Solutions #5.pdf - University of Toronto Scarborough Department of Computer Mathematical Sciences MAT A32H Winter 2018 Solutions#5 1 Section 11.3 16 If

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Unformatted text preview: University of Toronto Scarborough Department of Computer & Mathematical Sciences MAT A32H Winter 2018 Solutions #5 1. Section 11.3 16. If the cost function is c = c(q) = 0.1q 2 + 3q + 2, the marginal cost function is dc dc = c ′ (q) = 0.2q + 3. When q = 3, = c ′ (3) = (0.2)(3) + 3 = 3.6. dq dq q=3 2000 , the cost function is c = (c)(q) = q 5 q + 2000 and the marginal cost function is c ′ (q) = 5 for all q > 0. Hence c ′ (25) = c ′ (250) = 5.   1 2 1 q = 15q − q , the marginal 24. If the revenue function is r(q) = q 15 − 30 30 q 5 44 dr = r ′ (q) = 15 − . Now r ′ (5) = 15 − = , r ′ (15) = cost function is dq 15 15 3 15 150 15 − = 14 and r ′ (150) = 15 − = 5. 15 15 32. If the cost function is c = c(q) = 0.4q 2 + 4q + 5, the rate of change of c with dc = c ′ (q) = 0.8 q + 4. If q = 2, we have c ′ (2) = (0.8)(2) + 4 = 5.6. respect to q is dq △c c(3) − c(2) 20.6 − 14.6 Over the interval [2, 3] we have = = = 6. △q 3−2 1 36. We are given that y = 5 − 3x3 and x = 1. dy (a) The rate of change of y with respect to x is = y ′ = −9 x2 . dx y′ 1 dy −9 x2 (b) The relative rate of change of y with respect to x is = = = y y dx 5 − 3x3 9 . 3x3 − 5 (c) The rate of change of y with respect to x when x = 1 is y ′ (1) = −9(1)2 = −9.  ′ y (1) = (d) The relative rate of change y with respect to x when x = 1 is y −9(1)2 9 y ′ (1) = = − = −4.5. 3 y(1) 5 − 3(1) 2 (e)  The  percentage rate of change of y with respect to x when x = 1 is y′ (1) × 100% = −450 %. y 20. If the average cost function is c = 5 + MATA32H page 2 Solutions # 5 dc dc dq 1 1 dq which is given to be ; i.e, = =⇒ 46. The relative rate of change of c is c c q   q dc dc c and the average cost = = c. Hence the marginal cost function dq q dq function (c) are equal. 2. Section 11.4 2. f (x) = (3x − 1)(7x + 2), f (x) = (3)(7x + 2) + (3x − 1)(7) = 42x − 1. ′ 4. Q(x) = (x2 + 3x)(7x2 − 5), Q (x) = (2x + 3)(7x2 − 5) + (x2 + 3x)(14x) = 28x3 + 63x2 − 10x − 15. ′ C(I) = (2I 2 − 3)(3I 2 − 4I + 1), C ′ (I) = (4I)(3I 2 − 4I + 1) + (2I 2 − 3)(6I − 4) = 24I 3 − 24I 2 − 14I + 12 = 2(12I 3 − 12I 2 − 7I + 6). √ √ √ 1/2 3 x − 3 x) = (x1/2 + 5x − 2)(x1/3 16. g(x) =  ( x + 5x − 2)(    − 3x ), 1 −1/2 1 −2/3 3 −1/2 ′ 1/3 1/2 1/2 g (x) = = x + 5 (x − 3x ) + (x + 5x − 2) x − x 2 3 2 1 (−135x1/2 + 40x1/3 + 5x−1/6 + 18x−1/2 − 4x−2/3 − 18). 6 2x − 3 , 20. y = 4x + 1 2(4x + 1) − 2x − 3)(4) 14 y′ = = . 