local maximum for k < 1. (1 (a f x = ax and f y = by. Since a and b are nonzero, the derivatives both vanish only at the origin. f xx = a, f xy = 0, and f yy = b. If a and b have opposite sign, then the Hessian matrix has negative determinant, and the critical point is a saddle. If a and b are positive, then the Hessian matrix is positive definite, k 5
6 and the critical point is a local minimum. If a and b are negative, then the Hessian matrix is negative definition, and the critical point is a local maximum. (b f x = ax, f y = by and f z = cz. Since a, b, and c are nonzero, the only point where all three partial derivatives vanish is the origin. The Hessian matrix is diagonal, with entries a, b, and c. If a, b, and c are positive, then the matrix is positive definite, and the critical point is a local minimum. If a, b, c are negative, then the matrix is negative definite, and the critical point is a local maximum. Otherwise, the Hessian is indefinite, and the critical point is a saddle. (c For all i, f xi = a i x i with a i nonzero. This means the only place where all partial derivatives vanish is the origin, so the origin is the only critical point. We have three cases for the Hessian matrix: Suppose all a i are positive. Then for any nonzero vector h, the induced quadratic form is Q(h = i a ih i > 0, so the Hessian is positive-definite. Suppose all a i are negative. Then for any nonzero vector h, the induced quadratic form is Q(h = i a ih i < 0, so the Hessian is negative-definite. Suppose the a i have mixed signs, i.e., there is some a j < 0 and some a k > 0. Then we can choose a vector e j whose components are zero except for the jth index which is one, and Q(e j = a j < 0. Similarly, we can choose a vector e k whose components are zero except for the kth index which is one, and Q(e k = a k > 0. Therefore, the Hessian is indefinite, and the critical point is a saddle. (13 Since f is a product of two single-variable functions, and we are working on a rectangular domain, it suffices to find extrema of the pieces sin x and cos y on [0, π]. sin x achieves a local minimum of 0 at 0, a maximum of 1 at π/, a minimum of 1 at 3π/, and a local maximum of 0 at π. cos y achieves maxima of 1 at 0 and π and a minimum of 1 at π. Therefore the product achieves absolute maxima both functions are both 1 or both 1, i.e., at the points (π/, 0, (3π/, π, and (π/, π. The product achieves absolute minima when one function is 1 and the other is 1, i.e., at the points (π/, π, (3π/, 0, (3π/, π. There is a way to solve this using partial derivatives, but it involves a lot of work.
(14 We compute some partial derivatives: f x = 3ye x 3e 3x f y = 3e x 3y f xx = 3ye x 9e 3x f xy = 3e x f yy = 6y f x vanishes when y = e x. Substituting this into the equation f y = 0 yields e x = e 4x, which has a unique solution at (0, 1. The Hessian at this point is then 6 3 3 6 which has positive determinant, and f xx (0, 1 < 0. Thus, the Hessian is negative definite, and f has a local maximum at (0, 1. If we set x = 0 and let y decrease, we find that the asymptotic behavior of f is dominated by the y 3 term, which increases without bound. f can have no global maximum, since any value attained at a particular point is surpassed somewhere on the y-axis. 7