Matemáticas para la administración y la economía haeussler ernest, paul richard 12ed

Table of Contents Chapter Chapter 35 Chapter 54 Chapter 89 Chapter 132 Chapter 160 Chapter 177 Chapter 231 Chapter 295 Chapter 333 Chapter 10 357 Chapter 11 378 Chapter 12 423 Chapter 13 469 Chapter 14 539 Chapter 15 614 Chapter 16 658 Chapter 17 670 www.elsolucionario.net Chapter Problems 0.1 True; True; –13 is a negative integer ⎛ b ⎞ ab True, because a ⎜ ⎟ = ⎝c⎠ c True, because −2 and are integers and ≠ False, because the natural numbers are 1, 2, 3, and so on False; the left side is 5xy, but the right side is x y False, because = 10 True; by the associative and commutative properties, x(4y) = (x ⋅ 4)y = (4 ⋅ x)y = 4xy 5 True, because = 11 distributive 12 commutative False, since a rational number cannot have is not a number denominator of zero In fact, at all because we cannot divide by False, because integer True; 13 associative 14 definition of division 25 = 5, which is a positive 15 commutative and distributive 16 associative is an irrational real number 17 definition of subtraction False; we cannot divide by 18 commutative 10 False, because the natural numbers are 1, 2, 3, and so on, and lies between and 19 distributive 20 distributive 11 True 21 2x(y − 7) = (2x)y − (2x)7 = 2xy − (7)(2x) = 2xy − (7 · 2)x = 2xy − 14x 12 False, since the integer is neither positive nor negative 22 (a − b) + c = [a + (−b)] + c = a + (−b + c) = a + [c + (−b)] = a + (c − b) Problems 0.2 False, because does not have a reciprocal True, because x+2 x x = + = + 2 2 23 (x + y)(2) = 2(x + y) = 2x + 2y 21 ⋅ = = 21 24 2[27 + (x + y)] = 2[27 + (y + x)] = 2[(27 + y) + x] = 2[(y + 27) + x] 25 x[(2y + 1) + 3] = x[2y + (1 + 3)] = x[2y + 4] = x(2y) + x(4) = (x · 2)y + 4x = (2x)y + 4x = 2xy + 4x False; the negative of is −7 because + (−7) = False; 2(3 · 4) = 2(12) = 24, but (2 · 3)(2 · 4) = · = 48 26 (1 + a)(b + c) = 1(b + c) + a(b + c) = 1(b) + 1(c) + a(b) + a(c) = b + c + ab + ac False; –x + y = y + (–x) = y – x True; (x + 2)(4) = (x)(4) + (2)(4) = 4x + www.elsolucionario.net Chapter 0: Review of Algebra ISM: Introductory Mathematical Analysis 27 x(y − z + w) = x[(y − z) + w] = x(y − z) + x(w) = x[y + (−z)] + xw = x(y) + x(−z) + xw = xy − xz + xw 51 X(1) = X 28 –2 + (–4) = –6 53 4(5 + x) = 4(5) + 4(x) = 20 + 4x 29 –6 + = –4 54 –(x – 2) = –x + 30 + (–4) = 55 0(–x) = 31 – = ⎛ ⎞ ⋅1 = 56 ⎜ ⎟ = ⎝ 11 ⎠ 11 11 52 3(x – 4) = 3(x) – 3(4) = 3x – 12 32 – (–4) = + = 11 33 −5 − (−13) = −5 + 13 = 57 =5 58 14 x ⋅ ⋅ x x = = 21 y ⋅ ⋅ y y 59 3 = =− −2 x −(2 x) 2x 60 2 ⋅1 ⋅ = = x ⋅ x 3x 61 a a(3b) 3ab (3b) = = c c c 34 −a − (−b) = −a + b 35 (–2)(9) = –(2 · 9) = –18 36 7(–9) = –(7 · 9) = –63 37 (–2)(–12) = 2(12) = 24 38 19(−1) = (−1)19 = −(1 · 19) = −19 39 −1 ⎛ 9⎞ = −1⎜ − ⎟ = −9 ⎝ 1⎠ 40 –(–6 + x) = –(–6) – x = – x ⎛ ⎞ 62 (5a ) ⎜ ⎟ = ⎝ 5a ⎠ 41 –7(x) = –(7x) = –7x 42 –12(x – y) = (–12)x – (–12)(y) = –12x + 12y (or 12y – 12x) 63 −aby −a ⋅ by by = = −ax −a ⋅ x x −3 1⋅ =− =− =− 44 −3 ÷ 15 = 15 15 5⋅3 64 7 ⋅1 ⋅ = = y x y ⋅ x xy 45 −9 ÷ (−27) = −9 9 ⋅1 = = = −27 27 ⋅ 3 65 ⋅ 10 ⋅ = = x y x ⋅ y xy 46 (−a ) ÷ (−b) = −a a = −b b 66 1 3+ + = + = = 6 6 67 5 + 14 ⋅ 7 + = + = = = = 12 12 12 12 12 ⋅ 6 68 14 − 14 −5 ⋅1 − = − = = =− =− 10 15 30 30 30 30 5⋅6 43 –[–6 + (–y)] = –(–6) – (–y) = + y 47 2(–6 + 2) = 2(–4) = –8 48 3[–2(3) + 6(2)] = 3[–6 + 12] = 3[6] = 18 49 (–2)(–4)(–1) = 8(–1) = –8 50 (−12)(−12) = (12)(12) = 144 www.elsolucionario.net ISM: Introductory Mathematical Analysis 69 70 + 10 + = = =2 5 5 X − Y = X −Y 72 16 15 16 − 15 − = − = = 40 40 40 40 74 = 6÷ x y l m = a3⋅7 = (b )5 b 4⋅5 = 11 75 −x y2 z xy 76 is not defined (we cannot divide by 0) w4 s y4 x9 x5 = x −5 = x ⎛ 2a 12 ⎜ ⎜ 7b5 ⎝ z x xy x2 =− ÷ =− ⋅ =− yz y xy y2 z x ⎞ (2a )6 ⎟ = ⎟ (7b5 )6 ⎠ 26 ( a ) = 76 (b5 )6 = 77 =0 = 78 is not defined (we cannot divide by 0) 13 ( x3 )6 = x( x ) 79 · = Problems 0.3 14 (23 )(22 ) = 23+ = 25 (= 32) 64a 4⋅6 117, 649b5⋅6 64a 24 117, 649b30 x3⋅6 1+3 ( x )3 ( x ) ( x3 ) w4 w8 = w4+8 = w12 16 81 = z zz = z 3+1+ = z 17 −128 = −2 y y = x y 9+5 12⋅4 ( x ) = x 12 = x 19 48 0.04 = 0.2 www.elsolucionario.net = x18− = x14 x 2⋅3 x3⋅2 25 = 18 y14 =x = x18 x x12−12 = x = 15 x x = x x6 x9 = x6+9 = x15 b 20 ⎛ w2 s ⎞ ( w2 s ) ( w2 ) ( s3 ) w2⋅2 s3⋅2 = = = 10 ⎜ ⎟ ⎜ y2 ⎟ ( y )2 y 2⋅2 y4 ⎝ ⎠ x y 6y = 6⋅ = y x x 3+ a 21 (2 x y )3 = 23 ( x )3 ( y )3 = x 2⋅3 y 3⋅3 = x6 y9 l m l l ÷ = ⋅ = 3 m 3m = 1 18 18 − + 17 − + = − + = = 12 12 12 12 12 (a3 )7 ⎛ x2 ⎞ ( x )5 x 2⋅5 x10 ⎜ ⎟ = = = ⎜ y3 ⎟ ( y )5 y 3⋅5 y15 ⎝ ⎠ 71 73 Section 0.3 1 = = 16 16 x3⋅4 = x6 x6 x12 = x12 x12 Chapter 0: Review of Algebra 20 –8 = 27 3 –8 27 = ISM: Introductory Mathematical Analysis −2 =− 3 21 (49)1/ = 49 = 37 (9 z )1/ = z = 32 ( z )2 = 32 ( z ) = 3z 22 (64)1/ = 64 = 3/ 23 = 3 13 39 39 39 = ⋅ = = = 13 13 13 13 13 132 36 ( 9) 3 38 (16 y8 )3 / = ⎡ 16 y8 ⎤ = ⎡ (2 y )4 ⎤ = (2 y )3 ⎢⎣ ⎣⎢ ⎦⎥ ⎦⎥ = (3) = 27 = 8y 24 (9) −5 / = (9) 25 (32) –2 / = (32) 26 (0.09) = 10 –1/ = ⎛ ⎞ 27 ⎜ ⎟ ⎝ 32 ⎠ = 5/ = 2/5 ( 9) = = ( 32 ) = 243 = (2) = ⎛ 27t ⎞ 39 ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ = = 0.3 0.09 10 41 4/5 ⎛ 64 ⎞ 28 ⎜ − ⎟ ⎝ 27 ⎠ ⎛ ⎞ ⎛1⎞ = ⎜⎜ ⎟⎟ = ⎜ ⎟ = 16 ⎝2⎠ ⎝ 32 ⎠ 2/3 42 −3 / 2/3 c2 = a5 ⋅ b −3 ⋅ c2 9t ⎡ 3t ⎤ =⎢ ⎥ = ⎣2⎦ −3 / = a5 ⋅ ⎡4⎤ =⎢ ⎥ ⎣ x3 ⎦ ⋅ b3 c = x / y / z –10 / = –10 x y z −3 = = 4−3 ( x3 )−3 a5 b3 c x2 / y3 / z2 2 ⎛ 64 ⎞ 16 ⎛ 4⎞ = ⎜⎜ − ⎟⎟ = ⎜ − ⎟ = 27 ⎝ ⎠ ⎝ ⎠ 30 54 = 27 ⋅ = 27 31 x3 = 3 x =x = 33 32 4x = x = x 33 16 x = 16 x = x a5b −3 43 5m−2 m−7 = 5m −2+ ( −7) = 5m −9 = 50 = 25 ⋅ = 25 ⋅ = 29 ⎛ ⎡ 3t ⎤ ⎞ = ⎜⎢ ⎥ ⎟ ⎜⎣ ⎦ ⎟ ⎝ ⎠ ⎛ ⎡ ⎤4 ⎞ = ⎜⎢ ⎥ ⎟ ⎜ ⎣ x3 ⎦ ⎟ ⎝ ⎠ −3 9 x x = = = −9 64 x ⎛ 256 ⎞ 40 ⎜ ⎟ ⎝ x12 ⎠ (0.09)1/ 2/3 44 x + y –1 = x + 45 (3t ) –2 = (3t ) 46 (3 − z ) –4 = m9 y = 9t (3 − z )4 47 4 x x x 34 = = 16 16 5 x = (5 x )1/ = 51/ ( x )1/ = 51/ x / 48 ( X 3Y −3 )−3 = ( X )−3 (Y −3 )−3 = X −9Y 35 − 27 + 128 = ⋅ − ⋅ + 64 ⋅ = = ⋅ 2 − ⋅ 3 + 43 = − 15 + 4 www.elsolucionario.net Y9 X9 ISM: Introductory Mathematical Analysis x − y = x1/ − y1/ 49 50 Section 0.3 u −2 v −6 w3 vw−5 w3−( −5) = = u v1−( −6) w8 x 9/ 3/ 1/ y 52 = a −3 / 4b −1/ a5b −4 63 = a17 / b −9 / = a17 / b = 9/ 53 (2a − b + c) / = (2a − b + c)2 64 54 (ab c3 )3 / = (ab c3 )3 = a b c 55 x = –4 / = x4 / − = x /15 59 60 ] = = = = [x −4 / 1/ ] =x –4 / 30 51/ 81/ (3 x)1/ 3x = 33 y 3y2 / 66 18 = =3 =x = (2 y ) y 2y 2y = 1(3 x) / = (3x)1/ (3 x)2 / 3 (3x) 3x = ⋅ y1/ 3 y / ⋅ y1/ = y1/ 3 y = 3y 3y = 68 ⋅ 51/ 51/ ⋅ 51/ ⋅ 21/ 81/ ⋅ 21/ = = 5 34 16 = 34 2 = (24 a10b15 )1/ 20 = ab = 31/ = 20 31/ ⋅ 32 / 3 = (2334 )1/ 6 648 = 3 u / v1/ www.elsolucionario.net = 16a10 b15 ab 21/ ⋅ 32 / 69 x y –3 x = x6 y –3 = 70 ⋅ a1/ b3 / / 1/ a1/ 2b1/ ⋅ a1/ b3 / a b a b 21/ a1/ 2b3 / 24 / 20 a10 / 20b15 / 20 = = ab ab = 15 = (2 y ) 1/ 12 = 4=2 = = 1/ x2 3x 12 –2 /15 x = 65 67 w3 / (3w)3 / 3 = − = − 5 5 w (3w) w 27 w3 1/ x 57 3w−3 / − (3w) −3 / = 58 [( x ) (2 x) y (2 y )1/ = (2 y ) 56 x1/ − (2 y )1/ = x − y –4 1/ 1/ y = (2 x) 2x 2x = 1/ 2y = a −3b −2 a5b −4 = (a −3b −2 )1/ a5b −4 1/ 2 2x x 2y z 1/ 4(2 x)1/ = (2 x) y 62 = 2x = u v7 51 x xy –2 z = x ( xy –2 z )1/ = x x1/ y –2 / z / = 61 = 23 / 634 / x6 y3 ⋅ u1/ v1/ u / v1/ ⋅ u1/ v1/ = 3u1/ v1/ u 3v Chapter 0: Review of Algebra 243 71 243 = 81 = = ISM: Introductory Mathematical Analysis = [(5k ) 3]1/ = 5k 31/ 72 {[(3a3 ) ]−5 }−2 = {[32 a ]−5 }−2 = {3−10 a −30 }−2 83 = 320 a 60 73 20 = 64 y ⋅ x x 74 s 3/ = s 75 76 = (2 –2 x1/ y –2 )3 1/ ⋅x s 5/ s 2/3 –6 x3 / y –6 x3 / 85 64 y x x = = 84 26 y 6 1/ = 1/ 15 / s s 4/6 75k = (75k )1/ = [(25k )(3)]1/ 82 (ab −3c)8 (a −1c ) −3 = s11/ = x2 ÷ = (31/ )8 = 38 / = 32 = 87 − 77 32 (32) −2 / = 32 (25 ) −2 / 78 ⎛⎜ x y ⎞⎟ ⎝ ⎠ 2/5 b 24 x6 = x ÷ x6−12 = x ÷ x −6 12 x x6 = x ⋅ x = x8 (–6)(–6) = 36 = 86 8s –2 2s 88 a5 c14 ⎡ x3 ⎤ x6 ( x3 ) ÷⎢ = ÷ ⎥ x4 x ( x6 )2 ⎣⎢ ( x ) ⎦⎥ = x2 ÷ = 32 (2−2 ) = 32 ⋅ 22 = = (–6)2 ≠ −6 since –6 < Note that a c −6 ( x )3 (4 3) a8b −24 c8 = ⋅ = 73 = 7(49) x yz 3 xy = ( x yz )( xy ) = x3 y z = xyz = =− =− s s ( a5b−3 c ) s5 = (a5 )3 (b −3 )3 (c1/ )3 = a15b −9 c3 / = = [( x y )1/ ]2 / = ( x y )2 / 25 a15 c3 / b9 ⎛ x3 y ⎞ 89 (3x3 y ÷ y z −3 ) = ⎜ ⎟ ⎜ y z −3 ⎟ ⎝ ⎠ = x / 25 y / 25 4 79 (2 x –1 y )2 = 22 x –2 y = 80 3 y4 x = y1/ x1/ = 4y ⎛ x3 z ⎞ =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 3 (3x z ) = (2)4 x2 ⋅ y / x3 / y1/ x1/ ⋅ y / x3 / = 24 81x12 z12 = 16 x3 / y / = xy 81 x x y3 xy = x1/ ( x y )1/ ( xy )1/ =x 1/ ( xy 3/ 1/ )( x 34 x12 z12 90 y) = x y 5/ ( –2 2x 16 x3 www.elsolucionario.net ) = ( ) 1/ 2 = –2 (x ) (161/ )2 ( x3 ) –4 2x 16 x6 = 1 x10 = x10 ISM: Introductory Mathematical Analysis Section 0.4 18 −{−6a − 6b + + 10a + 15b − a[2b + 10]} = −{4a + 9b + − 2ab − 10a} = −{−6a + 9b + − 2ab} = 6a − 9b − + 2ab Problems 0.4 8x – 4y + + 3x + 2y – = 11x – 2y – x − 10 xy + + z − xy + 19 x + (4 + 5) x + 4(5) = x + x + 20 = x − 11xy + z + 20 u + (5 + 2)u + 2(5) = u + 7u + 10 8t − 6s + s − 2t + = 6t − s + x +2 x + x +3 x = x 21 ( w + 2)( w − 5) = w2 + (−5 + 2) x + 2(−5) = w2 − 3w − 10 a + 3b − c + 3b = a + 3b − c 22 z + (–7 − 3) z + (–7)(–3) = z − 10 