(Solution Manual) Applied Calculus for the Managerial, Life, and Social Sciences A Brief Approach

Tan, 10

Math & Statistics

Solution Manual

10th Edition

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Solution Manual for Applied Calculus for the Managerial, Life, and Social Sciences A Brief Approach, The downloadable file consists of 497 pages, and the total size of the file is 0.66 MB. Purchase now only for highly discounted price.

Author(s): Tan

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CONTENTS CHAPTER 1 Preliminaries 1 1.1 Precalculus Review I 1 1.2 Precalculus Review II 81.3 The Cartesian Coordinate System 15 1.4 Straight Lines 20 Chapter 1 Review 29 Chapter 1 Before Moving On 34Chapter 1 Explore & Discuss 35Chapter 1 Exploring with T echnology 36 CHAPTER 2 Functions, Limits, and the Derivative 37 2.1 Functions and Their Graphs 37 2.2 The Algebra of Functions 49 2.3 Functions and Mathematical Models 55 2.4 Limits 672.5 One-Sided Limits and Continuity 76 2.6 The Derivative 85 Chapter 2 Review 97 Chapter 2 Before Moving On 105Chapter 2 Explore & Discuss 106Chapter 2 Exploring with T echnology 108 CHAPTER 3 Differentiation 113 3.1 Basic Rules of Differentiation 1133.2 The Product and Quotient Rules 121 3.3 The Chain Rule 132 3.4 Marginal Functions in Economics 143 3.5 Higher-Order Derivatives 150 3.6 Implicit Differentiation and Related Rates 156 3.7 Differentials 164 Chapter 3 Review 170 Chapter 3 Before Moving On 179 vvi CONTENTS Chapter 3 Explore & Discuss 180 Chapter 3 Exploring with T echnology 181 CHAPTER 4 Applications of the Derivative 183 4.1 Applications of the First Derivative 183 4.2 Applications of the Second Derivative 1974.3 Curve Sketching 210 4.4 Optimization I 231 4.5 Optimization II 245 Chapter 4 Review 253 Chapter 4 Before Moving On 267 Chapter 4 Explore & Discuss 268 Chapter 4 Exploring with T echnology 270 CHAPTER 5 Exponential and Logarithmic Functions 273 5.1 Exponential Functions 273 5.2 Logarithmic Functions 2795.3 Compound Interest 2845.4 Differentiation of Exponential Functions 290 5.5 Differentiation of Logarithmic Functions 303 5.6 Exponential Functions as Mathematical Models 314 Chapter 5 Review 322 Chapter 5 Before Moving On 329 Chapter 5 Explore & Discuss 330Chapter 5 Exploring with T echnology 330 CHAPTER 6 Integration 333 6.1 Antiderivatives and the Rules of Integration 333 6.2 Integration by Substitution 3406.3 Area and the Definite Integral 3476.4 The Fundamental Theorem of Calculus 350 6.5 Evaluating Definite Integrals 356 6.6 Area Between Two Curves 363CONTENTS vii 6.7 Applications of the Definite Integral to Business and Economics 373 Chapter 6 Review 378 Chapter 6 Before Moving On 385Chapter 6 Explore & Discuss 386Chapter 6 Exploring with T echnology 388 CHAPTER 7 Additional Topics in Integration 391 7.1 Integration by Parts 391 7.2 Integration Using Tables of Integrals 3987.3 Numerical Integration 404 7.4 Improper Integrals 414 7.5 Applications of Calculus to Probability 419 Chapter 7 Review 426 Chapter 7 Before Moving On 431Chapter 7 Explore & Discuss 431Chapter 7 Exploring with T echnology 433 CHAPTER 8 Calculus of Several Variables 435 8.1 