Mathematical Applications for the Management Life and Social Sciences 11th Edition Harshbarger SOLUTIONS MANUAL Full clear download at: https://testbankreal.com/download/mathematical-applications-managementlife-social-sciences-11th-edition-harshbarger-solutions-manual/ Mathematical Applications for the Management Life and Social Sciences 11th Edition Harshbarger TEST BANK Full clear download at: https://testbankreal.com/download/mathematical-applications-managementlife-social-sciences-11th-edition-harshbarger-test-bank/ Eercises.1 1. 3 1 0. 5 3 3 0 3. ( y 1)( y ) y 3 y y 3 y 0. ( z 1)( z 3) 1 z z 0 5. 1 1 0 9. 0 ( 1) 0 Solution: = 0, 1 Never divide by a variable. A root is lost if you divide. 10. t t 3t 11. 0 t t 0 t (t ) t 0 or t 0 Solution: t = 0, t t 1 0 (t 1)(t 1) 0 t 1 0 6 1 0 ( 6) ( 6) 0 ( 6)( ) 0 6 0 or 0 Solution: = -, 6 6. 11 10 11 10 0 10 10 0 1. Solution: t 1 9z 1z 1 0 (7 z 1)(7 z 1) 0 7 z 1 0 7 z 1 1 Solution: z 7 ( 10)( 1) 0 10 0 or 1 0 Solution: = 1, 10 7. 9 0 (3 )(3 ) 0
13. a. 0 a = 1, b =, c = ( ) ( ) ( 1 ) ( ) ( 1 ) 3
3 0 or 3 0 b. Since 1.1, the solutions are 3 3 Solution:, 8. 5 16 0 (5 )(5 ) 0 5 0 or 5 0 approimately.83, 0.83. c. 6 7 0 a = 1, b = 6, c = 7 6 36 8 6 8 6 3 Solution:, 1.1, the solutions are 5 5 approimately.83, 0.83.
1. 6 7 0 a = 1, b = 6, c = 7. ( 1) 1 6 36 8 1 6 8 6 3 3. 5 1 1 0 a. 3, 3 7 3 0 b..1, 1.59 Solution: = 7, 3 15. w w 1 0. 17 8 1 a =, b = 1, c = 1 w 1 18 1 7 9 1 0 There are no real solutions. 7 0 Solution: = 7, 16. z z 0 a = 1, b =, c = 5. w w 0 8 17. 16 1 z No real solutions. y 7 y 7 w w 3 0 (w 8)(w ) 0 w 8 0 or w 0 Solution: w = 8, y 11 18. z 1 z 1 z 3 6. y 1 0 6 3 y 11y 6 0 (3 y )( y 3) 0 3 y 0 or y 3 0 19. 5 80 0. 16 3 75 5 Solution: y, 3 3 7. 16z 16z 1 0 a = 16, b =16, c = 1 z 16 56 13 3 16 0 3 7 5 3 or 7 3
1. ( ) 5 5 5 Solution: = 1, 9 Solution: z,
8. 10 y y 65 0 a = 10, b = 1, c = 65 33. 1 70 7 0 1 1 (600) y 0 1 601 151 50 5 or 0 0 0 0 5 13 Solution: y, 5 9. ( 1)( 5) 7 5 7 1 0 ( 6)( ) 0 Solution: = 6, 30. ( 3)(1 ) 1 3 3 1 0 ( )( ) 0 0 Solution: = 3. Divide by 7 and rearrange. 3 10 0 ( 5)( ) 0 Solution: =, 5 3 11 6 0 31. 5 6 or 5 6 0 a = 5, b =, c = 6 10 1 31 10 5 Solution: 1 31, 1 31 5 5 3. 3 6 3 6 0 a = 3, b = 6, c = 6 36 6 1 6 6 6 3 3 3 6 3 Solution: 3 3, 3 3 3 3 (3 )( 3) = 0 Solution:, 3 3
35. 300 0.01 0 a = 5.6, b = 16.1, c = 1.1 16.1 59.1 11.6 51. 16.1 371.85 51. 0.69 or 0.06 38. 6.8z.9z.6 0 a = 0.01, b =, c = 300 1 0.0 300 or 100 0.0 36. 9.6 0.1 0 6.8z.9z.6 0 z.9.01 70.7 13.6.9 9.73.9 9.73 13.6 13.6 1.08 or 0.36 9.6 0.1 0 0 96 0 ( 1)( 8) 0 Solution; = 1, 8 39. 8 9 8 9 9 8 0 ( 8)( 1) 0 Solution: = 1, 8 37. 5.6 16.1 1.1 0 0. 3 1 1 ( 1) 1 ( )( 1) 3( ) 3 6 Solution: = 8 8
1. 1 6. P 16 0.1 100 1 1 When P = 180 we have 3 1 0 ( 1)( 1) 0 ( ) 1 Solution: 1 1 is not a root since division by zero is not defined. 180 16 0.1 100 or 0.1 16 80 0 16 56 11 16 1 0. 0. 16 1 10 or 0 units 0.. 5 3 z z 61, 800 18 600 00 7. a. P 18 600 00 5( z ) 3( z ) ( z )( z ) z z 8z 3 z 6z 10 0 (z 5)( z 1) 0 z + 5 = 0 or z 1 = 0 Solution: z 5, 1 3. ( 8) 3( 8) 0 ( 8) ( 8) 1 0 ( 8) 0 or ( 8) 1 0 Solution: = 10, 9. 5. (s ) 5(s ) 0 (s ) 8(s ) 3 0 (s ) 8 0 or (s ) 3 0 Solution: s = 10, 1 P 90 00 100 90 00 18 600 6, 00 0 Factoring appears difficult, so let us apply the quadratic formula. 600 600 600 (18)(6, 00) 36 36, 81, 600 36 600 600 10 or 35.56 36 So, a profit of $61,800 is earned for 10 units or for 35.56 units. b. Yes. Maimum profit occurs at verte as seen using the graphing calculator. 8. a. P 50 300 0.01 When P = 50 we have 50 50 300 0.01 or 0.01 50 550 0. 50 500 0.0 50 9.78 11 or 989 units 0.0 0 90 100 0 ( 0)( 70) A profit of $100 is earned at = 0 units or = 70 units of production. 9. b. Yes. Try P(000) and P(5000). P(000) $50. S 100 96t 16t 100 100 96t 16t 0 96t 16t 16t 6 t The ball is 100 feet high 6 seconds later.
