# Mathematical Sciences

Carnegie Mellon University Department of Mathematical Sciences 21-256, Spring 2021 Homework 5 Instructions: Please show all your work. 1. Let R = [0; 1] [0; 1] : Approximate ZZ xy dA (x; y) R by using m = 2 and n = 2 sub-intervals on the x and y axes, and taking the midpoint of each sub-rectangle as the sampling point. 2. Let R = [0; 2] [1; 3] : Calculate ZZ x dA (x; y) Ry by using Fubini’s Theorem, and iterating the integrals in both orders. 3. Consider the rectangle Ra = [0; a] (a) Evaluate ZZ e x y [0; a] ; where a > 0: dA (x; y) Ra as an expression in a: (b) Now evaluate ZZ lim a !1 e x y dA (x; y) Ra 4. Let D be the region bounded by the curves y = x2 + 6 and y = 5x: Evaluate ZZ x2 y dA (x; y) D 5. Let D be the region in the …rst quadrant with x ZZ x dA (x; y) 1 + y D 6. Evaluate Z 4Z 0 8 e x2 y y 2 : Evaluate dx dy 2y by using Fubini’s Theorem to reverse the order of integration. Carnegie Mellon University Department of Mathematical Sciences 21-256, Spring 2021 Homework 4 Solutions Instructions: Please show all your work. 1. Let f (x; y; z) = x2 + y 2 + z 2 xyz 4 (a) Find all critical points of f: Solution: Set the partial derivatives equal to 0. D x x2 + y 2 + z 2 xyz 4 = 2x yz = 0 2 2 2 xyz 4 = 2y xz = 0 2 2 2 xyz 4 = 2z xy = 0 Dy x + y + z Dz x + y + z From the …rst equation x = 1 2 yz 2 2y 1 2 yz: = Substitute into the second equation and obtain 0 1 y (z 2) (z + 2) = 0 2 So either y = 0; or z = 2: If y = 0; then we have x = 0; and thus z = 0: If z = 2; then x = y: And the third equation becomes 4 = x2 = y 2 : Thus x = y = 2; or x = y = 2: Likewise, if z = 2; then x = y; and 4 = x2 = y 2 ; so x = 2 and y = 2; or x = 2 and y = 2: Altogether the critical points are f(0; 0; 0) ; (2; 2; 2) ; ( 2; 2; 2) ; (2; 2; 2) ; ( 2; 2; 2)g (b) Now use the second derivative test to classify each critical point as a local minimum, local maximum, saddle point, or inconclusive. Solution: We build the Hessian matrix 3 2 2 z y x 5 H=4 z 2 y x 2 At (0; 0; 0) ; 2 2 H (0; 0; 0) = 4 0 0 0 2 0 3 0 0 5 2 and the second derivative test yields det (H1 ) = 2 > 0, det (H2 ) = 4 > 0; and det (H3 ) = 8 > 0: So we have a local minimum. At (2; 2; 2) 2 3 2 2 2 2 5 H (2; 2; 2) = 4 2 2 2 2 2 with det 0 det @ 2 2 2 det [2] = 2 > 0 2 2 = 0 2 2 1 2 2 2 2 A = 32 < 0 2 2 So this point is a saddle point, as det (H3 ) < 0 implies it’s not a local minimum, and det (H1 ) > 0 implies it’s not a local maximum. At (2; 2; 2) 2 3 2 2 2 2 5 H (2; 2; 2) = 4 2 2 2 2 2 and det (H1 ) = 2 > 0; det (H3 ) = 32 < 0 which means we have a saddle point. Likewise we …nd saddle points at ( 2; 2; 2) and ( 2; 2; 2) : 2. Let f (x; y; z) = x2 + y 2 + z 2 axy cyz, where jaj < 2 and a2 + b2 + c2 + abc < 4 bxz (a) Show that (0; 0; 0) is the only critical point of f: Solution: We have Dx x2 + y 2 + z 2 axy bxz cyz = 2x ay bz 2 2 2 axy bxz cyz = 2y ax cz 2 2 2 axy bxz cyz = 2z bx cy Dy x + y + z Dz x + y + z So (0; 0; 0) is a critical point. To see that it’s the only one, consider the system of equations with coe¢ cient matrix 2 3 2 a b 4 a 2 c 5 b c 2 The determinant 2 2 det 4 a b is a 2 c 3 b c 5 = 2 = 2a2 2abc 2b2 2c2 + 8 2 a2 + b2 + c2 + abc 4 >0 so the matrix is invertible and (0; 0; 0) is the only solution. (b) Show that (0; 0; 0) is a local minimum of f . Solution: The Hessian is 3 2 2 a b c 5 H=4 a 2 b c 2 with det (H1 ) = 2>0 det (H2 ) = 4 det (H3 ) = = a2 > 0 2a2 2abc 2 2 2b2 2 2c2 + 8 2 a + b + c + abc 4 >0 So we have a local minimum. 