2 (4x + 1) (4x + 1)2 x−5 36. y = √ , 8 x   √ √ (1)(8 x) − (x − 5) √4x x + √5x x+5 ′ y = = = . 64x 16x 16 x3/2  x−5 1 1/2 Alternate solution: y = √ = x − 5x−1/2 , 8 8 x   1 1 x+5 1 5 1 5 −1/2 −3/2 ′ = = x + x y = + . 8 2 2 16 x1/2 x3/2 16 x3/2 6. 42. 48. 27x2 + 15x − 2 (9x − 1)(3x + 2) = , 4 − 5x 4 − 5x (54x + 15)(4 − 5x) − (27x2 + 15x − 2)(−5) −135x2 + 216x + 50 y′ = = . (4 − 5x)2 (4 − 5x)2 y= 1 + x−1 + a−1 x = 1 −1 −1 x −a − x (1)(a − x) − (a + x)(−1) 2a f ′ (x) = = . 2 (a − x) (a − x)2 For a 6= 0 a constant, f (x) = 1 a 1 a = a+x , a−x MATA32H Solutions # 5 page 3 √ √ 20 I + 0.5 I 3 − 0.4I √ 68. C = I +5 The marginal propensity to consume is   √ √ √ √ 1 10 3 √ √ ( I + 0.75 I − 0.4)( I + 5) − (20 I + 0.5 I − 0.4I) 2 I dC √ . = dI ( I + 5)2 d C When I = 100, we have ≈ 0.393. d I I=100 dC so, when I = 100, we have The marginal propensity to save is 1 − dI d S ≈ 1 − 0.393 ≈ 0.607. d I I=100 √ √ √ ( I + 2)( I − 4) √ I −2 I −8 √ = = I −4 70. S = √ I +2 I +2 The marginal propensity to save is dS 1 1 = I −1/2 = √ . dI 2 2 I 1 d S = √ ≈ 0.04082. When I = 150, we have d I I=150 2 150 1 The marginal propensity to consume is 1 − √ so, when I = 150, we have 2 I d C ≈ 1 − 0.04082 ≈ 0.9592. d I I=150 72. 3. dc d (c) = 0. Hence 0 = = Given a cost function c = f (q) we assume dq dq   q dd qc − (c)(1) dc dc c dc d c = − c = 0 =⇒ q = c =⇒ = = c. =⇒ q 2 dq q q dq dq dq q Therefore, the marginal cost function is the same as the average cost, as required. Section 11.5 y = 2u3 − 8u and u = 7x − x3 . d y Chain d y d u · = (6u2 − 8)(7 − 3x2 ) = (6(7x − x3 )2 − 8)(7 − 3x2 ) = 2(3x6 − = d x Rule d u d x 42x4 + 147x2 − 4)(7 − 3x2 ). √ 4. y = 4 z and z = x5 − x4 + 3. 1 5x4 − 4x3 5x4 − 4x3 d y Chain d y d z p · = z −3/4 (5x4 −4x3 ) = = . = d x Rule d z d x 4 4 (x5 − x4 + 3)3/4 4 4 (x5 − x4 + 3)3 √ 2 6. z = u2 + u + 9 and  u = 2s −1. 1 d z Chain d z d u · = 2u + √ (4s). = d s Rule d u d s 2 u   1 d z 2 = 2+ (−4) = −10. When s = −1, u = 2(−1) − 1 = 1 and d s s=−1 2 2. MATA32H page 4 Solutions # 5 √ y = 3x2 − 7 = (3x2 − 7)1/2 , 1 3x y ′ = (3x2 − 7)−1/2 (6x) = √ . 2 3x2 − 7 √ 3 22. y = 8x2 − 1 = (8x2 − 1)1/3 , 16 x 1 . y ′ = (8x2 − 1)−2/3 (16x) = 3 3 (8x2 − 1)2/3 −2/3 3 2 30. y = = 3 3x − x , 2/3 2 (3x  −x) −5/3 −2 (6x − 1) 2 . 3x2 − x (6x − 1) = y′ = 3 − 3 (3x2 − x)5/3 √ 36. y = 4 x3 1 − x2 = 4 x3 (1 − x2 )1/2 , 20. y ′ = 12 x2 (1 − x2 )1/2 + 2x3 (1 − x2 )−1/2 (−2x) = 12 x2 √ 4 x2 (3 − 4x2 ) √ . 1 − x2 4  2x 42. y = , x+2  3   128 x3 (2)(x + 2) − (2x)(1) 2x ′ = . y =4 x+2 (x + 2)2 (x + 2)5 r 1/3  2 2 8x − 3 3 8x − 3 44. y = , = x2 + 2 x2 + 2   −2/3  (16 x)(x2 + 2) − (8 x2 − 3)(2x) 1 8 x2 − 3 ′ y = 3 x2 + 2 (x2 + 2)2 38 x = . 