50. D(t ) 16t 10t 350 55. p 0.17t.61t 5.6 0 16t 10t 350 55 0.17t.61t 5.6 16t 10t 350 0 0.17t.61t.36 0 8t 5t 175 0 8t 35 t 5 0 The answer t 5 is the only one that makes sense in this case, so the ball hits the ground at 5 seconds. Using the quadratic formula or a graphing utility gives the positive value t 16.. In 016 the percent of high school seniors who will have tried marijuana is predicted by the function to reach 55%. 51. p 5 0.01s 56. a. y 0.0013 10 a. 0 5 0.01s 0.0013 10 0 9.873 or 779.10 5 0.1s 5 0.1s p 0 if 5 0.1s 0 or s 50. b. y 10 81 3 b. s 0. p 0 means there is no 5. S 100 particulate pollution. 81 3 10 0 115.01 or 7.01 53. a. 0 100 A dosage of 0 or 100 ml gives S 0. b. Dosage is effective if 0 100. t 0.0010.73 15.17 607.738 8.99 0.0010.73 15.17 607.738 8990 0.73 15.17 607.738 0 0.73 15.17 838.6 15.17 t (15.17) (0.73)(838.6) (0.73) t 96.996 or t 118.058 The positive answer is the one that makes sense here, 97.0 mph. 57. P Given that the distance is not negative, the first projectile travels further (approimately 779 feet versus the second projectile s approimately 115 feet). C C 100 We know that the selling price is $1 and that the selling price equals the profit plus the cost C to the store. C 1 C 100 100 C 100C 5. B 0.006t 0.033t 6.05 5 0.006t 0.033t 6.05 0.006t 0.033t 11.05 0 C 100C 100 0 C 180 or C 80 Using the quadratic formula or a graphing utility gives the positive
value t 5.6. The fund is projected to be $5 trillion in the red in the year 06. The cost C of the necklace to the store is not negative, so C = $80 is the amount the store paid for the necklace.
58. y 0.787 11.0 90 1000 0.787 11.0 90 0.787 11 710 0 Using the quadratic formula or a graphing c. 0 0.01 r r 0.01 r 0.1 utility gives the positive value 38. Spending is projected to reach $1000 billion in the year 08. 61. K 16v In this case the corpuscle is at the wall of the artery. 59. E 7.9 33. 190 5000 7.9 33. 190 7.9 33. 810 0 Using the quadratic formula or a graphing utility gives the positive value 16.8. In each case below only positive values of K are reported. a. K 16(0) 3 K 18 b. K 16(60) 96 60. The model predict these ependitures will reach $5 trillion in 0. v k R r v 0.01 r In each case below only nonnegative values of r are reported. K 31 c. Speed triples, but K changes only by a factor of 1.7. 6. Given that s 16t and s 1090t, t 1 t 3.9 t 3.9 t 1 16t 1 1090t 10903.9 t 1 1 a. 0.0 0.01 r 51 1090t 1 0.01 0.01 r 16t 1 1090t 1 51 0 r 0 r 0 b. 0.015 0.01 r 0.0075 0.01 r r 0.005 r 0.05 Using the quadratic formula or a graphing utility gives the positive value t 1 3.70. s 16t 16 3.70 19 The depth of the fissure is about 19 ft. 1
Eercises.
1. y 1 6. f 3 b a. b 1 1 a. 1 a 1 a (1 / ) y 1 (1) (1) 1 f 1 1 1 3 Verte is at 1, 1. b. a > 0, so verte is a minimum. Verte is at 1,. b. a > 0, so verte is a minimum. c. 1 d. c. 1 d. 1 1 7. y Verte is a maimum point since a < 0.. y b 1 V: b a (1 / ) a. 1 a When = 1, y = 1. The verte is (1, 1). b. a > 0, so verte is a minimum. c. 1 d. 1 3. y 8 y 1 () 1 Zeros: 1 0 1 1 0 a. b 1 0, y-intercept = 0 a (1) y 8 (1) (1) 9 Verte is at (1, 9). b. a < 0, so verte is a maimum. c. 1 d. 9. y 6 a. b 1 a When = 1, y = 8. The verte is ( 1, 8). b. a < 0, so verte is a maimum. c. 1 8. y 18 d. 8 Verte is a maimum since a < 0. 5. f () 6 V: b 18 9 a
a. b 6 3. 9 9 81 16 81 a y 18 f 3 6 3 3 9 Verte is at 3, 9. Zeros: 0 18 0 ( 9) b. a 0, so verte is a maimum. c. 3 d. 9 0 or 9 0 0 y-intercept = 0 9
y y 50 10 y = 6 + 9 5 5-5 5 10 6 y = + 18-5 9. y Verte is a minimum point since a > 0. 11. y 1 3 Verte is a minimum point since a > 0. V: b b 1 a (1) V: 1 y () () 0 a (1 / ) Zeros: ( )( ) 0 y 1 (1) (1) 3 7 y-intercept = Zeros: 1 3 0 6 0 y y 7 1 7 y-intercept = 3 - - 10. y 6 9 Verte is a minimum since a > 0. V: b 6 3 a y 3 6(3) 9 0 Zeros: 0 6 9 0 ( 3)( 3) 3 0 3 y-intercept = 9
1. y 5 y 5 1 y 1 5 Verte is a maimum since a < 0. 1 V: b 1 y a 1 1 1 1 1 5 1 8 Zeros: Using the quadratic formula, 1. y ( 10) 1 a. Graph is shifted 10 units to the right and 1 units up. b. 1 1. y-intercept 5 y y = 1 1 + 5 15. y ( ) a. Graph is shifted units to the left and units down. b. - - - - 13. y ( 3) 1 a. Graph is shifted 3 units to the right and 1 unit up. b. 16. y ( 1) 8 a. Graph is shifted 1 units to the left and 8 units down. b. y 0 10-30 -0 10-10
17. y 1 15 V: b (1) 1 a (1 / ) y 1 (1) 1 15 8 Zeros: 15 ( 5)( 3) 0 5, 3 0. y 5 18. From the graph, the verte is (1, ). There are no real zeros. Algebraic check: From the graph, the verte is approimately, 3.5. The zeros are approimately 8 and. Algebraic check: b V: -coordinate: 1 a y-coordinate: 1 (1) 5 The discriminant is negative, so no real zeros. 1. f () y 5 Average Rate of Change f (1) f (1) 1 (1) 6 10 5 V: -coordinate: b. f () y 8 3 0.5 a y-coordinate: 0.1( 8 3) = 3.6 So, actual verte is (, 3.6) Zeros: 0 3 ( 8)( ) 8, 19. y 1 3 1 Average Rate of Change 3. y 63 0. 0.01 0. V: 10 0.0 y 63 1 6 f () f () 8 16 1 6 V: b 3 6 Zeros: 0 6300 ( 90)( 70) 0 a 1 90, 70 y 1 (6) 3(6) 1 3 Zeros: 1 8 0 b ac 1 19 0
There are no zeros.