Functions of Several Variables 4358.2 Partial Derivatives 4418.3 Maxima and Minima of Functions of Several Variables 4498.4 The Method of Least Squares 459 8.5 Constrained Maxima and Minima and the Method of Lagrange Multipliers 4678.6 Double Integrals 478 Chapter 8 Review 482 Chapter 8 Before Moving On 491 Chapter 8 Explore & Discuss 491 Chapter 8 Exploring with T echnology 4941 PRELIMINARIES 1.1 Precalculus Review I Exercises page 13 1.The interval36is shown on the number line below. Note that this is an open interval indicated by “”a n d“”. x 0 3 6()2.The interval25]is shown on the number line below. x _2 5(] 3.The interval [14is shown on the number line below. Note that this is a half-open interval indicated by “[’’ (closed) and “ )”(open). x _1 4[)4.The closed interval 6 51 2 is shown on the number line below. x [] _6 5_1 2 5.The in finite interval0is shown on the number line below. x 0(6.The in finite interval5]is shown on the number line below. x 5] 7.2723 3323329. 8.8431 843 1 241 16. 9. 1 30 1. Recall that any number raised to the zeroth power is 1.10.71247427249. 11. 1 8132  1 22 224. 12. 1 323  1 93 93729. 13.7572 721  75221 711717. 14. 9 1612  16 912 4 3. 15.12523121252312125131 12513 1 5.16.3 2626132613224. 17. 32 8 32 8 42. 18.3 8 2738 3 272 3. 12 1PRELIMINARIES 19.16581612 16781658127816142. 20.9395 9212 93521294121 81. 21.16148132 1 813 21 21. 22.625619 6146251914625191462 36 23.True. 24.True. 32223226236. 25.False. x32x22x322x52x6. 26.False. 3332733034. 27.False.24x 13x24x 124x. 28.True. 22322492362 62264. 29.False.1 434364. 30.True.432 248 161 2. 31.False.12121212141. 32.True. 523252352352235235435225. 33.xy21 xy2. 34.3s13s733s13733s633s23 s2. 35.x13 x12x1312x561 x56. 36. x1 9x3x123x323x1232 3x23 x2. 37.120st311 st31 st3. 38.xy x1y1 xy1 x1 y xyyx xy xyxy xyx2y2 xy. 39.x73 x2x732x7363x133. 40. 49x2124912x2121 7x 41. x2y3 x5y3 x25y33x3y0x31 x3.42.5x6y3 2x2y75 2x62y375 2x4y45x4 2y4. 43.x34 x14x3414x44x. 44.x3y2 z22 x32y22 z22x6y4 z4. 45.x3 27y623 x323 1 2723 y623 x2 1 32 y49 x2y4.46.ex ex212 e[xx2]12e11 e. 47.x3 y22y x4 x32y4 y22x4y44 x46y8 x10. 48.rn4 r52nr4n52nr4n2n5r6n5.1.1 PRECALCULUS REVIEW I 3 49.3 x2 4x5x23412x52x23522 2x116.50. 81x6y48112x62y429x3 y2. 51.4 16x4y81614x44y842xy2. 52.3 x3abx3ab13xab3. 53.6 64x8y36416x86y362x43y12. 54.3 27r6 s2t42713r613s212t412 3r2st2. 55.2322212214142828. 56.812231223222122828. 57.93432343643323312 317325196.58.6122312212312 141417322449. 59.10321012103162103162. 60.1000321033210921041012 10000316231,620. 61.1025102101210031623162. 62.0000113104131043101013 1021542154. 63.3 2xxx3x 2x. 64.3xyxy xy3xy xy. 65.2y3y3y3y2y3y 3y23y 3. 66.5x2  3x 3x 3x5x2 3x 3x5x 3x 3. 67.1 3x3 x2 3 x23 x2 3 x33 x2 x. 68.2x y 2xyy y2xy y. 69.2x 3xx2x 3x. 70.3x 243 x2 3 x2x 243 x2. 71. 2y x2yx2y2y2y2xy. 72.32x 3y3 2x 33y3 2x2 3 2x22x 3 12x2y. 73.3 x2z y3 xz2 3 xz23 x3z3 y3 xz2xz y3 xz2. 74.3 x2y 2x3 xy2 3 xy2xy 2x3 xy2y 23 xy2. 75.7x22x52x25x47x22x52x25x49x23x1. 76. 3x25xy2y  43xy2x2 3x25xy2y43xy2x2x22xy2y4. 77. 5y22y1  y23y7 5y22y1y23y74y2y8. 78.32ab4b2a6a3b4b8a14a7b72ab.4 1PRELIMINARIES 79.x2x[x1x]x2x[x1x]x2x1x2x1x1. 80.3x2x21x[x2x1]23x2x21xx2x12 3x2 x21xx1 23x2 x21x2x 2 3x22x2x12x21x2x2x1. 