3 (8 x2 − 3)2/3 (x2 + 2)4/3 56. 1 − x2 − √ z = 2y 2 − 4y + 5, y = 6x − 5 and x = 2t. d z Chain d z d y d x · · = (4y − 4)(6)(2). = d t Rule d y d x d t d z When t = 1, x = 2 and y = 7 so = (24)(6)(2) = 288. dt 4 x4 = 1 − x2 t=1 68. 100 100 q and r = p q = . q + 10 q + 10 d r Chain d r d q · . The marginal revenue product is = d m Rule d q d m dr 100(q + 10) − 100 q (1) 1000 250 Now = = = . When m = 5, q = √ 2 2 dq (q + 10) (q + 10) 36 125 d r 360 =⇒ . = 3 d q m=5 √ 961 50 m2 + 11 − (50 m)9m2 + 11)(2m) 550 dq = = . When Also 2 2 dm m + 11 (m + 11)3/2 q=√ 50 m , m2 + 11 p= MATA32H page 5 Solutions # 5 d q 275 m = 5, . = d m m=5  108  d r d q d r = Hence d m dq d m m=5 m=5 m=5  = 2750 360 275 · = . 961 108 2883 k , k is a constant, and q = f (m). Since r = p q, the marginal revenue q d r Chain d r d q product is · . = d m Rule d q d m dr dr dq Since r = p q = k, = 0. Hence = 0· = 0. dq dm dm 74. We are given S = 340 E 2 − 4360 E + 42800, where E = years of education, E ≥ 7, and S = salary. 70. p= (a) dS = 680 E − 4360 so, when E = 16, dE d S = 6520. d E E=16 dS = 680 E − If salary is changing at the rate of $ 5000 per year, we have dE 9360 4360 = 5000 =⇒ 680 E = 9360 =⇒ E = ≈ 13.8. Hence it takes 680 about 13.8 years of education to achieve this rate of change in salary. 50 q . 80. q = 2 m (2 m + 1)3/2 , and r = √ 1000 + 3 q (a) When m = 12, q = 3000 and r = 1500. Hence the price per unit is r 1500 1 p= = = = 0.50. The price per unit is 50 cents. q 3000 2  √ (50)( 1000 + 3q) − 50 q 21 (1000 + 3q)−1/2 (3) dr = . (b) Marginal revenue is dq 1000 + 3q 2750 d r 11 = When m = 12, q = 3000, so = . d q q=3000 10, 000 40 dr dr dq dq (c) The marginal revenue product is = · . Now = (2)(2m + dm dq dm dm   d q 3 1)3/2 + (2m) (2m + 1)1/2 (2), so = 610. Therefore, using part 2 d m m=12 !     671 d r d q d r 11 (610) = = = = (b), we have d m m=12 d q q=3000 d m m=12 40 4 167.75. 324 5 19 + + . 82. c = c(q) = p q 2 + 35 q 18 ! 5 19 19 324 + + = . (a) The limiting value of the average cost is lim p q→∞ 18 q 2 + 35 q 18 324 q 19 q . (b) The cost function is c(q) = c · q = p +5+ 2 18 q + 35 (b) MATA32H Solutions # 5 page 6 The marginalp cost is  (324) q 2 + 35 − (324 q) 21 (q 2 + 35)−1/2 (2 q) 19 ′ + . c (q) = q 2 + 35 18 ′ When q = 17, c (17) = 3. When q = 17, the marginal cost is $ 300 per unit. (c) If production is increased from 17 to 18, the addition cost is about $ 300 (from part (b)), but revenue will only increase by $ 275, resulting in a loss of about $ 25. Hence, production should not be increased from 17 to 18. ...
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