. y 0. 16 10 7. f () 8 16 16 V: -coordinate: b 16 0 a (0.) a. b 16 1 and f 1 y-coordinate: 0.(0) 16(0) 10 180 Zeros: 0 0.( 80 700) 0.( 70)( 10) 70, 10 Graphing range: -min = 100 y-min = 00 -ma = 0 y-ma = 50 a 16 b. Graphical approimation gives 0.73,.73 8. f () 3 18 16 5. y 0.0001 0.01 V: 0 0 (0.0001) y 0 0.01 0.01 Zeros: 0.0001 0.01 0.010.01 1 0 0.01(0.1 1)(0.1 1) 0 10, 10 a. b 18 3 and f 3 11. a 6 b. Graphical approimation gives 1.085,.915 6. y 0.01 0.001 0.001(10 ) Zeros: = 0, = 10 9. f () 3 8 a. The TRACE gives as a solution. b. is a factor. c. 3 8 3 d. 3 0 0 or 3 0 V: -coordinate: b 0.01 5 Solution is, / 3. a 0.00 y-coordinate: 0.01(5) 0.001(5) = 0.05 Graphing range: -min = 5 y-min = 0.1 -ma = 15 y-ma = 0.1 30. f () 5 7 a. The TRACE gives 1 as a solution. b. 1 is the factor. c. 5 7 15 7
d. 1 5 7 0 1 0 or 5 7 0 Solution is 1, 7 / 5.
31. P 0.1 16 100 36. A 50 The verte coordinates are the answers to the A questions. 700 a. a = 0.1, b = 16 600 Maimum: 65 b 16 80 a 0. 00 300 Profit is maimized at a production l vel of 00 80 units. y = (50 ) 100 b. P(80) 0.1(80) 16(80) 100 $50 is the maimum profit. 500 5 50 -coordinate of the verte = b 50 5 3. P 80 0. 00 a a. -coordinate of verte = b 80 A length of 5 feet and width of 5 feet gives a 100 a 0.8 maimum area of 65 square feet. When = 100, P 80(100) 0.(100) 00 8000 000 00 $3800 33. Y 800 Opens down so maimum Y is at verte. V: 800 00 Maimum yield occurs at = 00 trees. 37. R 70 90 Maimum rate occurs at verte. 70 3 V: (lumens) (90) is the intensity for maimum rate. 3. y k y 5 k = k = k = 1 0 15 10 5 1 3 5 35. S 1000 Maimum sensitivity occurs at verte. V: 1000 500 The dosage for maimum sensitivity is 500. 38. s 11t 16t t-coordinate of the verte = b 11 3.5 seconds a 3 At t = 3.5, s 11(3.5) 16(3.5) 196 feet 39. a. y 0.0013 10 V: 1 38.6; 0.006 y 0.0013(38.6) 38.6 10 0.31
0. 1 b. y 10 81 3 V: 3 5 ; 81. a. Rent would be $600 $00 $800. Price No. of skaters Total Revenue 1 y (5) (5) 10 6 b. The revenue increases. 81 3 c. R() 1 0.550 5 where is the number Projectile a. goes 0.31 6 = 156.31 feet higher. of each additional 5 skaters. d. R( ).5 35 600. Maimum v 1960(h 10) revenue is at 35 7, or 85 skaters. v 5 h 10 1 11 10 50 60 70 $600 $660 $700 1960 0 15 h 1 v 10 1960 h h = 1 v 10 1960 5. a. A quadratic function or parabola. b. a 0 because the graph opens downward. c. The verte occurs after 00 (or when 0 ), so b 0. Hence with a 0 we a must have b 0. The value c f (0) or 10 5 0 5 10 100 00 v 15 Verte: b a, 0 6. the y-value during 00 which is positive. y a b c Zeros: (0, 0) and (0, 0) 1. a. From b to c. The average rate of change is the same as the slope of the segment. The (0, 0): 0 a(0) b(0) c 0 c segment from b to c is steeper. So, y a b b. Needs to satisfy d b to make the segment from a to d have a greater slope.. a. From b to c. The average rate of change is the same as the slope of the segment. The segment from b to c has a negative slope. b. Needs to satisfy d b to make the segment from a to d have a greater slope. (0, 0): 0 1600a 0b b 0a So, y a 0a 3. a. No. of Apts Rent Total Revenue 50 $600 $30,000 00a 0 9 $60 $30,380 1 a and b 10 8 $60 $30,70 b. Revenue increases $70 The equation is y 0a. So, -coordinate of the verte = a 0 When = 0, y = 0 0 a(0) 0(a)(0) 0 00a 800a 1
c. R 50 600 0 10 d. R 0 00 30, 000 R is maimized at 00 10. 0
7. y 0.61 116. 706 b. Using the equation, we identify the For 010, 10 gives y 8303. For 015, 15 gives y 10, 97.5. For 00, 0 gives y 13, 3. Average rate of change from 010 to 015: 10, 97.5 8303 398.85 15 10 Average rate of change from 015 to 00: 13, 3 10, 97.5 60.95 0 15 To the nearest dollar, the projected average rate of change of U.S. per capita health care costs from 010 to 015 will be $399/year, and from 015 to 00 it will be $605/year. 9. maimum point by computing t b 0.38 5.78 a 0.0036 p 5.78 8.65 The maimum point is (5.78, 8.65). c. According to this model, the maimum percentage of women in the workforce occurs in the year 1970 53 03. 8. a. 50. The graphing calculator gives a minimum point at 9.3, 36.6. Eercises.3 1. C ( ) 0 000 R 130 0 000 130 90 000 0 ( 0)( 50) 0 = 0 or = 50 Break-even values are at = 0 and 50 units. 3. C ( ) 15, 000 35 0.1 R( ) 385 0.9 15, 000 35 0.1 385 0.9 350 15, 000 0 ( 300)( 50) 0 = 300 or = 50. At the break-even point, R() = C(). 3600 5 1 175 1. At the break-even points, R() = C(). 1600 1600 1500 0 100 1600 150 3600 0 ( 10)( 30) 0 = 10 or = 30 units 0 ( 0)( 80) = 0 or = 80 units
5. P( ) 11.5 0.1 150 At the break-even points, P() = 0. 0 11.5 0.1 150 0.1 11.5 150 0 ( 15)( 100) 0 Since production < 75 units, = 15. 6. P( ) 1100 10 11. 1. P( ) 110 1000 Maimum profit is at the verte or when 110 55. P 55 $05. At the break-even points, P() = 0. a P( ) 88 100 The -coordinate giving the maimum profit is b 88. 0 1100 10 10 1100 0 ( 110)( 10) 0 Since production < 100 units, = 10. 13. a. P() 88() () 100 $736 7. R( ) 385 0.9 a = 0.9, b = 385 Maimum revenue is at the verte. V: 385 13.89 or 1 total units 1.8 R(1) 385(1) 0.9(1) $1,173.60 8. R( ) 1600 Maimum occurs at the verte. b. 00, 9000 is the maimum c. positive d. negative e. closer to 0 -coordinate = 1600 800 R(800) 1600(800) (800) $60, 000 9. R( ) (175 0.50 ) 175 0.5 a = 0.50, b = 175 Revenue is a maimum at 175 175. 1 Price that will maimize revenue is p = 175 87.50 = $87.50. 10. D: p 1600 1600 p Revenue: R p p(1600 p) R 1600 p p Ma. revenue for p 1600 $800.