3. Let f (x; y) = x2 + 4y 2 : Use the method of Lagrange Multipliers to determine the absolute minimum value of f (x; y) subject to the constraint 2x + y = 5: Solution: We set rf = 2x = 8y = 2x + y = rg (2) (1) 5 So x = ;y = 8 Thus 2 + 8 17 8 = 5 = 5 = 40 17 Hence 40 5 ;y = 17 17 This is a local minimum, with value x= 2 40 17 5 17 +4 2 : 100 17 We can check, by writing h (x) = f (x; 5 2x) 2 2 = x + 4 (5 = 17×2 2x) 80x + 100 with h0 (x) = 34x 40 h0 = 0 17 40 100 h = 17 17 80 4. Let f (x; y; z) = 3x y + 2z: Use the method of Lagrange Multipliers to determine the absolute maximum and absolute minimum values of f (x; y; z) subject to the constraint x2 + y 2 + z 2 = 25: Solution: We set rf 2 2 = rg 3 = 1 = (2y) 2 = (2z) 2 = x +y +z (2x) 25 We have x= 3 ;y = 2 1 2 ;z = 2 2 so 3 2 2 + 1 2 2 + 2 2 2 7 2 2 2 = 25 = 25 = = 7 50p 14 10 So the critical points are f 15 p 14; 14 5p 5p 14; 14 14 7 15 p 3 14 14 p = 5 14 = 5p 14 + 2 14 5p 14 7 the absolute maximum, and 5p 15 p 14; 14; 14 14 f 5p 14 7 = = 3 15 p 14 14 p 5 14 5p 14 + 2 14 5p 14 7 which is the absolute minimum. 5. Let f (x; y; z) = 2x+y+z: Use the method of Lagrange Multipliers to determine the absolute maximum and absolute minimum values of f (x; y; z) subject to the constraints x2 + y 2 = 4 and x y + z = 3: Solution: We set rf 2 rg + rh 2 = 1 = (2y) + ( 1) 1 (2x) + (1) = (0) + (1) 2 = 4 y+z = 3 x +y x = Immediately, = 1: So 1 = (2x) 2 = (2y) Thus 1 2 2 2 x = y = We have now 1 2 2 2 2 2 + 5 4 2 2 = 4 = 4 = = 5 16p 5 4 Thus we have 1 x = 2 y = 5 4 2 = 2 z p 3 p 5 4 = 2p 5; 5 = 4p 5; 5 2p 4p 2p 5+ 5= 5+3 5 5 5 and x 1 = 2 y p 5 4 2 = 2 p 5 4 = 2p 5; 5 = 4p 5; 5 2p 5 5 3 4p 5=3 5 z = f 2p 4p 2p 5; 5; 5+3 5 5 5 2p 5 5 So 2p 4p 2p 5 + 5+ 5+3 5 5 5 = 2 = p 2 5+3 is the absolute maximum, and f 2p 5; 5 4p 5; 5 2p 5+3 5 = 2 = 3 2p 5 5 p 2 5 4p 5 5 2p 5+3 5 is the absolute minimum. 6. Use the Lagrangian function and the Bordered Hessian matrix to determine any local minimum or maximum of f (x1 ; x2 ; x3 ; x4 ) = x21 + x22 + x23 + x24 subject to the constraints x1 + x2 + x3 + x4 = 1; x1 x2 + x3 x4 = 0; and x1 + x2 x3 x4 = 0: Solution: We de…ne L = x21 + x22 + x23 + x24 1 (x1 + x2 + x3 + x4 1) 2 so (x1 + x2 + x3 + x4 (x1 x2 + x3 (x1 + x2 x3 2×1 1 2×2 1 2×3 2×4 We solve the system 2 0 6 0 6 6 0 6 6 1 6 6 1 6 4 1 1 2 1 0 6 0 1 6 6 0 0 6 ! 