SOLUTIONS to Problems for Review Chapter 15 McCallum, Hughes, Gleason, et al. ISBN by Vladimir A. Dobrushkin

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1 SOLUTIONS to Problems for Review Chapter 1 McCallum, Hughes, Gleason, et al. ISBN 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

2 4. fx,y) = x + 10xy + 0y xy The first derivatives are f x = y = 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/ , 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 = ], f x y yy = 160 xy, [ we define Dx,y) def = f xx f yy fxy = 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.

3 = 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) = x + 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, ) =,

4 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

5 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 =, ) 1/ 1 which we dismiss. So we conclude that the function has two points, ± 6 ± , 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, 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 ).

6 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 = and y = 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/ This point, 4, 1000 ) is minimum because at this point the function has the minimum value 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 = fx,y) = x y + 0xy, x + y = x + 0y = λ, 4y + 0x = λ, x + y =

7 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, ) = 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 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 = 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

8 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/, / ) , ) and O0, 0). Since fa) = / 4/ and fo) f0, 0) = 0, the point A 1/, / ) , ) 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

9 . 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 = 0, impossible. Hence the function φx) has no critical points, so is fx,y) subject to the given constraint. 9