1. a. 17. a. C( ) 8, 000 5 15. b. 15,115 is the maimum c. positive d. negative e. closer to 0 R( ) 385 0.9 C ( ) 15, 000 35 0.1 a. P( ) 385 0.9 (15, 000 35 0.1 ) 350 15, 000 5 8, 000 R( ) 150 3 150 3 5 5 (The key is per unit. ) R( ) C ( ) 150 3 8, 000 5 5 108 8, 000 0 ( 1000)( 8) 0 Break-even values are at = 8 and = 1000. b. Maimum revenue occurs at 150 10 (rounded). 6 At the verte we have So, P(175) = $15,65. 350 175. b. No. More units are required to maimize revenue. c. The break-even values and zeros of P() are c. the same. 5 R 10 $651, 01.60 is the maimum revenue. P( ) 150 3 8, 000 5 5 108 8, 000 16. a. P( ) R( ) C( ) 1600 (1600 1500) 100 1600 -coordinate of ma is 100 50 P(50) 100(50) (50) 1600 $900 b. No. More units are required to maimize revenue. Maimum profit is at 108 51. P 51 $36,196 is the maimum profit. d. Price that will maimize profit is 3 p 150 (51) $91.60. 5 c. 0 100 1600 100 1600 0 ( 80)( 0) 0 The -coordinates are the same.
18. a. C( ) 300 3 160 d. The model fits the data quite well. 300 3 160 R( ) 1500 1 1500 1 At break-even points C() = R(). 300 3 160 1500 1 0 300 0 ( 30)( 10) 0 = 30 or = 10 b. Maimum revenue: b 1500 -coordinate: 3000 0. R(t ) 0.031t 0.776t 0.179 a. Maimum occurs at the verte. The t- 0.776 coordinate of the verte is 1.5. 0.06 Maimum revenue occurred during 016. The maimum revenue predicted by the model is R(1.5) $5.035 million. b. The entry in the table for 016 is $.789 million, so the values are close. However, the 013 revenues were greater than this. c. a 1 R(3000) 1500(3000) 1 (3000) $, 50, 000 P( ) R( ) C ( ) c. 1500 1 300 3 160 0 300 Maimum profit: d. Although there are differences, the model appears to be a good quadratic fit for the data. 1. a. p t b. 011 c. 0.019t 0.8t 0.56 -coordinate: b 0 0 a P(0) 0(0) (0) 300 $100 d. 19. a. Selling price 1500 1. When = 0, p 1500 1 (0) $195 t 5.1, in 01; R $60.79 billion b. The data show a smaller revenue, R = $60.7 billion in 011. c. d. The model projects decreasing profits, and, ecept for 015, the data support this. e. Management would be interested in increasing revenues or reducing costs (or both) to improve profit.
. a. Supply: p q 8q 16 (see below). a. Supply: p q 8q (see below) 300 p S Demand: p p 198 q 1 q (see below) 00 100 E D 300 00 E S 8 1 16 0 q 100 D Demand: p 16 q (see below) b. See E on the graph. c. Supply = Demand q 8q 16 16 q q 10q 00 0 (q 10)(q 0) 0 q = 10 (only positive value) p = 16 (10) = 196 q = 10, p = $196 8 1 16 0 b. See E on the graph. c. Supply = Demand q 8q 198 q 1 q 5q 8q 70 0 (5q 88)(q 8) 0 q = 8 (only positive value) q 3. a. Supply: p 1 q 10 (see below) Demand: b. See E on graph. c. 1 q 10 86 6q 3q p 86 6q 3q (see below) 5. When q = 8, p (8) 8(8) p 150 So, E = (8, 150). p q 8q 16 p 3q 6q 36 q 8q 16 3q 6q 36 q q 0 0 q q 10 0 (q 1)(q 10) 0 q 10 p 10 8(10) 16 196 E: (10, 196) q 0 3 q 1q 6. S: p q 8q 0 0 13q q 30 0 (q )(13q 76) q = must be positive. p 1 () 10 1 D: 100 q q p q 8q 0 100 q q E: (, 1)
q 1q 80 0 (q 10)(q ) 0 q = (only positive value) When q =, p 8() 0 $68 Equilibrium point: (, 68)
7. p q 1600 300 p q 0 (300 q) q 1600 6q 1300 q 16 3 p 1300 1600 or p 6 9. p q 10 or q p 10 q p 10 100 q 100 p 10 100 p 10 p 10 p 10 p 10 100 p 30 p 100 100 733.33 or p 7.08 p 30 p 000 0 E: 16, 7.08 p 15 p 1000 0 3 8. S: p q = or q = p D: ( p )q 100 or q 100 p p 100 p p 3 p 8 100 p 3 p 18 0 ( p 39)( p 8) 0 p = 8 (only positive value) When p = $8, q = (8) = 70 Equilibrium point: (70, 8) p 0 p 5 0 p 0 or p 5 (only the positive answer makes sense here) q 0 10 30 E : 30, 0 30. S: p q + 6 = 0 or q = p + 6 D: (p + q)(q + 10) = 3696 Substitute p + 6 for q in D and solve for p. (3 p 6)( p 16) 3696 6 p 60 p 3600 0 3. S: p q 50 or p D: pq 100 0q or p q 50 100 0q q 50 100 0q q p 10 p 600 0 q ( p 30)( p 0) 0 p = 0 (only positive value) When p = 0, q = (0) + 6 = 6. Equilibrium point: (6, 0) p 50 31. p q 10 = 0 (p + 10)(q + 30) = 700 So, ( p 10)( p 10 30) 700 p 0 p 100 3600 p 0 p 3500 0 ( p 70)( p 50) 0
q 50q 00 0q q 10q 00 0 (q 0)(q 10) 0 q = (50) 10 = 90 E: (q, p) = (90, 50) q = 10 (only positive value) When q = 10, p = 30 Equilibrium point: (10, 30)
33. p 1 q 5 1 q 7 So, 1 q 7 10 (q 30) 700 (q 7)(q 30) 1, 00 q 10q 1,180 0 (q 17)(q 70) 0 p 1 (70) 7 6 E: (70, 6) 3. S: p q 50 1.50 100 0q q D: p q 50 1.50 100 0q When q = 5, Equilibrium point: (5, 0) q q 75q 00 0q q 35q 00 0 (q 0)(q 5) 0 q = 5 (only positive value) Eercises. 1. b. g 3. f. h 5. j 16. 17. 18. 19. y 3 1 : c y 16 ( )( ) : b y 3 ( )( 1) : h y 7 ( 7) : d 6. e 0. y 7 7 : a 7. k 1. y 3 : g 1 8. d 9. a 10. i 11. c 1. l 13. a. cubic b. quartic 1. a. quartic b. cubic 15.. 3. y 1 3 : f 5 5 5 5 y y ( 1)( 3)( 1)
