SOLUTIONS to Problems for Review Chapter 1 McCallum, Hughes, Gleason, et al. ISBN 978-0470-118-9 by Vladimir A. Dobrushkin For Exercises 1, find the critical points of the given function and classify them as local maxima, local minima, saddle points, or none of these. 1. fx,y) = xy x y + 1 Since f x = y x = y x) and f y = 4xy y = 4yx 1), we obtain three critical points: O0, 0), A1, 1), and B1, 1). Second derivatives are f xx = < 0, f xy = 4y, f yy = 4x 1). = f xx f yy fxy = 8[x 1) + y ], then DA) D1, 1) = D1, 1) = 16 < 0, so two critical points A1, 1) and B1, 1) are saddle points On the other hand, D0, 0) = 8 > 0. Since f xx 0, 0) = < 0, O0, 0) is maximum. fx,y) = x x y + 6x 6y Since f x = 6x 6xy+1x = 6xx y+) and f y = x 1y = x +4y), we obtain the critical point: O0, 0). Second derivatives are f xx = 1x 6y + 1, f xy = 6y, f yy = 1. = f xx f yy fxy = 6[4 + 4x y y ], then D0, 0) = 6 4 < 0; hence 0, 0) is a saddle point. fx,y) = x y + y xy + 6 Since f x = xy y = yx 1) and f y = x + 4y x = 4y + xx ), we obtain three critical points: O0, 0), A, 0), and B 1, 1 ). Second derivatives are 4 f xx = y, f xy = x 1), f yy = 4. = f xx f yy fxy = 4[y x 1) ], then D0, 0) = 4 < 0, so the origin is the saddle point. D1, 1) = > 0, and, since f 4 xx1, 1) = 1 > 0, the point 1, 1 ) is a local minimum 4 4 On the other hand, D, 0) = 4 < 0, so, 0) is the saddle point
4. fx,y) = 80 + 10x + 10xy + 0y xy The first derivatives are f x = 80 +10+10y = 10 x y [ 1 + y 8 x y ], f y = 80 [ +0+10x = 10 + x 8 ] xy xy Equating them to zero, we obtain the following system of equations: 1 + y 8 x y = 0, + x 8 xy. Adding them, we obtain the relation y = x. Hence, we get x y + y ) 4y y + y ) = 8 or y + y 4 =. This fourth order equation has two real roots: y = 1 and 17 + ) 1/ 1 17 + ) 1/ 1.4689, and two complex conjugate roots which we discard). Therefore, the given function fx, y) has two critical points: A, 1) and B.08, 1.4). The second derivatives are f xx = 160 [ x y, f xy = 10 1 + 8 ], f x y yy = 160 xy, [ we define Dx,y) def = f xx f yy fxy = 100 19 1 16 x 4 y 4 x y ]. Since DB) 1.6, the critical point B.08, 1.4) is the saddle point while D, 1) = 700. Since f xx, 1) = 0, the point A, 1) is a local minimum. fx,y) = sin x + siny + sinx + y), 0 < x < π, 0 < y < π. Since f x = cosx + cosx + y) and f y = cos y + cosx + y), we equate these expressions to zero. This leads to the equation cosx = cos y or x = y. Hence cos x+cos x = 0. Remember that cos x = cos x 1, we obtain a quadratic equation with respect to cosx: { cos x + cos 1, x 1 = 0 cosx = The equation cosx = 1 has no solution within 0,π), so we need to solve another equation cos x = 1. This yields x = π, and we obtain the critical point: Aπ/,π/). Second derivatives are f xx = sin x sinx + y), f xy = sinx + y), f yy = sin y sinx + y). 1.