81.1 31e  1 31e1 1 31e1 311 e2 3e1 e3e22e3 3e. 82.3 4y1 4x1001 2x1 4y1203 4y1 4y1 4x1 2x1001201 2y1 4x20. 83.3 882y1 2x3 4y3 881 2x11 4y6 281 2x11 4y. 84.8 9x22 3x16 3x216 3x2x28x26x48x248x18x18 956x260x18 9 2 928x230x9. 85.x8x2xx28x2x22x8x16x26x16. 86.5x23x45x3x423x415x220x6x815x214x8. 87.a52a5a5aa55a5a25a5a25a210a25. 88.3a4b23a4b3a4b3a3a4b4b3a4b9a212ab12ab16b2 9a224ab16b2. 89.x2y2x2yx2yxx2y2yx2yx22xy2yx4y2x24xy4y2. 90.63x263x63x663x3x63x3618x18x9x23636x9x2. 91.2xy2xy2x2xyy2xy4x22xy2xyy24x2y2. 92.3x223x3x23x223x6x9x246x9x24. 93.2x213x2x23 4x6x43x24x312x6x44x33x212xx6x34x23x12. 94. x21 2xx22x2x32x2x32x. 95.6x 1 2 2x23124x6 2x23123 2x2312 x4x2 2x23 64x23  2x2312. 96. x121 1 2x12  x121 1 2x12 1 2x12 x121  x121 1 2x1221 x121x. 97.10010te01t100e01t100010te01t. 98.2tt22t22tttt2t22t22ttt2t22t24tt2t2t2 4tt2t2t 2t1 .1.1 PRECALCULUS REVIEW I 5 99.4x512x46x32x32x26x3. 100. 4x2y2z2x5y26x3y2z22x2y2 2zx33xz2 . 101. 7a442a2b249a3b7a2 a27ab6b2 . 102. 3x232x13x133x132. 103. exxexex1x. 104. 2yexy22xy3exy22yexy21xy2. 105. 2x523 2x321 2x5243x. 106.1 2 2 3u322u12 1 22 3u12u31 3u12u3. 107. 6ac3bc4ad2bd3c2ab2d2ab2ab3c2d. 108. 3x3x23x1x23x113x1 x21 3x1. 109. 4a2b22ab2ab, a difference of two squares. 110. 12x23y234x2y232xy2xy. 111. 1014x12x226x27x523x52x1. 112. x22x15x5x3. 113. 3x26x243 x22x8 3x4x2. 114. 3x24x43x2x2. 115. 12x22x302 6x2x15 23x52x3. 116.xy21xy1xy1. 117. 9x216y23x24y23x4y3x4y. 118. 8a22ab6b22 4a2ab3b2 2ab4a3b. 119. x6125x2353x25x45x225. 120. x327x333x3x23x9. 121. x2y2 xxy2yx3xy22xy2x3xy2. 122. 2krRrkr22kRr2kr2kr22kRr3kr2kr2R3r.6 1PRELIMINARIES 123. 2x12x23[4x12x2]2x12x234x42x2 2x12x236x24x13x12x23 32x13x1x13. 124. 5x2 3x2146x 3x2152x2x 3x214 15x2 3x21 2x 3x214 18x21 . 125. 4x122x2322x242x12x12x23[4x12x2] 2x12x236x24x13x12x23 32x13x1x13. 126.x214x33x22xx4x3x2 2x4x53x42x34x33x22x2x52x42x3 2x5x44x33x22x. 127.x222 5x2223 2xx2225x44x243 2x2xx2225x420x217. 128.x24x24 2x8x28x44x3x416 2x84x532x416x3 2x58x432x1284x532x416x32x524x416x332x128 2 x512x48x316x64 . 129. We factor the left-hand side of x2x120 to obtainx4x30, so x4o rx3. We conclude that the roots are x4a n d x3. 130. We factor the left-hand side of 3 x2x40 to obtain3x4x10. Thus, 3 x4o rx1, and we conclude that the roots are x4 3andx1. 131. 4t22t22t12t20. Thus, the roots are t1 2andt1. 132.6x2x123x42x30. Thus, x4 3andx3 2are the roots of the equation. 133.1 4x2x1 1 2x1 1 2x1 0. Thus1 2x1, and so x2 is a double root of the equation. 134.1 2a2a12a22a24a6a40. Thus, a6a n d a4 are the roots of the equation. 135. We use the quadratic formula to solve the equation 4 x25x60. In this case, a4,b5, and c6. Therefore, xb b24ac 2a5 52446 245 121 8511 8. Thus, x16 82 andx6 83 4are the roots of the equation. 136. We use the quadratic formula to solve the equation 3 x24x10. Here a3,b 4, and c1, so xb b24ac 2a4 42431 234 4 6. Thus, x6 61a n d x2 61 3are the roots of the equation.1.1 PRECALCULUS REVIEW I 7 137. We use the quadratic formula to solve the equation 8 x28x30. Here a8,b8, and c3, so xb b24ac 2a8 82483 288 160 1684 10 162 10 4. Thus, x1 21 4 10 and x1 21 4 10 are the roots of the equation. 138. We use the quadratic formula to solve the equation x26x60. Here a1,b6, and c6. Therefore, xb b24ac 2a6 62416 2162 3 23 3. Thus, the roots are 3  3a n d 3 3. 