y 3 ( 1)( 1) : e 5 5
. 5 y y 3 8 19 1 8. y 5 ; ; y ; ; 5 5 5 5. 6. 5 55 1 9. F ( ) y 5 1 1 1 9 8 y 1 a. F 3 1 3 3 100 1 99 5 5 b. F (10) 10 10 1 5 F ( ) 1 1 c. F 1 9 8 3 1 3 3 y 5 y 1 d. F (10) 100 1 99 10 10 5 5 e. F (0.001) 0.000001 1 0.001 0.999999 0.001 5 7. 30. H () 1 999.999 f. F (0) is not defined division by zero. 5 y 3 ; 3 ; 1 1 5 y y 1; 1; 11 5 31. a. H( 1) = b. H(1) = 0 c. H(0) = 1 d. No f () 3 / 5 b. c. a. d. f (16) ( 16) 3 6 f (1) ( 1) 3 1 f (100) ( 100) 3 1000 f (0.09) ( 0.09) 3 0.07
3. k ( ) if 0 36. f ( ) 3 3 if 0 a. a. k 0.1 0.1. b. k 0.1 0.1 3.9 3.9 8 y c. k 3.9 3.9 0.1 0.1 d. k.1 is undefined if 0 - - 33. k ( ) if 0 1-8 1 if 1 b. polynomial a. k 5 = since < 0. b. k 0 0 c. d. k 1 11 0 k 0.001 since < 0. 0.5 if 0 37. c. no asymptotes d. turning point at = 3 y 1 a. 3. g ( ) if 0 0 if a. g 0.5 b. g 1 1 3 c. g 7 0 d. g 3.9 3.9 0.1 35. y 1.6 0.1 a. b. rational c. vertical: = 1 horizontal: y = d. no turning points b. polynomial c. no asymptotes d. turning points at = 0 and approimately =.8 and =.8
38. f ( ) 3 0. f ( ) 1 if 1 a. y a. if 1 y 8 6-8 -6 - - 6 - - - b. rational c. vertical: = horizontal: = 1 d. no turning points 1. - b. piecewise c. no asymptotes d. no turning point (there is a jump at = 1). V V () (108 ) 39. ) a. f ( 5 if 0 if 0 a. V(10) = 100(68) = 6800 cubic inches V(0) = 00(8) = 11,00 cubic inches b. 108 0 108 0 7. FC 000 3000 000 1000 FC = 000 b. piecewise c. no asymptotes 0 0 60 80 100 d. turning point at = 0. 3. f () 15.875 1.18 a. upward b. c. Intersecting the graphs of y 15.875 1.18 and y 1150 gives 37.7. Global
spending is epected to reach $1,150,000,000,000 ($1150 billion) in 1980 + 38 = 018.
. C 0.11W 1.5 a. W 1.5 b. C is close to W. The graph is turning up. 10 8 6 5 10 15 0 c. C 10 0.1110 1.5 3.81 grams 5. C ( p) 7300 p 100 p a. 0 p 100 W b. C (5) 7300 5 $597.73 100 5 c. C (90) 7300 90 $65, 700 100 90 d. C (99) 7300 99 $7, 700 100 99 e. C(99.6) 7300(99.6) $1, 817, 700 100 99.6 f. To remove p% of the pollution would cost C(p). Note how cost increases as p (the percent of pollution removed) increases. 6. C 50, 000 105 a. C 3000 50, 000 105 3000 $11.67 3000 b. C 500 00 300 00 100 1000 000 3000 000 c. Yes. C ( ) 50, 000 105
7. A = A() = (50 ) a. A() 8 96 square feet A(30) 30 0 600 square feet b. 0 < < 50 in order to have a rectangle. 58 if 0 0 8. f ( ) 58 0.( 0) if 0 a. f 0.3 $58 b. f 30 58 0. 30 0 $6 c. f 0 58 0. 0 0 $66 d. 100 f () 58 if 0 0 f ( ) 58 0.( 0) if 0 50 9. 10 0 30 0 50 5.59 93.5 633 for 0 55 y 6.56 519 0, 900 for 55 90 a. b. y 50 5.59 50 93.5 50 633 $9933 billion ($9.933 trillion) c. y 75 6.56 75 519 75 0, 900 $18, 875 billion ($18.875 trillion) 50. a. C(5) = 7.5 + 0.1079(5) = $8.06 b. C(6) = 19. + 0.1079(6) = $19.87 c. C 3000 131.35 0.0313000 $7.65 9 if 0 1 51. a. 70 if 1 P( ) 91 if 3 11 if 3 b. P 1. 70; it costs 70 cents to mail a 1.-oz letter. c. Domain: 0 ; Range: 9, 70, 91,11 d. The postage for a -ounce letter is 70 cents; for a 01-ounce letter, it is 91 cents.
0.10 if 0 16, 750 5. a. b. c. d. T ( ) 0.15( 16, 750) 1, 675 if 16, 750 68, 000 0.5( 68, 000) 9, 36.50 if 68,000 137, 300 T 70, 000 0.570, 000 68, 000 9, 36.50 $9, 86.50 T 50, 000 0.1550, 000 16, 750 1, 675 $6, 66.50 T 68, 000 0.1568, 000 16, 750 1, 675 $9, 36.50 T 68, 001 0.568, 001 68, 000 9, 36.50 $9, 36.75 Jack s ta went up $0.5 for the etra dollar earned. He is only charged 5% on the money he earns above $68,000. 53. p a. 00 0.1 55. C ( ) 30( 1) 3000 10 a. 5. b. No 1 y, positive integers 1 y 1.0 0.8 0.6 0. 0. b. A turning point indicates a minimum or maimum cost. c. This is the fied cost of production. 0 6 8 10 Eercises.5 1. Linear: The points are in a straight line. 6. Cubic. 3. Power Quadratic: The points appear to fit a parabola.. Linear 5. Quartic: The graph crosses the -ais four times. Also there are three bends. 7. Quadratic: There is one bend. A parabola is the best fit. 8. Cubic
9. y = 3 is the best fit. 15. y 1/ is the best fit. 10. y = 1.5 is the best fit. 16. y 3 3 / is the best fit. 11. y 1.5 is the best fit. 17. a. 1. y 3 is the best fit. 18. a. b. linear c. y = 5 3 13. y 3 3 is the best fit. 19. a. b. linear c. y = + 1. y 3 3 is the best fit b. quadratic c. y 0.0959 0.656 1.758.