= f xx f yy fxy = [sinx + sinx + y)] [sin y + sinx + y)] sin x+y) = sinx sin y + sin x sinx +y) + siny sinx +y). Hence Dπ/,π/) = 9/4 because sin π = sin π =. Since f xx = < 0, the point π/,π/) is a local maximum with fa) = /. For Exercises 6 9, find the local maxima, local minima, and saddle points of the function. Decide if the local maxima or minima are global maxima or minima. Explain. 6. fx,y) = 10 + 1x + 6y x y Equating the first derivatives, f x = 1 6x = 6 x) and f y = 6 y = y), to zero, we obtain the critical point, ). Second derivatives are f xx = 6, f xy = 0, f yy =. = f xx f yy fxy = 1 > 0; since f xx < 0, the point, ) is the global maximum 7. fx,y) = x + y xy Equating the first derivatives, f x = x y and f y = y x = y x), to zero, we obtain the equation xy ) = 0, which leads to two critical points: 0, 0) and 9, ). 4 Second derivatives are f xx =, f xy =, f yy = 6y. = f xx f yy fxy = 1y 9 = 4y ). At the origin we have D0, 0) = 9, so the origin is a saddle point D 9, ) = 18 > 0, so the point 4 9, ) is local minimum 4 8. fx,y) = x + y + 1 x + 4 y Equating the first derivatives, f x = 1 1 and f x y = 1 4, to zero, we find y four critical points: A1, ), B1, ), C 1, ), and D 1, ). Second derivatives are f xx = x, f xy = 0, f yy = 8 y. = f xx f yy fxy = 16. Then x y D1, ) =, f1, ) =, D1, ) =, D 1, ) =, D 1, ) =,
f1, ) = ; hence points B1, ) and C 1, ) are saddle points The point 1,) is local minimum while the point 1, ) is local maximum 9. fx,y) = xy + ln x + y 10, x > 0. Equating the first derivatives, f x = y + 1 and f x y = x + y, to zero, we find two critical points: A, 1/ ) and B, 1/ ); however, the point B does not belong to the domain x > 0). Second derivatives are f xx = 1 x, f xy = 1, f yy =. = f xx f yy fxy = 1 1 < 0. Therefore, the point x, 1/ ) is a saddle points For Exercises 10 1, use Lagrange multipliers to find the maximum and minimum values of f subject to the constraint. 10. fx,y) = x 4y, x + y =. = λx, 4 = λy, x + y =. From the first two equations, we find that y = 4 x. Substitution into the constraint, we obtain two critical points: A, 4 ) and B 4, ). Since fa) = and fb) =, the point, 4 ) is maximum and, 4 ) is minimum 11. fx,y) = x + y, x 4 + y 4 =. x = λ4x, y = λ4y, x 4 + y 4 =. From the first two equations, we find that either x = 0 or y = 0 or y = x. So we get the following 6 critical points: A 0, 1/4), B 0, 1/4), C 1/4, 0 ), D 1/4, 0 ), E1, 1), F 1, 1). Evaluation at these points leads to fa) = fb) = fc) = fd) = 1.414, and fe) = ff) =, so at the points A, B, C, and D the function attains its minimum and the points E and F it attains it maximum 4
1. fx,y) = x + y, 4x y = 1. x = λ4, y = λ ), 4x y = 1. Hence x = y and from the constraint, we get the critical point, ), which is minimum We can also check the answer by substituting y = x 1 from the constraint) into the function f to obtain φ def = fx, x 1) = x 0x + ) 1. Its critical point, x = is minimum. 1. fx,y) = x xy + y, x y = 1. x y = λx, x + y = λ y), x y = 1. Eliminating λ, we obtain the relation: x + y 4xy = 0. From the constraint, we find x = 1 + y, which, upon substitution, yields 1 + y 4y 1 + y = 0. Since the root has two values, we get two equations to determine the critical points: 1 + y 4y 1 + y = 0 and 1 + y + 4y 1 + y. To solve these equations, we rearrange them by isolating the root, and then square both sides. This leads to fourth order equation 1 = 1y +1y 4. Solving for y, we obtain two roots, one of them is positive: y = + 1 and another one is negative y =, 6 6 + ) 1/ 1 which we dismiss. So we conclude that the function has two points, ± 6 ±0.78119168, where the function f has minimum but not maximum. Therefore, there are two points where the function f attains 0.866, the minimum: ) ) 1 + 1, 1 1 1 and + 1, 1 1. To check the answer, we substitute x = 1 + y from the constraint) into the given function to obtain φy) def = f± 1 + y,y) = 1 + y y 1 + y Taking the derivative and equating the result to zero, we get 4y 1 + y y 1 + y = 0 = 4y 1 + y = ±1 + y ).