z + 21 3a + 7b − − 5a − 9b − 21 = −2a − 2b − 30 23 (2 x)(5 x ) + [(2)(2) + (3)(5)]x + 3(2) = 10 x + 19 x + x –10 xy + − z + xy − = x − xy − z + − 24 (t)(2t) + [(1)(7) + (−5)(2)]t + (−5)(7) = 2t − 3t − 35 x +2 x − x −3 x = − x x + y − x − 3z = y − 3z 25 X + 2( X )(2Y ) + (2Y )2 = X + XY + 4Y 26 (2 x)2 − 2(2 x)(1) + 12 = x − x + 10 8z – 4w – 3w + 6z = 14z – 7w 11 9x + 9y – 21 – 24x + 6y – = –15x + 15y – 27 27 x − 2(5) x + 52 = x − 10 x + 25 12 u − 3v − 5u − 4v + u − = −3u − 7v − 13 x − y + xy − x − xy − 28 y 2 28 (1 ⋅ 2) 2 + [(1)(5) + (–1)(2)] x + (–1)(5) = 2x + x − = x − 33 y − xy 29 14 – [3 + 4s – 12] = – [4s – 9] = – 4s + = 11 – 4s ( 3x ) +2 ( ) x (5) + (5)2 = 3x + 10 3x + 25 15 2{3[3x + − x + 10]} = 2{3[ x + 16]} 30 = 2{3x + 48} = x + 96 ( x) ( y) − 32 = y − 31 (2 s )2 − 12 = s − 16 4{3t + 15 – t[1 – t – 1]} = 4{3t + 15 – t[–t]} = 4{3t + 15 + t } = 4t + 12t + 60 32 ( z )2 − (3w)2 = z − 9w2 17 −5(8 x3 + x − 2( x − + x)) 33 x ( x + 4) − 3( x + 4) = −5(8 x3 + x − x + 10 − x) = x3 + x − 3x − 12 = −5(8 x3 + x − x + 10) = −40 x3 − 30 x + 20 x − 50 34 x( x + x + 3) + 1( x + x + 3) = x3 + x + x + x + x + = x3 + x + x + www.elsolucionario.net Chapter 0: Review of Algebra ISM: Introductory Mathematical Analysis 35 x (3 x + x − 1) − 4(3x + x − 1) 46 = 3x + x3 − x − 12 x − x + = 3x + x3 − 13x − x + 47 36 y (4 y + y − y ) − 2(4 y + y − y ) 3 2 x3 x 4 − + = x2 − + x x x x x5 2x = 12 y + y − y − y − y + y 48 = 12 y − y − 13 y + y 37 x{2( x − x − 35) + 4[2 x − 12 x]} = x{2 x − x − 70 + x − 48 x} = x{10 x − 52 x − 70} = 10 x3 − 52 x − 70 x + x3 2x − 2x = x3 + x − 3y − − y − 3y −6 y − = 3y −6 y = − 3y 3y = −2 − y 38 [(2 z )2 − 12 ](4 z + 1) = [4 z − 1](4 z + 1) x = (4 z ) − 12 = 16 z − 49 x + x + x − x2 + 5x 39 x(3x + 2y – 4) + y(3x + 2y – 4) + 2(3x + 2y – 4) −3 = 3x + xy − x + 3xy + y − y + x + y − = 3x + y + xy + x − Answer: x + 40 [ x + ( x + 1)]2 = ( x ) + x ( x + 1) + ( x + 1) 2 −3 x+5 x −1 50 x − x − x + x2 − x –x + –x + Answer: x – = x + x3 + x + x + x + = x + x3 + x + x + 41 (2a )3 + 3(2a )2 (3) + 3(2a )(3) + (3)3 = 8a3 + 36a + 54a + 27 x − x + 17 51 x + 3x3 − x + x − 42 (3 y )3 − 3(3 y )2 (2) + 3(3 y )(2)2 − (2)3 = 27 y3 − 54 y + 36 y − x3 + x –8 x + x 43 (2 x)3 − 3(2 x)2 (3) + 3(2 x)(3)2 − 33 –8 x − 16 x 17 x − 17 x + 34 – 37 = x3 − 36 x + 54 x − 27 44 x3 + 3x (2 y ) + x(2 y ) + (2 y )3 = x3 + x y + 12 xy + y Answer: 3x − x + 17 + 45 z 18 z − = z − 18 z z www.elsolucionario.net –37 x+2 2x2 Chapter 17: Multivariable Calculus 11 ISM: Introductory Mathematical Analysis f ( x, y , z ) = xy z , x + y + z = 1, x − y + z = (xyz ≠ 0) Since there are two constraints, two Lagrange multipliers are used F ( x, y, z , λ1 , λ2 ) = xy z − λ1 ( x + y + z − 1) − λ2 ( x − y + z ) ⎧ F = y z − λ1 − λ2 = ⎪ x ⎪ Fy = xyz − λ1 + λ2 = ⎪ ⎨ Fz = xy − λ1 − λ2 = ⎪F = − x − y − z + = ⎪ λ1 ⎪⎩ Fλ2 = − x + y − z = (1) (2) (3) (4) (5) Subtracting (3) from (1) gives y z − xy = 0, so x = z (since xy z ≠ 0) Subtracting (5) from (4) gives 1 1 −2y + = 0, so y = Substituting z = x and y = in (5) gives −2 x + = 0, so x = Thus, z = Critical 2 4 ⎛1 1⎞ point of f: ⎜ , , ⎟ ⎝4 4⎠ 12 f ( x, y, z , w) = x + y + z − 5w2 , x + 6y + 3z + 2w = F ( x, y , z , w, λ ) = 3x + y + z − 5w2 − λ ( x + y + z + w − 4) ⎧ Fx ⎪F ⎪ y ⎨ Fz ⎪ Fw ⎪F ⎪⎩ λ = 6x − λ = = y − 6λ = = z − 3λ = = −10 w − 2λ = = − x − y − 3z − 2w + = 3λ λ and w = − , y = 3λ , z = 240 40 720 180 48 Substituting into the last equation gives λ = and w = − Thus x = , y= ,z= 1201 1201 1201 1201 1201 48 240 ⎞ ⎛ 40 720 180 Critical point of F: ⎜ , , ,− , ⎟ 1201 1201 1201 1201 1201 ⎠ ⎝ 48 ⎞ ⎛ 40 720 180 Critical point of f: ⎜ , , ,− ⎟ 1201 1201 1201 1201 ⎝ ⎠ Solving the first four equations for x, y, z, and w in terms of λ gives x = λ 13 We minimize c = f ( q1 , q2 ) = 0.1q12 + 7q1 + 15q2 + 1000 subject to the constraint q1 + q2 = 100 F ( q1 , q2 , λ ) = 0.1q12 + q1 + 15q2 +1000 − λ ( q1 + q2 − 100 ) ⎧ Fq1 = 0.2q1 + − λ = ⎪ ⎨ Fq2 = 15 − λ = ⎪ F = − q − q + 100 = ⎩ λ (1) (2) (3) From (2), λ = 15 Substituting λ = 15 into (1) gives 0.2q1 + − 15 = , so q1 = 40 Substituting q1 = 40 into (3) gives −40 − q2 + 100 = , so q2 = 60 Thus λ = 15 , q1 = 40, and q = 60 Thus plant should produce 40 units and plant should produce 60 units 704 www.elsolucionario.net ISM: Introductory Mathematical Analysis Section 17.8 14 We minimize c = 3q12 + q1q2 + 2q22 subject to the constraint q1 + q2 = 200 F ( q1 , q2 , λ ) = 3q12 + q1q2 + 2q22 − λ ( q1 + q2 − 200 ) ⎧ Fq1 = 6q1 + q2 − λ = ⎪ ⎨ Fq2 = q1 + 4q2 − λ = ⎪ F = −q − q + 200 = ⎩ λ (1) (2) (3) 3 Eliminating λ from (1) and (2) yields q1 = q2 Substituting q1 = q2 into (3) yields q2 = 125 and thus 5 q1 = 75 Thus plant should produce 75 units and plant should produce 125 units 15 We maximize f (l , k ) = 12l + 20k − l − 2k subject to the constraint 4l + 8k = 88 F (l , k , λ ) = 12l + 20k − l − 2k −λ (4l + 8k − 88) (1) ⎧ Fl = 12 − 2l − 4λ = ⎪ (2) ⎨ Fk = 20 − 4k − 8λ = ⎪⎩ Fλ = −4l − 8k + 88 = (3) Eliminating λ from (1) and (2) yields k = l – Substituting k = l – into (3) yields l = 8, so k = Therefore the greatest output is f(8, 7) = 74 units (when l = 8, k = 7) 16 We maximize f (l , k ) = 20l + 25k − l − 3k subject to the constraint 2l + 4k = 50 F (l , k , λ ) = 20l + 25k − l − 3k − λ (2l + 4k − 50) ⎧ Fl = 20 − 2l − 2λ = ⎪ ⎨ Fk = 25 − 6k − 4λ = ⎪⎩ Fλ = −2l − 4k + 50 = (1) (2) (3) 25 − λ Substituting these expressions for l and k into (3) yields 20 90 85 ⎛ 90 85 ⎞ 3725 ≈ 133 units (when λ = − Thus l = and k = Therefore the greatest output is f ⎜ , ⎟= 28 7 14 ⎝ 14 ⎠ 90 85 ⎞ l = , k = ⎟ 14 ⎠ From (1), l = 10 − λ and from (2), k = 17 We maximize P( x, y ) = x y − x − y subject to the constraint x + y = 60,000 F ( x, y, λ ) = x y − x − y − λ ( x + y − 60, 000) − 34 34 ⎧ ⎪ Fx = x y − − λ = ⎪ 27 14 − 14 ⎪ x y −1− λ = ⎨ Fy = ⎪ ⎪ Fλ = − x − y + 60, 000 = ⎪⎩ (1) (2) (3) − 34 34 27 14 − 14 − 3 27 14 − 14 x y − x y = , x y4 = x y , y = 3x 4 4 Substituting for y in (3) gives –4x + 60,000 = 0, so x = 15,000, from which y = 45,000 Thus each month $15,000 should be spent on newspaper advertising and $45,000 on TV advertising Solving (2) for λ and substituting in (1) gives 705 www.elsolucionario.net Chapter 17: Multivariable Calculus ISM: Introductory Mathematical Analysis 18 We maximize f (l , k ) = 6l k subject to the constraint 25l + 69k = 25,875 F (l , k , λ ) = 6l k − λ (25l + 69k − 25,875) 12 − 35 35 ⎧ ⎪ Fl = l k − 25λ = ⎪ 18 − ⎪ ⎨ Fk = l k − 69λ = ⎪ ⎪ Fλ = −25l − 69k + 25,875 = ⎪⎩ 12 − 53 53 18 52 − 52 From the first two equations, l k = 25λ and l k = 69λ Thus, 5 12 l − k 5 2 18 l k − 5 = 25λ 25 , from which = 69λ 29 25 l Substituting this for k in the third equation and solving for l gives l = 414 so k = 225 46 414 units of labor and 225 units of capital should be invested k= 19 We minimize B( x, y, z ) = x + y + z subject to x + y = 20 and y + z = 20 Since there are two constraints, two Lagrange multipliers are used F ( x, y, z , λ1 , λ2 ) = x + y + z −λ1 ( x + y − 20) − λ2 ( y + z − 20) ⎧ Fx = x − λ1 = ⎪ Fy = y − λ1 − λ2 = ⎪⎪ ⎨ Fz = z − λ2 = ⎪ Fλ1 = − x − y + 20 = ⎪ ⎪⎩ Fλ2 = − y − z + 20 = (1) (2) (3) (4) (5) Eliminating y from (4) and (5) gives x = z From (1) and (3), λ1 = 2x and λ2 = 4z Substituting in (2) we have 2y – 2x – 4z = 0, 2y – 2x – 4x = 0, 2y – 6x = 0, y = 3x Substituting in (5) gives –(3x) – x + 20 = 0, so x = Thus z = and y = 15 Therefore, x = 5, y = 15, z = 20 a P = TR – TC = 64q – (8l + 16k) ⎡ 65 − 4(l − 4)2 − 2(k − 5) ⎤ = 64 ⎢ ⎥ − 8l − 16k 16 ⎢⎣ ⎥⎦ P = −196 − 16l + 120l − 8k + 64k b 15 Pk = −16k + 64 = ⇒ k = Pl = −32l + 120 = ⇒ l = ⎛ 15 ⎞ Thus there is one critical point: (l , k ) = ⎜ , ⎟ ⎝ ⎠ Second-Derivative Test: Pll = −32, Pkk = −16, Plk = ⎛ 15 ⎞ ⎛ 15 ⎞ Thus D(l , k ) = Pll Pkk − [ Plk ] = (−32)(−16) − 02 = 512 At ⎜ , ⎟ , D ⎜ , ⎟ = 512 > and Pll = −32 < ⎝ ⎠ ⎝ ⎠ 15 Thus there is a relative maximum at l = , k = Substituting these values into the profit function gives a relative maximum profit of $157.00 706 www.elsolucionario.net ISM: Introductory Mathematical Analysis c Section 17.8 F (l , k , q, λ ) = 64q − 8l − 16k − λ ⎡16q − 65 + 4(l − 4)2 + 2(k − 5)2 ⎤ ⎣ ⎦ F = − − λ ( l − 4) = (1) ⎧ l ⎪ Fk = −16 − 4λ (k − 5) = (2) ⎪ ⎨ Fq = 64 − 16λ = (3) ⎪ ⎪⎩ Fλ = −16q + 65 − 4(l − 4)2 − 2(k − 5)2 = (4) 15 Similarly, from (2) 251 251 ⎞ ⎛ 15 we get k = Substituting for l and k in (4) gives q = Thus (l , k , q ) = ⎜ , 4, ⎟ 64 ⎠ 64 ⎝ From (3), λ = Substituting λ = into (1) gives –8 – 32(l – 4) = 0, so l = ( 21 U = x3 y , px = , p y = , I = 48 x3 y ≠ ) We want to maximize U = x3 y subject to 2x + 3y = 48 F ( x, y, λ ) = x3 y − λ (2 x + y − 48) ⎧ Fx = 3x y3 − 2λ = (1) ⎪⎪ (2) ⎨ Fy = x y − 3λ = ⎪ F = −2 x − y + 48 = (3) ⎪⎩ λ 3 From (1), λ = x y3 and from (2), λ = x3 y Thus x y = x3 y , so x = y 2 ⎛ ⎞ Substituting this expression for x into (3) yields y = Hence x = ⎜ ⎟ = 12 ⎝2⎠ 22 U = 40 x − x + y − y , px = 2, p y = 3, I = 10 We want to maximize U = 40 x − x + y − y subject to 2x + 3y = 10 ⎧ F = 40 − 10 x − 2λ = ⎪ x F ( x, y, λ ) = 40 x − x + y − y − λ (2 x + y − 10) ⎨ Fy = − y − 3λ = ⎪ F = −2 x − y + 10 = ⎩ λ λ 3λ 20 208 38 Substituting these values into the third equation gives λ = Thus x = and y = 53 53 53 From the first equation, x = − and from the second equation y = − 23 U = f(x, y, z) = xyz px = p y = p z = 1, I = 100 (xyz ≠ 0) We want to maximize U = xyz subject to x + y + z = 100 F ( x, y, z, λ ) = xyz − λ ( x + y + z − 100) ⎧ Fx ⎪⎪ Fy ⎨ ⎪ Fz ⎩⎪ Fλ = yz − λ = = xz − λ = = xy − λ = = − x − y − z + 100 = (1) (2) (3) (4) 707 www.elsolucionario.net Chapter 17: Multivariable Calculus ISM: Introductory Mathematical Analysis From (1) and (2), yz = xz , so y = x Similarly, from (1) and (3), z = x Substituting 100 Thus y = x and z = x into (4) yields x = 100 100 y= and z = 3 Problems 17.9 n = 6, Σxi = 21 , Σyi = 18.6 , Σxi yi = 75.7 , Σxi2 = 91 a = 0.98 b = 0.61 Thus yˆ = 0.98 + 0.61x When x = 3.5, then yˆ = 3.12 24 To maximize U = f(x, y) subject to the constraint xp x + yp y = I , we consider ( ) F ( x, y, λ ) = f ( x, y ) − λ xp x + yp y − I For maximum satisfaction, Fx = f x ( x, y ) − λ px = and Fy = f y ( x, y ) − λ p y = y (1) (2) x f x ( x, y ) and from (2), px f y ( x, y ) f ( x, y ) f y ( x , y ) λ= Thus λ = x = py px py From (1), λ = n = 7, Σxi = 28 , Σyi = 29.3 , Σxi yi = 154.1 , Σxi2 = 140 a = −1.09 , b = 1.32 Thus yˆ = −1.09 + 1.32 x When x = 3.5, then yˆ = 3.53 Since f x ( x, y ) represents change in total utility from a one unit change in X (which costs px ), 10 y f x ( x, y ) is the marginal utility of a dollar’s px f y ( x, y ) worth of X Likewise is the marginal py then utility of a dollar’s worth of Y Thus maximum satisfaction is obtained when the consumer allocates the budget so that the marginal utility of a dollar’s worth of X is equal to the marginal utility of a dollars worth of Y Similarly, for U = f(x, y, z, w) subject to the constraint xp x + yp y + zpz + wpw = I , U is maximized x n = 5, Σxi = 22 , Σyi = 37 , Σxi yi = 189 , Σxi2 = 112.5 a = 0.057 , b = 1.67 Thus yˆ = 0.057 + 1.67 x When x = 3.5, then yˆ = 5.90 when f ( x, y, z , w) f y ( x, y, z , w) λ= x = px py 15 y f z ( x, y, z, w) f w ( x, y, z , w) = pz pw That is, U is maximized when the marginal utility of a dollar’s worth of each of the products is the same = x 10 708 www.elsolucionario.net ISM: Introductory Mathematical Analysis Section 17.10 n = 6, Σxi = 27, Σyi = 21.6, Σxi yi = 105.8 , 10 Σxi2 = 139 a = 1.39 , b = 0.49 Thus yˆ = 1.39 + 0.49 x When x = 3.5, then yˆ = 3.12 Year ( x) Index ( y ) 77 100 126 134 n = 4, Σxi = 16, Σyi = 437, Σxi yi = 1945 , Σxi2 = 84 a = 69.85, b = 9.85 Thus yˆ = 69.85 + 9.85 x y 11 a Year ( x) Quantity ( y ) 35 31 26 24 26 n = 5, Σxi = 15, Σyi = 142, Σxi yi = 401 , x Σxi2 = 55 a = 35.9, b = −2.5 Thus yˆ = 35.9 − 2.5 x n = 6, Σpi = 250, Σqi = 322, Σpi qi = 11, 690 , Σpi2 = 13,100 a = 80.5 b = −0.643 Thus qˆ = 80.5 − 0.643 p n = 4, ∑ xi2 b n = 5, Σxi = 0, Σyi = 142, Σxi yi = −25 , Σxi2 = 10 a = ∑ xi = 80, ∑ yi = 23.9, ∑ xi yi = 498.4, b= = 1920, a = 4.7, b = 0.06 Thus yˆ = 4.7 + 0.06 x When x = 20, then yˆ = 5.9 12 Σxi2 = 7040 a = 100, b = 0.13 Thus yˆ = 100 + 0.13x When x = 40, then yˆ = 105.2 ∑ xi = 539, a = 1.95, Year ( x) Production ( y ) 10 15 16 18 21 n = 5, Σxi = 15, Σyi = 80, Σxi yi = 265, a = 8.5 b = 2.5 Thus yˆ = 8.5 + 2.5 x Σxi yi Σxi2 Σyi = 407.2 and n = 26.4 Thus yˆ = 407.2 + 26.4 x Problems 17.10 Σxi2 = −2.5 Thus yˆ = 28.4 − 2.5 x Year ( x) − −1 Index ( y ) 357 380 403 434 462 b= b = 1.04 Thus yˆ = 1.95 + 1.04 x Σxi2 Σxi2 = 10 a = ∑ yi = 569, ∑ xi yi = 76, 736, ∑ xi2 = 72, 691, Σxi yi Σyi = 28.4 and n n = 5, Σxi = 0, Σyi = 2036, Σxi yi = 264 , n = 4, Σxi = 160, Σyi = 420.8, Σxi yi = 16,915.2 , n = 4, Year ( x) −2 −1 Quantity ( y ) 35 31 26 24 26 = 55 4 ∫0 ∫0 ∫1 ∫0 3 0 0 x dy dx = ∫ xy dx = ∫ x dx = x y dy dx = ∫ 1 1 x2 y ∫0 ∫0 xy dx dy = ∫0 x dy dx 0 ∫ ∫ 709 www.elsolucionario.net y2 2 dx = ∫ 49 dx = 9x 27 = 2 y y2 dy = ∫ dy = 02 = 18 = 2 0 0 = ∫ x y dx = ∫ x dx = x3 =8 Chapter 17: Multivariable Calculus ∫1 ∫1 ( =∫ ISM: Introductory Mathematical Analysis ⎞ x − y dx dy = ∫ ⎜ − xy ⎟ dy ⎟ 1⎜ ⎝ ⎠1 ⎛ x3 ) 11 ⎡⎛ 3⎛ ⎞ ⎛1 ⎞⎤ ⎞ ⎢⎜ − y ⎟ − ⎜ − y ⎟ ⎥ dy = ∫1 ⎜ − y ⎟ dy ⎠ ⎝ ⎠⎦ ⎝ ⎠ ⎣⎝ x2 ∫0 ∫3x 1 ( = ∫ x − 63 x ) ⎛7 y2 ⎞ 9⎞ ⎛7 1⎞ ⎛ =⎜ y− ⎟ = ⎜7 − ⎟ −⎜ − ⎟ = ⎜3 ⎟ ⎠ 2⎠ ⎝3 2⎠ ⎝ ⎝ 12 2 x2 ∫0 ∫0 ⎛ y3 ⎞ ∫ ∫ ( y − xy )dy dx = ∫ ⎜ − xy ⎟ dx ⎟ −2 −2 ⎜ ⎝ ⎠0 ⎡⎛ 8 ⎤ ⎞ ⎛ ⎞ = ∫ ⎢⎜ − x ⎟ − ⎥ dx = ∫ ⎜ − x ⎟ dx −2 ⎣⎝ −2 ⎝ ⎠ ⎦ ⎠ =∫ 13 ⎛8 ⎞ ⎛ 16 ⎞ = ⎜ x − x ⎟ = (8 − 18) − ⎜ − − ⎟ ⎝3 ⎠ −2 ⎝ ⎠ 10 = 3 ∫0 ∫0 9− x x6 12 2 = ) 3x dx dx = x2 ⎛ 63 x5 ⎞ 58 dx = ⎜ x7 − ⎟ =− ⎜ ⎟ ⎠ ⎝ xy dy dx = ∫ x5 ( 14 x y dy dx = ∫ x y xy 2 x2 dx 16 y dy dx = ∫ 9− x y2 dx ⎞ ⎛ − x2 =∫ ⎜ − ⎟ dx = ∫ (9 − x )dx ⎟ 0⎜ 2 ⎝ ⎠ 1⎛ ⎝ ( ∫0 ∫0 ( x =∫ ∫1 ∫0 ⎞ ⎟ dx ⎟ ⎠0 )0 =3 14 x3 x3 ⎞ x4 + ⎟ dx = ∫ dx = 3 ⎟⎠ y dy dx = ∫ y2 5x dx = ∫ = 27 x dx 25 525 = x = 10 x −1 ∫1 ∫0 y − y ) dy 1− x 1 2 y dy dx = ∫ y = ∫ ( x − 1)2 dx = 1− x ⎛ y2 ⎞ 15 ∫ ∫ 3( x + y )dy dx = ∫ ⎜ xy + dx ⎟ −1 x −1 ⎜ ⎟⎠ ⎝ x 2 ⎡ ⎤ ⎛ ⎞ (1 − x) x = ∫ ⎢ x(1 − x) + − ⎜ x2 + ⎟ ⎥ dx ⎜ ⎟⎥ −1 ⎢ 2 ⎝ ⎠⎦ ⎣ ⎡ x (1 − x) ⎤ = ∫ 3⎢x − + ⎥ dx −1 2 ⎦⎥ ⎣⎢ 25 1 y ∫0 ∫ y y dx dy = ∫0 xy y dy = ∫0 ( y ⎛ y3 y ⎞ 1 =⎜ − ⎟ = − = ⎜ ⎟ ⎝ ⎠ 12 x ) ⎝ 5x x3 ⎞ 1⎛ = ⎜ x − ⎟ = (27 − 9) − = ⎜⎝ ⎟⎠ 2 3⎛ y3 ⎞ x + y dy dx = ∫ ⎜ x y + ⎟ dx 0⎜ ⎟⎠ ⎝ 3⎛ ⎜x 0⎜ y ∫0 ∫0 ( x + y)dy dx = ∫0 ⎜⎜ xy + = ∫ (2 x + 2)dx = x + x ( x − 1) ⎡ x x3 (1 − x)3 ⎤ = 3⎢ − − ⎥ 6 ⎦⎥ ⎣⎢ −1 x −1 dx = ⎡1 ⎤ ⎡1 4⎤ = ⎢ − − ⎥ − ⎢ + − ⎥ = −1 ⎣2 ⎦ ⎣2 3⎦ 710 www.elsolucionario.net ISM: Introductory Mathematical Analysis 16 3y ∫0 ∫ y x dx dy = ∫ ⎛ 45 y =∫ ⎜ 0⎜ ⎝ 3y 5x2 Section 17.10 21 dy y 5y ⎞ − ⎟ dy ⎟⎠ y y x+ y e dx dy = e x + y dy 0 0 ⎡ e2 y ⎤ e2 ⎛1 y ∫ 1 y− x ∫0 ∫0 e ( =∫ e 1 0 2y −e y ) dy 22 0 2 9y −1 ∫−1 z dx dy dz = ∫ =∫ ∫ 0 2 −1 −1 z dy dz = ∫ y z −1 x x+ y x 3x −1 ∫−1 27 z = ∫ 27 z dz = −1 ∫0 ∫0 ∫0 3x y e x y e x ∫ y dy dx = ∫ e y2 x ln x dx = ∫ e x2 − (ln x)2 dx e e3 e ⎛ ⎞ e3 e − − ⎜ − 1⎟ = − + ⎝6 ⎠ 6 2 −( x + y ) ∫ e dx dy 2 2 = ∫ −e− ( x + y ) dy = ∫ ⎡ −e−(2+ y ) + e− y ⎤ dy ⎣ ⎦ 1 y z dy dz = ⎡e − (2+ y ) − e− y ⎤ = e−4 − e−2 − e−3 + e−1 ⎣ ⎦1 dz dx = 51 3x 10 ∫ ∫ 12e−4 x −3 y dx dy −3e−4 x −3 y ) dy 2( −16 −3 y = ∫ ( −3 + 3e−12−3 y ) dy = ( e −16−3 y − e−12−3 y ) =∫ 0 = ∫ x x+ y x z dy dx x x 24 P (3 ≤ x ≤ 4, ≤ y ≤ 6) = ∫ 0 [ x ( x + y ) − 0]dy dx = ∫ dx x2 23 P (0 ≤ x ≤ 2, ≤ y ≤ 2) = ∫ ( x + x y )dy dx 1⎛ ⎡⎛ x2 y ⎞ x4 ⎞ ⎤ = ∫ ⎜ x3 y + ⎟ dx = ∫ ⎢⎜ x + ⎟ − ⎥ dx 0⎜ ⎢⎜ ⎟⎠ ⎟⎠ ⎦⎥ ⎝ ⎝ ⎣ =∫ ⎡ x3 ⎤ = ⎢ − ( x ln x − x ln x + x) ⎥ ⎢⎣ ⎥⎦ 27 =− x dz dy dx = ∫ x 0 20 xy dy dx 2 xy dy dx = ∫ ln x −1 ∫−1 ∫−1 ∫1 xy =∫ ∫ e x =∫ = (−e0 + e1 ) − (−e−1 + e0 ) = −1 + e + e−1 − 0 x2 ∫1 ∫ln x ∫0 dz dy dx = ∫1 ∫ln x z dy dx = = ∫ (−e y −1 + e y )dy = (−e y −1 + e y ) 19 ∫ z ⎡ x x6 ⎤ x5 ⎤ = ∫ ⎢ − ⎥ dx = ⎢ − ⎥ = 2 ⎦⎥ ⎢⎣ ⎣⎢ 12 ⎥⎦ 24 dx dy = ∫ −e y − x dy = −2 + e + e x xy x ⎡ x3 ⎞ e =⎢ −e ⎥ = − e − ⎜ − 1⎟ = −e+ ⎝2 ⎠ ⎣⎢ ⎦⎥ 18 dz dy dx = ∫ x2 ∫∫ ∫0 ∫x ∫0 =∫ ⎛ 15 y y ⎞ 405 243 =⎜ − − = 81 ⎟ = ⎜ ⎟ ⎠ 2 ⎝ 17 xy x = e −34 − e −30 − e−22 + e−18 1 1⎞ ⎛ 25 P ⎜ x ≥ , y ≥ ⎟ = ∫ ∫ dx dy 1/ 1/ 2 3⎠ ⎝ 1 ⎛ 1⎞ = ∫ x 1/ dy = ∫ ⎜ − ⎟ dy 1/ 1/ ⎝ 2⎠ 1 1 1⎛ 1⎞ 1⎛ 2⎞ dy = y =∫ = ⎜1 − ⎟ = ⎜ ⎟ = 1/ 2 1/ ⎝ ⎠ ⎝ ⎠ 3(1) −0 = 10 10 26 11 ∫0 ∫0 711 www.elsolucionario.net dx dy = ∫ 1x 08 dy = ∫ 11 08 dy = y = 80 Chapter 17: Multivariable Calculus ISM: Introductory Mathematical Analysis Chapter 17 Review Problems x + z = The x,z-trace is x + z = , which is a circle For any fixed value of y, we obtain x + y + z = can be put in the form Ax + By + Cz + D = 0, so the graph is a plane The intercepts are (1, 0, 0), (0, 1, 0), and (0, 0, 1) the circle x + z = z z y 1 x y 1 x f x ( x, y ) = 4(2 x) + 6(1) y + − = x + y f y ( x, y ) = + x(1) + y − = x + y z = x can be put in the form Ax + By + Cz + D = 0, so the graph is a plane Every point on the y-axis is an intercept The x,z-trace is z = x, which is a line For any fixed value of y, we obtain the line z = x z y ∂P = 3l + − (1)k = 3l − k ∂l ∂P = + 3k − l (1) = 3k − l ∂k ∂ z ( x + y )(1) − x(1) y = = ∂x ( x + y) ( x + y )2 Because z = x( x + y )−1 , ∂z x = x ⎡(−1)( x + y ) −2 (1) ⎤ = − ⎣ ⎦ ∂y ( x + y )2 x z = y The y,z-trace is z = y , which is a parabola For any fixed value of x, we obtain the curve z = y2 z f pB ( pA , pB ) = + 5(1 − 0) = y 10 w = x ( ) ln x + y 2 y ∂ 1 [ f ( x, y )] = ⋅ (2 y ) = 2 x2 + y ∂y x + y2 f ( x, y ) = ln x + y = x x2 + y = x( x + y ) − 12 ∂w xy −3 ⎡ ⎤ = x ⎢ − ( x + y ) (2 y ) ⎥ = − ∂y ⎣ ⎦ (x + y2 ) 11 wx ( x, y, z ) = xyze x yz ( ) 2 wxy ( x, y, z ) = xz ⎡⎢ y e x yz ⋅ x z + e x yz ⋅1⎤⎥ ⎣ ⎦ ( ) = xze x yz x yz + 712 www.elsolucionario.net ISM: Introductory Mathematical Analysis 12 Chapter 17 Review ⎡ ⎛ ⎤ ⎞ f x ( x, y ) = y ⎢ x ⎜ ⋅ y ⎟ + ln( xy ) ⋅1⎥ = y[1 + ln( xy )] ⎣ ⎝ xy ⎠ ⎦ ⎡1 ⎤ f xy ( x, y ) = y ⎢ ⋅ x ⎥ + [1 + ln( xy )] ⋅1 = + + ln( xy ) = + ln(xy) ⎣ xy ⎦ 13 ∂ [ f ( x, y, z )] = ( x + y + z )(2 z ) + ( x + y + z )(1) = 3z + z ( x + y ) + x + y ∂z ∂2 ∂ z2 ( [ f ( x, y, z )] = z + 2( x + y ) = x + y + z 14 z = x − y )( y − 2xy ) = x2 y − x3 y − y3 + xy2 ∂z = x y − x3 − y + xy ∂y ∂ 2z ∂ y2 15 = 2x2 − y + x w = e x + y + z ln xyz = e x + y + z (ln x + ln y + ln z ) ⎛1⎞ ∂w = e x + y + z (ln x + ln y + ln z ) + e x + y + z ⎜ ⎟ ∂y ⎝ y⎠ ⎡ 1⎤ = e x + y + z ⎢ ln xyz + ⎥ y⎦ ⎣ By symmetry, ∂w = ex+ y+ z ∂x ∂2 w = e x+ y+ z ∂z∂x 1⎤ ⎡ ⎢ ln xyz + x ⎥ ⎣ ⎦ ⎤ x+ y + z ⎡ ⎢ ln xyz + x ⎥ + e ⎣ ⎦ 1⎤ ⎡ = e x + y + z ⎢ ln xyz + + ⎥ x z⎦ ⎣ 16 17 ⎡1⎤ ⎢z⎥ ⎣ ⎦ ∂P = 100 ⎡ (0.11)l 0.11−1 ⎤ k 0.89 = 11l −0.89 k 0.89 ⎣ ⎦ ∂l ∂ P = 11l −0.89 ⎡ (0.89)k 0.89−1 ⎤ = 9.79l −0.89 k −0.11 ⎣ ⎦ ∂ k∂ l x+ y y = + xz z xz y f x ( x, y , z ) = − x2 z f xy ( x, y, z ) = − x2 z f xyz ( x, y, z ) = 2 x z 1 f xyz (2, 7, 4) = = 2 64 ⋅4 f ( x, y , z ) = 713 www.elsolucionario.net Chapter 17: Multivariable Calculus 18 f x ( x , y , z ) = 6e y ISM: Introductory Mathematical Analysis ln( z +1) f xy ( x, y, z ) = 12 y ln( z + 1)e y ln( z +1) ⎡ y ⎪⎫ y ln( z +1) ⎤ ⎪⎧ ⋅ f xyz ( x, y, z ) = 12 y ⎢ ln( z + 1) ⎨e y ln( z +1) ⋅ ⎥ ⎬+e z + ⎪⎭ z + ⎥⎦ ⎢⎣ ⎪⎩ f xyz (0, 1, 0) = 12[0 + 1] = 12 19 ( ) ∂w ∂w∂x ∂w∂ y ⎛ ⎞ = + = (2 x + y ) er + (2 x + y ) ⎜ ⎟ ∂r ∂x ∂r ∂ y ∂r ⎝r+s⎠ 2( x + y ) r+s ∂w ∂w∂x ∂w∂ y ⎛ ⎞ = + = (2 x + y )(0) + (2 x + y ) ⎜ ⎟ ∂s ∂x ∂s ∂ y ∂s ⎝r+s⎠ = 2( x + y )er + = 20 2( x + y ) r+s ⎛ ⎞ ∂z ∂z ∂x ∂z ∂ y ⎛1 ⎞ = + = + ye xy − y ⎟ ( s ) + ⎜ − + xe xy − x ⎟ (2(r + s)) ∂ s ∂ x ∂ s ∂ y ∂ s ⎜⎝ x y ⎠ ⎝ ⎠ 21 x + y − z (−4 z + x) ∂z ⎡ ∂z ⎤ + x + z (1) ⎥ + = ∂ x ⎢⎣ ∂ x ⎦ ∂z = −(2 x + y + z ) ∂x ∂ z −(2 x + y + z ) x + y + z = = ∂x −4 z + x 4z − x 22 z + ln( yz ) + ln z + x + z = or z + ln y + ln z + ln z + x + z = 0, thus z + ln y + ln z + x + z = ∂z 1 ∂z ∂z + =0 2z + + ⋅ ∂y y z ∂y ∂y ∂z ⎡ ⎤ z + + 1⎥ = − ∂y ⎢⎣ z ⎦ y ∂z ⎡ z + + z ⎤ ⎢ ⎥=− ∂y ⎣⎢ z y ⎦⎥ ∂z z =− ∂y y (2 z + + z ) 714 www.elsolucionario.net ISM: Introductory Mathematical Analysis Chapter 17 Review 23 P = 20l 0.7 k 0.3 Marginal productivity functions ∂P = 20(0.7)l −0.3 k 0.3 and are given by ∂l ∂P ∂P = 20(0.3)l 0.7 k −0.7 Thus = 14l −0.3 k 0.3 ∂k ∂l ∂P and = 6l 0.7 k −0.7 ∂k 29 z xyz = 32 (xyz ≠ 0) Let S be the amount of cardboard used S = xy + 2yz + 2xz ⎡ 32 ⎤ ⎡ 32 ⎤ = xy + y ⎢ ⎥ + x ⎢ ⎥ xy ⎣ ⎦ ⎣ xy ⎦ 24 c = 3x + 0.05xy + 9y + 500 Marginal cost with respect to x is ∂c = + 0.05 y When x = 50 and y = 100, then ∂x ∂c = ∂x = xy + The critical point occurs when x = 4, y = 4, and z = 2, which gives a minimum The dimensions are ft by ft by ft ∂ qA ∂ qB = > and = > , A and B ∂ pB ∂ pA 30 ∂α ∂α = 0.530; = −0.027 ∂P ∂S 27 f ( x, y ) = x + y − xy − y + f y ( x, y ) = 2by + cx − 20 At (1, 2), f x ( x, y ) = and f y ( x, y ) = Thus 2a + 2c − 10 = (1) and 4b + c − 20 = (2) From Eq (1), a = – c; from Eq (2), 20 − c b= f xx ( x, y ) = 2a, f yy ( x, y ) = 2b, f xy ( x, y ) = c ⎧ f x ( x, y ) = x − y = ⎨ f ( x, y ) = y − x − = ⎩ y Critical point: (2, 2) f xx ( x, y ) = 2, f yy ( x, y ) = 4, f xy ( x, y ) = −2 At (2, 2), D = (2)(4) − (−2) = > and f xx ( x, y ) = > ; thus relative minimum at (2, 2) 28 f ( x, y ) = ax + by + cxy − 10 x − 20 y f x ( x, y ) = 2ax + cy − 10 ; are competitive products 26 64 64 + x y ∂S 64 ∂ S 64 = y− , = x− ∂y ∂x x y2 25 qA = 100 − pA + pB , qB = 150 + pA − pB Since y x At (1, 2), D = (2a )(2b) − c = ⎛ 20 − c ⎞ Thus 4ab − c = , 4(5 − c) ⎜ ⎟−c = , ⎝ ⎠ f ( w, z ) = 2w + z − 6wz + 100 − 25c + c − c = , 100 = 25c, or c = So a = – c = – = and 20 − c 20 − b= = = 4 Thus a = 1, b = 4, c = ⎧⎪ f w ( w, z ) = w2 − z = ⎨ ⎪⎩ f z ( w, z ) = z − w = Critical points: (0, 0), (1, 1) f ww ( w, z ) = 12 w, f zz ( w, z ) = 12 z , f wz ( w, z ) = −6 At (0, 0), D = (0)(0) − (−6) = −36 < ; thus neither relative maximum nor relative minimum At (1, 1), D = (12)(12) − (−6) = 108 > and f ww ( w, z ) = 12 > ; thus relative minimum at (1, 1) 715 www.elsolucionario.net Chapter 17: Multivariable Calculus ISM: Introductory Mathematical Analysis 31 Profit = P = ( pA − 50 ) qA + ( pB − 60 ) qB P = ( pA − 50 ) ⎡⎣ 250 ( pB − pA ) ⎤⎦ + ( pB − 60 ) ⎡⎣32, 000 + 250 ( pA − pB ) ⎤⎦ ∂P = ( pA − 50 ) (−250) + ⎡⎣ 250 ( pB − pA ) ⎤⎦ (1) + 250 ( pB − 60 ) ∂ pA = −500 pA + 500 pB − 250(10) = 500 ( − pA + pB − ) ∂P Also, = ( pA − 50 ) (250) + ( pB − 60 ) (−500) + ⎡⎣32, 000 + 250 ( pA − pB ) ⎤⎦ (1) ∂ pB = 500 pA − 1000 pB + 49,500 = 500 ( pA − pB + 99 ) Setting ∂P ∂P = and = gives ∂ pA ∂ pB − pA + pB − = (1) and pA − pB + 99 = (2) Adding Equations (1) and (2) gives − pB + 94 = So pB = 94 From Equation (1), pA = pB − , so pA = 94 − = 89 At pA = 89 and pB = 94 , D = ∂ 2P ∂ pA2 ∂ 2P ∂ 2P ∂ pA2 ∂ pB2 − ∂ 2P = (−500)(−1000) − (500)2 > and ∂ pB∂ pA = −500 < Thus there is a relative maximum profit when the price of A is 89 cents per pound and the price of