0. a. 5. a. y 15.0 35,860 b. y 7 15.0 7 35,860 0, 018 The projected population of females under age 18 in 037 is 0,018,000. c. 5, 000 15.0 35860 59.35 b. power This population will reach 5,000,000 in 010 + 60 = 070 according to this model. c. 1. a. y 3 1/ 6. a. b. c. y 18.96 31.5 y 1 18.96 1 31.5 586.9 million metric tons m 18.96; each year since 010, carbon dioide emissions in the U.S. are epected to change by 18.96 million metric tons. 7. a. A linear function is best; y 37.6 9591 b. quadratic b. y 17 37.617 9591 $15,160 billion c.. a. y 5 1 c. m 37.6 means the U.S. disposable income is increasing at the rate of about $37.6 billion per year. 8. a. b. y 0.65 1.0 y 18 0.6518 1.0 0.% 3. a. b. quadratic c. 3 3 1 c. 5 0.65 1.0 8 This model predicts that the percent of U.S. adults with diabetes will each 5% in 000 + 8 = 08. 9. a. y 0.005 0.6 15 b 0.6 b. a 59.6 0.005 c. No, it is unreasonable to feel warmer for winds greater than 60 mph. 30. a. y 0.07.6 1.1 b. cubic b. A maimum occurs at approimately c.. a. y 3 5 1 8.0, 8.9. The model predicts that in the year 000 + 8 = 08, developing economies reach their maimum share, 8.9%, of the GDP.
b. cubic c. y 3 3
31. a. c. b. c. d. y 106 870 8, 500 y 1.70 3 7.9 1970 570 d. The fits look to be equally close. 33. a. b. c. y 0.0157.01 y 0.00105 0.367 1.9 The cubic model fits better. 3. a. b. y 0.00336 0.017.7 y 0.0000537 3 0.00738 0.0609.63 d. The quadratic model is a slightly better fit.
3. a. b. c. y 1.03 88.1 y 0.0005 3 0.0178 0.0756 87.8 d. Using the coefficient values reported by the calculator, the model estimates the median income to be $56,50 at age 57. 36. a. It appears that both quadratic and power functions would make good models for these data. b. power: y 0.0315.7 quadratic: y 1.76 71.0 679 c. power: y 70 $3661 billion d. The cubic model indicates that the percent quadratic: y 70 $335 billion 35. a. of energy use may increase after 035. d. The quadratic model more accurately approimates the data point for 00. y 75 $557 billion; $557 billion is the national health-care ependiture predicted by the model for 05. 37. a. b. y 0.051.73 A cubic model looks best because of the two bends. b. c. y 0.86 3 18 6610 6, 600 c. y 30 $56 billion
38. a. b. Possible models are linear: y.53 16. quadratic: y 0.00100.633 160.7 cubic: y 0.0000756 3 0.01030.191 163.5 c. linear: y 90 390.08 quadratic: y 90 389.36 cubic: y 90 389.77 The linear model most accurately approimates the data point for the year 00. d. Replacing y with 5 in the linear model gives 103.8. The U.S. population is predicted to reach 5 million in 1950 + 10 = 05.
Chapter Review Eercises 1. 3 10 5 3 5 0 (3 5) 0 0 or. 3 0 ( 3) 0 0 or 3 5 3 3. 5 6 0 ( 3)( ) 0 = 3 or = 9. 7.07 0.0 10. 11. 0.0.07 7 0 a = 0.0, b =.07, c = 7.07.89 0.56.07 1.93 0.0 0.0 100 or 3.5 6.3 117 0.5 0 a = 0.5, b = 6.3, c = 117 6.3 13.69 (3) 6.3 3.7 90 or.6 z 5 0 1 1. 1110 0 a =, b = 10, c = 11 10 100 88 5 7 5. ( 1)( 3) 8 3 8 5 0 b ac 0 No real solution 1. z 5 0 The sum of squares cannot be factored. There are no real solutions. f (z) z 6z 7 6. 3 3 3 3 7. 0 3 0 15 35 3 0 0 (7 5)(5 ) 0 5 or 7 5 8. 8 8 1 8 16 8 1 0 a = 16, b = 8, c = 1 8 6 6 1 3 From the graph, the zeros are 9 and 3. Algebraic solution: z( z 6) 7 z 6z 7 0 ( z 9)( z 3) 0 z = 9 or z = 3
13. 3 18 8 0 3( 6 16) 0 3( 8)( ) 0 =, = 8 18. 19. 1.03.0 1.015 0 a = 1.03, b =.0, c = 1.015.0.080.1818.0.87.06.06.38 or 0.1 y 1 a > 0, thus verte is a minimum. 1. f () 3 6 9 3 6 9 0 3 3 0 3 3 1 0 V: 1 y 1 () () 3, 1 Zeros: 1 0 1 0 = 0, 15. a b 0 To apply the quadratic formula we have a = 1, b = a, and c = b. a a b 16. r ar c 0 To solve for r, use the quadratic formula with a =, b = a, and c c. r a 16a ( c) a 16a 3 c 0. y 1 a a 3 c V: -coordinate = 0 a a 3 c y-coordinate = (0, ) is a maimum point Zeros are. 17. 0.00 1.1 3.1 0
1.1 198.81 0.188 1.1 1.107 0.00 0.00 7051.6, 1.6, or 1.75 (using 1.107)
. y 5 V: -coordinate y-coordinate () 5 1 (, 1) is a minimum point. Zeros: Since the minimum point is above the - ais, there are no zeros. 1. y 6 a < 0, thus verte is a maimum. V: 1 1 (1) y 6 1 1 5 6 Zeros: 6 0 (3 )( ) 0 =, 3 3. y 6 9 a > 0, thus verte is a minimum. V: 3 (1) Zeros: y (3) 6(3) 9 0 6 9 0 ( 3)( 3) 0 3. y 1 9 V: -coordinate 1 3 8 y-coordinate 1 3 9 3 0
3, 0 is a maimum point. 7. y 5 Zeros: From the verte we have that only zero. 3 is the V: ( 1, ) There are no real zeros. 8. y 10 7 5. y 1 3 3 V: (0, 3) Verte: 7, 9 maimum Zeros: 7 10 0 ( 5)( ) 0 = 5 or = Graph using -min = 0 y-min = 5 -ma = 8 y-ma = 5 Zeros: 1 3 0 3 9 3 6. y 1 Verte: (0, ) minimum No zeros. The graph using -min = y-min = 0 -ma = y-ma = 6 9. y 0 0.1 Zeros: (0 0.1) = 0 = 0, 00 (This is an alternative method of getting the verte.) The -coordinate of the verte is halfway between the zeros. V: (100, 1000) is shown below.