139. We use the quadratic formula to solve 2 x24x30. Here a2,b4, and c 3, so xb b24ac 2a4 42423 224 40 442 10 42 10 2. Thus, x11 2 10 and x11 2 10 are the roots of the equation. 140. We use the quadratic formula to solve the equation 2 x27x150. Then a2,b7, and c 15. Therefore, xb b24ac 2a7 724215 227 169 4713 4. We conclude that x3 2andx5 are the roots of the equation. 141. The total revenue is given by02t2150t05t2200t07t2350tthousand dollars tmonths from now, where 0t12. 142. In month t, the revenue of the second gas station will exceed that of the first gas station by05t2200t02t2150t03t250tthousand dollars, where 0 t12. 143. a. f30,000561011 30,00015107,772, or 107,772 families. b.f60,000 561011 60,0001538,103, or 38,103 families. c.f150,000 561011 150,000159639, or 9639 families. 144.t36t215ttt26t15. 145. 8000 x100x2100x80x. 146. True. The two real roots areb b24ac 2a. 147. True. If b24ac0, then b24acis not a real number. 148. True, becauseabbab2a2.8 1PRELIMINARIES 1.2 Precalculus Review II Exercises page 23 1.x2x2 x24x2x1 x2x2x1 x2. 2.2a23ab9b2 2ab23b32a3ba3b b22a3ba3b b2. 3.12t212t3 4t213 4t24t1 4t2132t12t1 2t12t132t1 2t1. 4.x32x23x 2x2x3xx22x3  2x2x3xx3x1 2x3x1xx3 2x3. 5.4x133x14 4x1212x312x4 4x127 4x12. 6.1x2222x21x2 2x  1x241x2 21x24x2  1x241x2 23x21  1x24213x2  1x23. 7.2a22b2 ba4a4b a22abb22abab4ab ababab8. 8.x26x9 x2x63x6 2x27x33x32x2 x3x22x1x33 2x1. 9.3x22x1 2x6x21 x22x33x1x1 2x3x3x1 x1x13x1 2. 10.3x24xy4y2 x2y2yx2 x3y3x2yx2y x2yx3y 2yx2yxx3x2y x2y. 11.58 33t21 3583t2 33t23t60 33t2t20 3t2. 12.a1 3ab2 5b5ba13ab2 15ab5ab5b3ab6a 15ab6a8ab5b 15ab. 13.2x 2x13x 2x52x2x53x2x1 2x12x54x210x6x23x 2x12x52x213x 2x12x5 x2x13 2x12x5. 14.xex x1exxexx1ex x1xexxexex x1ex x1. 15.4 x295 x26x94 x3x35 x324x35x3 x32x3x27 x32x3.1.2 PRECALCULUS REVIEW II 9 16.x 1x2x3 x21xx12x3 x1x1x2x2x3 x21x2x3 x21. 17.11 x 11 xx1 x x1 xx1 xx x1x1 x1. 18.1 x1 y 11 xyxy xy xy1 xyxy xyxy xy1xy xy1. 19.4x2 2 2x27 2x274x22 2x27 2x27 2 2x274x24x214 2 2x274x27 2x27. 20.62x12 x2x2x14 2 x2x62x12 x2x2 x2x2x14 2 x2x 2x12 12 x2x 4x24x1 2 x2x2x12 16x216x1 2 x2x. 21.5 t21 1t2t t212 5 t212t2 t2125 1t2 t2125 t21 t212. 22.2xx112x112 x2x1122xx1 x2x112x1 x2x1 x2x1. 23.x21222x2x21 2x  x2142x21x214x2  x21423x21  x213. 24. x21122x2 x2112 1x2 x2112 x212x2 1x2 x2112 x21 1x21 x21. 25.32x1 3x223x222x13 3x22 32x126x46x3 3x2432x12 3x24. 26.2x112x22x112 2x12x1122x1x2 2x12x112x1 2x1x1 2x132. 27.100t220t100 2t10t210t100 2t20  t220t1002 100 2t340t2200t10t2200t10002t320t2200t20t2200t2000 t220t1002 100 10t21000 t220t10021000t10 t103.10 1PRELIMINARIES 28.22x313x12x323 2x3232x323[22x3x1] 2x3232x3234x6x1 2x323 3x5 2x343. 29.1 31 31 31 31 31 31 2. 30.1x5x5x5x5 x25. 31.1xyxyxyxy xy. 32.a 1a1a 1aa1a 1a. 33.a b a ba b a ba b2 ab. 34.2a b 2a b2a b 2a b 2a b2 4ab. 35.x 3xxx 3x. 36.3y x3 y2 3 y2y x3 y2. 37.1 3 31 3 1 312 32 3 1 32 3 1 3. 38.x1 xx1x1x1 xx1. 39.1x2x21x2 1x21x2x21x2x1x21x2. 40.x3x 3x3x x3xx3x 3 x3x1 x3x. 41.The statement is false because 3 is greater than 20. See the number line below. x _20 0 _3 42.The statement is true because 5 is equal to 5. 