Therefore, critical points of the function φy) are roots of this equation. Square both sides, we obtain 16y 1 + y ) = 1 + 4y 4 + 4y = 1y 4 + 1y 1 = 0. It has two roots with respect to y : latter one and get two roots: y = 1 + and y = 1. We dismiss the y1 = 1 + 1 0.78119168 and y = 1 + 1 0.78119168. At these two points, the function φy) attains its minimum 14. fx,y) = x + y, x + y = 00. x = λ, 4y = λ, x + y = 00. Substituting x = 6y/ into the constraint equation, we get 18 y + y = 00 or y = 1000/4.8; hence x = 100/4 7.90697. 100 This point, 4, 1000 ) is minimum because at this point the function has the minimum value 440000 1860.46. 4 1849 1. fx,y) = xy, x + 4y = 100. y = λ, x = λ4, x + 4y = 100. Substituting y = x/4 into the constraint equation, we get x + 4 x = 100 or x = 10; 4 hence y = /. Therefore the point 10, ) is maximum because the function attains the maximum value of 0. To check the answer, we substitute y = x/4 from the given constraint) into the function f to obtain φx) def = fx, x/) = 0x x, which attaints its maximum at x = 10. 16. fx,y) = x y + 0xy, x + y = 100. 6x + 0y = λ, 4y + 0x = λ, x + y = 100. 6
Eliminating λ, we get two equations 6x + 0y = 4y + 0x, x + y = 100. Solving these equations, we get x = 48 and y =, which is maximum having the value f48, ) = 7600. 17. fx,y,z) = x y + z, x + y + z = 1. The Lagrange equations have the following solutions: x = λx, 1 = λy, z = zλ a) λ = 1 = x = x, 1 = y, z = z; this yields the critical point 0, 1, 0). b) λ = 1 = x = x, y = 1, z = z; this yields the critical point 0, 1, 0). c) λ = = x = x, 1 = y, z = zp; this yields the critical points 0, 1 ), and 0, 1 ),. Evaluation of the function f at these points yields f0, 1, 0) =, f0, 1, 0) =, f 0, 1 ), = f 0, 1 ), = =.. Hence the point 0, 1, 0) is minimum and the points 0, 1 ), and 0, 1 ), are maxima Note that 0.86604040. To check our conclusion, we substitute x from the constraint, x = 1 y z into the function f to obtain φy,z) = f1 y z,y,z) = 1 + z y y. Its critical points are roots of the vector equation φ = 0, which leads to y = 1 and z = 0. 18. fx,y,z) = x y z, x + y = z. The Lagrange equations x = λx, y = λy, = λ have the following solutions: λ =, so x = 0 and y = 0 from equations x = x and y = y). Hence z also must be zero. The point 0, 0, 0) is maximum 7
19. z = 4x xy + 4y, x + y. Since z x = 8x y, g x = x, z y = x + 8y, and g y = y, the Lagrange equations become 8x y = xλ, x + 8y = yλ. Elimination λ, we obtain 8x y x + 8y = x y. Substitution y = kx yields 8 k)k = 1 + 8k or k = 1; so y = ±x. Substituting into the constraint, we get four critical points: A1, 1), B1, 1), C 1, 1), and C 1, 1). Also, the function z has one critical point, O0, 0). Evaluation of z at these points, we obtain za) = zd) = 7, z0, 0) = 0, zb) = zc) = 9. Therefore, the points B and C are maxima and the point O0,0) is minimum 0. fx,y) = x y, x y. become Since f x = x, g x = y, f y = y, and g y = 1, the Lagrange equations x = yλ, y = λ. Therefore, x = y. Substitution into the constraint yields two critical points A 1/, / ) 0.7970060, 0.699600) and O0, 0). Since fa) = / 4/ 0.11060 and fo) f0, 0) = 0, the point A 1/, / ) 0.7970060, 0.699600) is maximum, and O0,0) is minimum. 1. fx,y) = x + y, x + y 1. The domain x + y 1 is unbounded, so the function fx,y) = x + y is neither bounded from above nor from below. On the boundary, x + y = 1, we have φx) def = x + 1 x, so it has no limits. In Exercises, does fx,y) = x y have a maximum, a minimum, neither, or both when subject to the constraint?. x = 10 Since f = xi yj and g = i, from the Lagrange equation f = λ g, it follows that the point 10, 0) is maximum 8
. y = 10 The Lagrange equations x = λ 0, y = λ have one solution 0, 10), which is minimum 4. x + y = 10 The Lagrange equations x = λx, y = λy have either x = 0 or y = 0. Substitution into the constraint, x + y = 10, yields the following four critical points: A0, 10), B0, 10), C 10, 0), D 10, 0). Since fa) = fb) = 10 and fc) = fd) = 10, we claim that points C and D are maxima and the points A and B are minima. xy = 10. The Lagrange equations x = λy, y = λx = x = y have no solutions. To check the answer, we substitute y = 10/x into the function fx,y) = x y to obtain φx) def = fx, 10/x) = x 100 = φ x) = x + 00 x x, which is equal zero when x 4 +100 = 0, impossible. Hence the function φx) has no critical points, so is fx,y) subject to the given constraint. 9