30. y 50 1.5 0.01 Verte: (75, 6.5) minimum 37. a. f () Zeros: 0.01 1.5 50 0 0.01( 150 5000) 0 0.01( 50)( 100) 0 = 50 or =100 Graph using -min = 0 y-min = 10 -ma = 15 y-ma = 10 b. f ( ) 1 31. f (50) f (30) 500 100 00 0 50 30 0 0 3. f (50) f (10) 10 178 100 30 50 10 0 0 33. a. The verte is halfway between the zeros. So, the verte is 1, 1. b. The zeros are where the graph crosses the -ais. =,. c. The graph matches B. c. f () 1/ 3. From the graph, a. Verte is (0, 9) b. Zeros are 7. c. Matches with D. 35. a. The verte is halfway between the zeros. So, the verte is (7,.5). 38. b. Zeros are = 0, 1. c. The graph matches A. 36. From the graph, a. Verte is ( 1, 9). b. Zeros are = and =. c. Matches with C. if 0 f ( ) 1 if 0 a. f (0) (0 ) 0 b. f (0.0001) c. f (5) (5) 5 d. f (10) 39. f ( ) 1 10, 000 0.0001 1 0.1 10 if 1 3 if 1 a. f () b. f (0) 0
c. f (1) 1 d. f () 3 0. f ( ) if 1 3 if 1 y 3. y 3 9 Using -min =.7, -ma =.7, y-min = 15, y-ma = 15, the turning points are at = 1.73. - - - Note: Your turning points in 3. may vary depending on your scale. 1. a. - f () ( ) 1. y There is a vertical asymptote =. There is a horizontal asymptote y = 0. b. f () ( 1) 3 5. 1 y 1 3 1 3 Vertical asymptote is = 3. Horizontal asymptote is y =.. y 3 3 9 Using -min = 10, -ma = 10, y-min = 10, y-ma = 35, the turning points are at = 3 and 1. 6. a. b. y =.1786 + 159.8571 is a good fit to the data.
c. 7. a. y 0.0818 0.13 153.3095 is a slightly better fit. 5. b. y =.113 + 3.3913 is a good fit to the data. 53. c. y.766 0.59 is a slightly better fit. 8. S 96 3t 16t a. 16(6 t t ) 0 t t 1.65 or t 3.65 b. t 0 Use t = 3.65 c. After 3.65 seconds 9. P( ) 0.10 8 1600 (0.10 80)( 0) 0 Break-even at = 0, 800 5. a. 50. E(t) 0.005t 0.080t 1 a. The employment is a maimum at t b 0.080 7.69 a 0.005 f 7.69 1.3; the maimum employment in manufacturing in the U.S. is predicted to be 1.3 million in 010 + 8 = 018. b. 11.5 0.005t 0.080t 1 The quadratic formula gives t.8 or t 0.. The employment in manufacturing in the U.S. will be 11.5 million in 010 + 1 = 031. 51. A 3 300 a. V: 300 00 ft 3 b. 0.1q 1 85 0.q 0.1q 0.q 0.q 8 0 0.(q q 0) 0 0.(q 0)(q 1) 0 q = 0 (only positive value) p 0.1(0) 1 1 b. A 3 (00) 300(00) 30, 000 sq ft
55. p q 300 p q 10 q 300 q 10 q q 110 0 (q 11)(q 10) 0 q 10 59. R() 100 V: 100 50 R(50) 100(50) 50 $500 ma revenue P( ) (100 ) (900 5) p = 10 + 10 = 00 75 900 So, E: (10, 00). V: 75 37.5 56. D: p 5q 00 p 00 5q P(37.5) $506.5 ma profit S: 0 p 3q 0 Substitute 00 5q for p in the second 60. P() 1.3 0.01 30 equation and solve for q. 0 (00 5q) 3q 0 160 8q q 0 1.3 -coordinate of the verte 65 0.0 P(65) 1.3(65) 0.01(65) 30 1.5 ma Break-even points: p 00 5(0) 0 1.3 0.01 30 p 100 or p 10 0 0.01( 130 3000) 57. R( ) 100 0. C( ) 1760 8 0.6 100 0. 1760 8 0.6 9 1760 0 0 0.01( 30)( 100) 30 or 100 9 1 6 89 6.87, 7.13 ( 1 16 89) 58. C ( ) 900 5 R( ) 100 61. P( ) (50 0. ) (360 10 0. ) 900 5 100 0. 0 360 75 900 0 ( 60)( 15) 0 = 60 or = 15 R(60) = 00; R(15) = 175 (60, 00) and (15, 175) 0 V: 50 units for maimum profit. 0.8 P(50) 0.(50) 0(50) 360 $60 maimum profit. 6. a. b.