43.The statement is false because2 34 6is less than5 6. x 02 356 44.The statement is false because 5 610 12is greater than 11 12. 45.We are given 2 x48. Add4 to each side of the inequality to obtain 2 x4, then multiply each side of the inequality by1 2to obtain x2. We write this in interval notation as 2. 46.We are given 645x.A d d4 to each side of the inequality to obtain 645x,s o105x.D i v i d i n g by 2, we obtain 2x,s ox2. We write this in interval notation as 2. 47.We are given the inequality 4x20. Multiply both sides of the inequality by 1 4and reverse the sign of the inequality to obtain x5. We write this in interval notation as 5].1.2 PRECALCULUS REVIEW II 11 48.123x4x,o rx4. We write this in interval notation as 4]. 49.We are given the inequality 6x24. First add 2 to each member of the inequality to obtain 62x42a n d4x6, so the solution set is the open interval 46. 50.We add1 to each member of the given double inequality 0 x14t oo b t a i n 1x3, and the solution set is [13]. 51.We want to find the values of xthat satisfy at least one of the inequalities x14a n d x21. Adding 1t o both sides of the first inequality, we obtain x1141, so x3. Similarly, adding 2 to both sides of the second inequality, we obtain x2212, so x3. Therefore, the solution set is 33. 52.We want to find the values of xthat satisfy at least one of the inequalities x12a n d x12. Solving these inequalities, we find that x1o rx1, and the solution set is 11. 53.We want to find the values of xthat satisfy the inequalities x31a n d x21. Adding 3 to both sides of the first inequality, we obtain x3313, or x2. Similarly, adding 2 to each side of the second inequality, we obtain x2212, so x3. Because both inequalities must be satis fied, the solution set is 23. 54.We want to find the values of xthat satisfy the inequalities x41a n d x32. Solving these inequalities, we find that x5a n d x1 ,a n dt h es o l u t i o ns e ti s 15]. 55.We want to find the values of xthat satisfy the inequality x3x50 . From the sign diagram, we see that the given inequality is satis fied when3x5, that is, when the signs of the two factors are different or when one of the factors is equal to zero.x _3 0 5Sign of x+3Sign of x-5 ++++++++0_+0_________+ __ ++ 56.We want to find the values of xthat satisfy the inequality 2x4x20. From the sign diagram, we see that the given inequality is satis fied when x2o rx2; that is, when the signs of both factors are the same or one of the factors is equal to zero.x _2 0 2Sign of x+2Sign of 2x-4 +++++++0__++0________+ 57.We want to find the values of xthat satisfy the inequality 2x3x10. From the sign diagram, we see that the given inequality is satis fied when x1o rx3 2; that is,when the signs of both factors are the same, or one of the factors is equal to zero.x 0Sign of x-1Sign of 2x-3 ++++0 __0_______ __ 13 2___++ 58.We want to find the values of xthat satisfy the inequalities 3x42x20. From the sign diagram, we see that the given inequality is satis fied when1x4 3, that is, when the signs of the two factors are different or when one of the factors is equal to zero.x _1Sign of 2x+2Sign of 3x-4 ++++ __+0 _______+ ++ 4 30 0+++__12 1PRELIMINARIES 59.We want to find the values of