6 6 0 0 6 0 0 6 4 0 0 0 0 0 0 0 1 1 1 1 1) = 0 x4 ) = 0 x4 ) = 0 2 = 0 + 2 = 0 1 2 = 0 1+ 2 = 0 0 0 0 1 1 1 1 1 1 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 1 1 0 2 0 0 1 2 0 0 1 4 1 4 1 4 1 4 1 1 1 0 0 2 0 3 7 7 7 7 7 7 7 7 5 1 1 1 0 0 0 2 1 0 0 0 0 0 0 3 7 7 7 7 7 7 7 7 5 (x1 x2 + x3 x4 ) 3 (x1 + x2 x3 x4 ) So the critical point is when x1 = x2 = x3 = x4 = 41 : The Bordered Hessian is 3 2 0 0 0 1 1 1 1 6 0 0 0 1 1 1 1 7 7 6 6 0 0 0 1 1 1 1 7 7 6 1 1 2 0 0 0 7 H=6 7 6 1 6 1 1 1 0 2 0 0 7 7 6 4 1 1 1 0 0 2 0 5 1 1 1 0 0 0 2 With n = 4; and n = 3; we check the last 02 0 0 0 B6 0 0 0 B6 B6 0 0 0 B6 6 B 1 1 1 det (H7 ) = det B6 B6 1 1 1 B6 @4 1 1 1 1 1 1 As this has the sign of ( 1) m 3 = ( 1) = 4 1 1 1 2 0 0 0 3 = 1 principal minor: 31 1 1 1 C 1 1 1 7 7C C 1 1 1 7 7C C 0 0 0 7 7C = 128 < 0 C 2 0 0 7 7C 0 2 0 5A 0 0 2 1 < 0; we have a local minimum at the critical point. Carnegie Mellon University Department of Mathematical Sciences 21256, Spring 2021 Exam 2, Part 1, Lecture 1 Solutions PART 1 1. Find the absolute maximum and minimum values of f (x; y) = 2×2 + 4xy on the rectangle [0; 2] 3y 2 5x [0; 2] : Solution: Obtain critical points inside the domain: fx (x; y) = 4x + 4y fy (x; y) = 4x 5 = 0 6y = 0 5 6y Solve 4x + 4y 4x and we get x = 34 ; y = 21 ; which is inside the rectangle. Evaluate: f 3 1 ; 4 2 15 8 = Boundary: x = 0; f (0; y) = f (0; 0) = f (0; 2) = 3y 2 0 12 x = 2; f (2; y) d dy 3y 2 + 8y 2 = 8 4 = 3 f (2; 0) = 4 3 10 3 2 f (2; 2) 2 y f 3y 2 + 8y = = 2; = y=0 f (x; 0) d 2×2 dx 5x x f 5 ;0 4 = 2×2 = 4x = 5 4 = 25 8 5x 5 6y 2 y=2 f (x; 2) d 2×2 + 3x dx 12 = 2×2 + 3x = 4x + 3 12 So we have values of 3 1 ; = 4 2 f (0; 0) = f 15 8 0 f (0; 2) = 5 f ;0 = 4 4 f 2; = 3 12, abs min. 25 8 10 , abs. max 3 Graph of :y = 2×2 + 4xy 3y 2 5x 0 z -10 0 0 1 1 2 y 2 x 2. Let f (x; y) = 4xy x3 y xy 3 : Find all critical points and classify them as local maximum, local minimum, or saddle points. Solution: Take the partials: fx (x; y) fy (x; y) = = 4y 3×2 y 4x 3 y3 3xy 2 x Solve to obtain y 3×2 + y 2 2 x x + 3y 2 4 = 0 4 = 0 In the case x = 0; or y = 0 we obtain points (0; 0) ; ( 2; 0) ; (0; 2) : Otherwise y 2 = 4 x2 + 3 4 3×2 4 = 0 8 2 = 0 x = 8x 1; y = 1 3×2 ; so So D (x; y) = fxx (x; y) fyy (x; y) = ( 6xy) ( 6xy) = 36×2 y 2 (fxy (x; y)) 3×2 4 3×2 4 2 3y 2 2 2 3y 2 At the various critical points D (0; 0) = 16 < 0; saddle point D ( 2; 0) = 64 < 0; saddle points D (0; 2) = 64 < 0; saddle points D (1; 1) = 32 > 0, fxx (1; 1) = 6 < 0, local max D ( 1; 1) = 32 > 0, fxx (1; 1) = 6 < 0, local max D (1; 1) = 32 > 0, fxx (1; 1) = 6 > 0,local min D ( 1; 1) = 32 > 0, fxx ( 1; 1) = 6 > 0,local min x3 y Graph: f (x; y) = 4xy xy 3 -2 z 2 1 0 -1 -2 -1 0 -2 1 -1 0 y 1 x 2 2 1 : Find all critical points and classify them as local maximum, local 1 + + y2 + z2 minimum, or saddle points. 3. Let f (x; y; z) = x2 Solution: Critical points, set the partials equal to 0 : fx (x; y; z) = fy (x; y; z) = fz (x; y; z) = 2x (1 + x2 2 =0 2 =0 2 =0 + y2 + z2 ) 2y (1 + x2 + y 2 + z 2 ) 2z (1 + x2 + y 2 + z 2 ) So (0; 0; 0) is the critical point. The Hessian is 2 6×2 2y 2 2z 2 2 1 4 8xy 3 (x2 + y 2 + z 2 + 1) 8xz 8xy 2×2 + 6y 2 2z 2 8yz 2 2x 2 8xz 8yz 2y 2 + 6z 2 2 3 5 So H (0; 0; 0) = det (H1 ) = 2 4 2 0 0 0 2 0 3 0 0 5 2 2; det (H2 ) = 4; det (H3 ) = so we have a local maximum. 8

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