C( ) 15, 000 (10 0.0 ) 15, 000 10 0.0 R( ) (300 0.06) 300 0.06 15, 000 10 0.0 300 0.06 0.10 160 15, 000 0 0.1( 1600 150, 000) 0 0.1( 100)( 1500) 0 = 100 or = 1500
c. Maimum revenue: 300 -coordinate: 0.1 d. P( ) R( ) C( ) 500 0.10 160 15, 000 160 -coordinate of ma = 800 0.0 67. a. Linear, quadratic, cubic, and power functions are each reasonable. b. y 3.779 0.55 c. 63. e. P(500) = $0,000 loss P(800) = $9,000 profit D t.95t 0.95 a. power function b. D 0 1.8% d. f (5) 3.779(5) 0.55 55 mph c..; in 05 about.% of U.S. adults are epected to have diabetes. e. Use the TRACE KEY. It will take 9.9 seconds. 6. a. 68. a. b. y 0 (6 ) Domain: 0 6 65. C ( p) 800 p 100 p a. rational function b. Domain: 0 p 100 c. C(0) = 0 means that there is no cost if no pollution is removed. d. C(99) 800(99) $75, 00 100 99.557 0 100 b. A quadratic model could be used. 66. C( ) 55.70.0( 100) 100 1000 a 7.70 180 0, 870; 091.7 1.689( 1000) a. C(1) =.557(1) = $30.68 1000 A 0.079 0.9815.5 c. 00 data: $63,676 b. C(85) = 55.70 +.0(85 100) = $173.70 a 10 $63,66 closer; A10 61.56 ($61,56) 050 data: 0.5 ($0,500) a 0 $189, 9; A0 00.13 ($00,13 closer) d. a 150, 000 when 3.5, in 03; A 150, 000 when 31.9, in 0
69. a. b. c. O 0.73 6.97 S 0.6.57 F 0.73 6.97. 0.6.57 This is called a rational function and measures the fraction of obese adults who are severely obese. 70. a. d. horizontal asymptote: y 0.73 0.355. 0.6 This means that if this model remains valid far into the future, then the long-term projection is that about 0.355, or 35.5%, of obese adults will be severely obese. A quadratic function could be used to model each set of data. W 0.00903 1.8 1 O 0.0065 1.0 0.0 b. At 91.1, W O 166. In 1970 + 9 = 16, these population segments are predicted to be equal (at about 166. million each). Chapter Test 1. a. f () b. g()
c. h() = 1 b. f () ( ) 3 1 d. k () 5. f () 3 ( ). a. and b. are the cubic choices. f() < 0 if 0. Answer: b 1 8 if 0 6. f ( ) if 0 6 if. figure b is the graph for b > 1. figure a is the graph for 0 < b < 1. 3. f () a b c and a < 0 is a parabola opening downward. a. f(16) = 6 16 = 10 b. f () 8() 1 16 1 c. f(13) = 6 13 = 7 7. g ( ) if 1 if 1. a. f () ( 1) 1 y 1 1 10 8 6-8 -6 - - - 8. f () 1 (7 )(3 ) Verte: b () a (1) Point: (, 5) Zeros: f() = 0 at = 7 or 3.
f (0) f (10) 30 90 160 1. 0 10 30 30 15. a. quartic b. cubic 16. a. f() = 0.3577 + 19.97 9. 3 7 10. 3 7 0 (3 1)( ) 0 3 1 = 0 or = 0 1, 3 6 9 0 6 36 7 6 6 3 3 3 3 17. S: b. f(0) = 5.6 c. f() = 0 if 19.97 55.7 0.3577 D: p 1 q 30 6 p 30, 000 0 q 1 q 30 30, 000 0 6q 11. 1 1 1 3 3 3 6 3 3 6 0 (3 ) 0 18. 6 q q 180q 180, 000 10q q 300q 180, 000 0 (q 600)(q 300) 0 E q : q 300 E p : p 50 30 80 R( ) 85 0.9 is the only solution. C( ) 15, 000 35 0.1 1. 13. 3 3( ) g ( ) Vertical asymptote at =. g() = 0 Answer: c f ( ) 8 10 a. P( ) 85 0.9 (15, 000 35 0.1 ) 50 15, 000 (100 )( 150) b. Maimum profit is at verte. 50 15 (1) Maimum profit = P(15) = $65 c. Break-even means P() = 0. Horizontal: y 0 From a., = 100, 150. Vertical: 10 0 10
5
19. a. Use middle rule for s = 15. f(15) = 19.5 means that when the air temperature is 0ºF and the wind speed is 15 mph, then the air temperature feels like 19.5 ºF. In winter, the TV weather report usually gives the wind chill temperature. b. f(8) = 31.ºF c. Break-even means P() = 0. From a., = 100, 150. 0. a. b. Linear: y 6.8 695; Cubic: c. Linear: Cubic: y 0.175 3 5.7 65.7 65 y 1 $158; y 1 $136 The cubic model is quite accurate, but both models are fairly close. d. The linear model increases steadily, but the cubic model rises rapidly for years past 01.
Chapter Etended Applications & Group Projects I. Body Mass Inde (Modeling) 1. Eight points in the table correspond to a BMI of 30. Converting heights to inches, we have: Height (in.) Weight (lb) 61 160 63 170 65 180 67 190 68 00 69 00 7 0 73 30. A linear model seems best as there appears to be roughly a constant rate of change of weight vs. height. 3. y 5.700 189.5. We note that y 61 158.1, lb close to the actual value of 160 lb, and y 7 0.8, close to the actual value of 0 lb. The model seems to fit the data. 5. To test for obesity, substitute the person s height in inches for in the model, computing y. If the person s weight is larger than y, then the person is considered obese. For a 5-foot-tall person, y 60 15., so 15. lb is the obesity threshold for someone who is 5 feet tall. For a 6-feet--inches-tall person, y 7 3., so 3. lb is the obesity threshold for someone who is 6 foot. 6. The Centers for Disease Control and Prevention (CDC) post the BMI formula weight (lb) BMI 703 height (in) at their website (http://www.cdc.gov/healthyweight/assessing/bmi/adult_bmi/inde.html, accessed September 15, 01). 7. First solving the CDC formula for weight given a BMI of 30 gives weight 30 height. 703 Height (in.) Weight (lb) from model Weight (lb) from CDC definition Weight (lb) from table 61 158 159 160 6 16 16 63 170 169 170 6 175 175 65 180 180 180 66 187 186 67 19 19 190 68 198 197 00 69 0 03 00 70 09 09 71 15 15 7 1 1 0 73 7 7 30
II. Operating Leverage and Business Risk 1. R p. a. C 100 10, 000 b. C is a linear function. 3. An equation that describes the break-even point is p 100 10, 000. a. p 100 10, 000 10, 000 p 100 b. The solution is.a. is a rational function. c. The domain is all real numbers, p 100. d. The domain in the contet of this problem is p 100. 5. a. b. The function decreases as p increases. 6. A price of $1100 would increase the revenue for each unit but demand would decrease. 7. A price of $101 per unit would increase demand but perhaps such a demand could not be met. 8. a. Increasing fied costs gives a higher operating leverage. Using modern equipment would give the higher operating leverage. b. To find the break-even point with current costs we have 00 100 10, 000 100. To find the break-even point with modern equipment we have 00 50 30, 000 00. The higher the break-even point the greater the business risk. The cost with the modern equipment creates a higher business risk. c. In this case, higher operating leverage and higher business event together. This higher risk might give greater profits for increases in sales. It might also give a greater loss of sales fall. Mathematical Applications for the Management Life and Social Sciences 11th Edition Harshbarger SOLUTIONS MANUAL Full clear download at: https://testbankreal.com/download/mathematical-applications-managementlife-social-sciences-11th-edition-harshbarger-solutions-manual/ Mathematical Applications for the Management Life and Social Sciences 11th Edition Harshbarger TEST BANK Full clear download at:
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