xthat satisfy the inequalities x3 x20. From the sign diagram, we see that the given inequality is satis fied when x3o rx2, that is, when the signs of the two factors are the same. Notice that x2 is not included because the inequality is not de fined at that value of x.x _3Sign of x-2Sign of x+3 0+0__+ ++ _Inequality not defined 0 2+++++ ______ 60.We want to find the values of xthat satisfy the inequality 2x3 x14. We rewrite the inequality as2x3 x140, 2x34x4 x10, and2x7 x10. From the sign diagram, we see that the given inequality is satis fied when7 2x1;xSign of x+1Sign of _2x-7 +_0 + _Inequality not defined _1____0 7 20 _+ _+ ____ ______ ++ that is, when the signs of the two factors are the same. Notice that x1 is not included because the inequality is not de fined at that value of x. 61.We want to find the values of xthat satisfy the inequality x2 x12. Subtracting 2 from each side of the given inequality and simplifying givesx2 x120, xSign of x-1Sign of _x +0+ + _Inequality not defined _____ 1+ _+ ____ ______ 0+ 0 x22x1 x10, andx x10. From the sign diagram, we see that the given inequality is satis fied when x0o rx1; that is, when the signs of the two factors differ. Notice that x1 is not included because the inequality is unde fined at that value of x. 62.We want to find the values of xthat satisfy the inequality2x1 x24Subtracting 4 from each side of the given inequality and simplifying gives2x1 x240, 2x14x2 x20,2x14x8 x20, and finallyxSign of x+2Sign of _2x-9 _Inequality not defined __0 0_0 ____ ____ + _2 _9 2+ _+ ____ +++ 2x9 x20. From the sign diagram, we see that the given inequality is satis fied when x9 2orx2. 63.624. 64.44448. 65.124 16128 42. 66.0214 162412 0815. 67. 323 3 323 35 3. 68.1 2212 2. 69.π12π12π1. 70.π636π33π. 71. 213 2 213 22.1.2 PRECALCULUS REVIEW II 13 72.2 33 342 33 4 3 3 37. 73.False. If ab,t h e nab,abbb,a n d ba0. 74.False. Let a2a n d b3. Then ab2 32 31. 75.False. Let a 2a n d b 3. Then a24a n d b29, and 49. Note that we need only to provide a counterexample to show that the statement is not always true. 76.False. Let a2a n d b3. Then1 a1 2and1 b1 3,a n d1 21 3. 77.True. There are three possible cases. Case 1: Ifa0a n d b0, then a3b3,s i n c e a3b3ab a2abb2 0. Case 2: Ifa0a n d b0, then a30a n d b30, and it follows that a3b3. Case 3: Ifa0a n d b0, then a3b3aba2abb20, and we see that a3b3. (Note that ab0a n d ab0.) 78.True. If ab, then it follows that abbecause an inequality symbol is reversed when both sides of the inequality are multiplied by a negative number. 79.False. If we take a2, thena222a. 80.True. If b0, then b20, andb2b2. 81.True. If a40, thena44a4a.I fa40, then4aa4a4. 82.False. If we let a2, thena12111213. 83.False. If we take a3a n d b1, thenab312ab314. 84.False. If we take a3a n d b1, thenab4ab312. 85.If the car is driven in the city, then it can be expected to cover 18120362miles galgal, or 362 miles, on a full tank. If the car is driven on the highway, then it can be expected to cover 181274887miles galgal, or 4887 miles, on a full tank. Thus, the driving range of the car may be described by the interval [3624887]. 