B is 94 cents per pound 32 f ( x, y, z ) = xy z , x + y + z − = F ( x, y, z , λ ) = xy z − λ ( x + y + z − 1) ⎧ Fx ⎪ ⎪ Fy ⎨ ⎪ Fz ⎪⎩ Fλ = y2 z − λ = = xyz − λ = = xy − λ = = −x − y − z +1 = From the first and third equations, we have λ = y z = xy , so x = z (since xyz ≠ 0) With x = z in the second equation, we have λ = x y Combining this with λ = xy , we get y = 2x Substituting y = 2x and z = x in the 1 fourth equation gives −x − 2x − x + = so 4x = and x = Thus y = and z = The critical point is 4 ⎛1 1⎞ ⎜ , , ⎟ ⎝4 4⎠ 716 www.elsolucionario.net ISM: Introductory Mathematical Analysis 33 Mathematical Snapshot Chapter 17 f ( x, y, z ) = x + y + z , 3x + 2y + z = 14 F ( x, y , z , λ ) 37 = x + y + z − λ (3x + y + z − 14) ⎧ Fx ⎪⎪ Fy ⎨ ⎪ Fz ⎪⎩ Fλ = x − 3λ = = y − 2λ = = 2z − λ = = −3x − y − z + 14 = From (1), x = λ z= y2 ∫0 ∫0 xy dx dy = ∫ (1) (2) (3) (4) y6 = ⋅ 38 = y dy dx = ∫ ∫1 ∫x2 = −0.43 39 x2 ∫0 ∫ x Year ( x) Expenditures ( y ) 15 22 21 26 27 34 nΣxi yi − ( Σxi )( Σyi ) nΣxi2 − ( Σxi ) Thus yˆ = 12.67 + 3.29 x ∫ ∫ = y 18 = =∫ = 3.29 dx x ⎤ ⎥ ⎥ ⎦0 Mathematical Snapshot Chapter 17 y x3 y ) ) x2 ⎡ x x x x x3 x ⎢ =7 + − − − + ⎢5 7 ⎣ ⎡1 1 2 ⎤ = 7⎢ + − − − + ⎥−0 = 7 30 ⎣ ⎦ Σxi2 ) ( Σyi ) − ( Σxi )( Σxi yi ) ( a= = 12.67 y 2 x y dx dy ( ( Σxi2 = 91 36 ) 1⎡ ⎛ ⎞⎤ = ∫ ⎢ x + x5 − x6 − ⎜ x + x − x ⎟ ⎥dx 0⎣ ⎝ ⎠⎦ n = 6, Σxi = 21, Σyi = 145, Σxi yi = 565, b= ( x + xy − y dy dx = ∫ x y + xy − y3 nΣxi2 − ( Σxi ) x ⎛ x3 x5 ⎞ =⎜ − ⎟ ⎜ 10 ⎟⎠ ⎝ ⎛ 128 512 ⎞ ⎛ ⎞ =⎜ − ⎟−⎜ − ⎟ ⎠ ⎝ 10 ⎠ ⎝ 603 =− 10 Thus pˆ = 85.15 − 0.43t 35 dx x4 ⎞ = ∫ ⎜ x − ⎟ dx ⎜ ⎟⎠ ⎝ Σti2 ) ( Σpi ) − ( Σti )( Σti pi ) ( a= = 85.15 2x 4⎛ Σti = 104, Σpi = 381, Σti pi = 7482, Σti2 = 3192 nΣti2 − ( Σti ) 1 ⋅ −0 = 12 y2 2x 34 n = 5, nΣti pi − ( Σti )( Σpi ) dy ⎞ y4 ⋅ y 1 =∫ ⎜ − ⎟ dy = ∫ y dy ⎟ 0⎜ 2 ⎝ ⎠ Substituting into (4) gives λ ⎛ 3λ ⎞ −3 ⎜ ⎟ − 2λ − + 14 = , from which λ = ⎝ ⎠ Thus x = 3, y = 2, and z = Critical point of F: (3, 2, 1, 2), so the critical point of f is (3, 2, 1) b= y2 1⎛ 3λ ; from (2), y = λ , from (3), nΣti2 − ( Σti ) x2 y dy = ∫ y5 20 y dy 64 63 − = = 18 18 18 x 10 ax y = Ceax + 5, y − = Ce , ln( y − 5) = ax + ln C 717 www.elsolucionario.net Chapter 17: Multivariable Calculus x y 15 12 7 10 y −5 10 ISM: Introductory Mathematical Analysis ln( y − 5) 2.30259 1.94591 1.38629 0.69315 0.00000 dT = k (T − a ) , where dt dT dT = k dt , = k (T − 45) , a = 45 Thus T − 45 dt dT ∫ T − 45 = ∫ k dt , ln T − 45 = kt + C Because T – 45 > 0, ln(T – 45) = kt + C Thus Newton’s law of cooling: n = 5, Σxi = 22, Σ ln ( yi − ) = 6.32794 , Σ ⎡⎣ xi ln ( yi − ) ⎤⎦ = 12.34312, Σxi2 = 166 a= = T − 45 = ekt +C , or T = e kt +C + 45 = eC e kt + 45 = C1e kt + 45 , where nΣ ⎡⎣ xi ln ( yi − ) ⎤⎦ − ( Σxi ) ⎡⎣Σ ln ( yi − ) ⎤⎦ ( ) n Σxi2 − ( Σxi ) 5(12.34312) − 22(6.32794) 5(166) − (22) C1 = eC So T = C1e kt + 45 When t = 0, then T = 124 Hence 124 = C1 + 45, or C1 = 79 Thus ≈ −0.22399 Σxi2 ) ⎡⎣Σ ln ( yi − ) ⎤⎦ − ( Σxi ) {Σ ⎡⎣ xi ln ( yi − ) ⎤⎦} ( ln C = n ( Σxi2 ) − ( Σxi ) = 166(6.32794) − 22(12.34312) 5(166) − (22)2 ≈ 2.25112 C , ln y = ln C − ln x r , x xr ln y = ln(C) – r ln x Thus r(ln x) + (ln y) – ln(C) = Since ln C and r are constants, ln x and ln y are linearly related r ln y = ln ( ) ln 19 19 79 ,k= ≈ −0.01113 Thus 79 128 Using a graphics calculator on the points displayed in Fig 17.25 produces the same result as in the snapshot Performing an exponential regression on the points shown in Fig 17.24, however, does not produce the right curve, because the exponential model lacks a constant term This difficulty can be overcome by subtracting 45 (the long-term temperature) from every temperature value, running the exponential regression, and then adding a constant term of 45 to the resulting model Thus y = 9.50e−0.22399 x + C 128k = ln T = 79e −0.01113t + 45 C ≈ e 2.25112 ≈ 9.50 y = T = 79ekt + 45 When t = 128, then T = 64, so 19 64 = 79e128k + 45 , 19 = 79e128k , e128k = , 79 718 www.elsolucionario.net ... x = –5 61 y? ??6 y+ 6 − = y y y? ??6 Multiplying both sides by y( y − 6) gives 66 ( y − 6)2 − 6( y − 6) = y ( y + 6) y − 12 y + 36 − y + 36 = y + y y−2 y? ??2 = y+ 2 y+ 3 x+5 = ( y + y? ??6 = y −4 y= 2 63 x x... ⎛ ⎜ −∞, ⎥ 3⎦ ⎝ –2 34 29 y y y + > y+ 1 5y + 1 0y > 3 0y + 6y 2 5y > 3 6y > 1 1y 0 >y y y + −6 > y −3 > y y < −3 (−∞, −3) –3 35 12(50)... Multiplying both sides by gives 15x + x – 25 = + 25x 16x – 25 = + 25x –9x = 26 26 x=− 35 3x + 41 y y y y + − = 6 0y – 3 0y + 2 0y – 1 5y = 1 2y 3 5y = 1 2y 2 3y = y= 0 36 y − 37 38 42 2y − 6y + = Multiplying