86.Simplifying 5C2517525C, we obtain 5 C12517525C,5C25C175125, 25C12675, and finally C507. Therefore, the minimum cost is $50 70. 87.6P25004P2400can be rewritten as 6 P15,0004P9600, 2 P24,600, or P12,300. Therefore, the maximum pro fit is $12,300. 88. a. We want to find a formula for converting Centigrade temperatures to Fahrenheit temperatures. Thus, C5 9F325 9F160 9. Therefore,5 9FC160 9,5F9C160, and F9 5C32. Calculating the lower temperature range, we have F9 515325, or 5 degrees. Calculating the upper temperature range, F9 553223, or 23 degrees. Therefore, the temperature range is 5F23.14 1PRELIMINARIES b.For the lower temperature range, C5 96332155 9172, or 172 degrees. For the upper temperature range, C5 980325 948267, or 267 degrees. Therefore, the temperature range is 172C267. 89.Letxrepresent the salesman’s monthly sales in dollars. Then 0 15x12,0006000, 15x12,000600,000, 15 x180,000600,000, 15 x780,000, and x52,000. We conclude that the salesman must have sales of at least $52,000 to reach his goal. 90.Letxrepresent the wholesale price of the car. ThenSelling price Wholesale price1Markup; that is,11,200 x1030, whence11,200 x130, 13x11,200, and x861538. We conclude that the maximum wholesale price is $861538. 91.The rod is acceptable if 0 49x051 or001x05001. This gives the required inequality, x05001. 92.x01001 is equivalent to 001x01001 or 009x011. Therefore, the smallest diameter a ball bearing in the batch can have is 0 09 inch, and the largest diameter is 0 11 inch. 93.We want to solve the inequality 6x230x1014. (Remember that xis expressed in thousands.) Adding 14 to both sides of this inequality, we have 6x230x10141414, or6x230x240. Dividing both sides of the inequality by 6 (which reverses the sign of the inequality), we have x25x40. Factoring this last expression, we have x4x10. From the sign diagram, we see that xmust lie between 1 and 4. (The inequality is satis fied only when the two factors have opposite signs.) Because xis expressed in thousands of units, we see that the manufacturer must produce between 1000 and 4000 units of thecommodity.xSign of x-1Sign of x-4 0_____ __0 + 0___ +++ 1 4+ +++++ _ 94.We solve the inequality02t t21008, obtaining 0 08t200802t,008t202t0080, 2t25t20, and2t1t20. From the sign diagram, we see that the required solution is 1 22 , so the concentration of the drug is greater than or equal to 008 mgcc between1 2hr and 2 hr after injection.tSign of 2t-1Sign of t-2 0_____ __0 + 0___ +++ 2+ +++++ _ 1 2132 95.We solve the inequalities 25 05x 100x30, obtaining 2500 25x05x300030x, which is equivalent to 250025x05xand 05x300030x. Simplifying further, 25 5x2500 and 305x3000, so x2500 2559804 and x3000 3059836. Thus, the city could expect to remove between 98 04% and 9836% of the toxic pollutant.1.3 THE CARTESIAN COORDINATE SYSTEM 15 96.We simplify the inequality 20 t40t5035 to 20 t40t150( 1 ) . L e t ut.T h e n u2t,s ow e have 20 u240u150, 4u28u30, and2u32u10. From the sign diagram, we see that we must have uin 1 23 2 . Because tu2, we see that the solution to Equation (1) is 1 49 4 . Thus, the average speed of a vehicle is less than or equal to 35 miles per hour between 6:15 a.m. and 8:15 a.m.uSign of 2u-1Sign of 2u-3 0____0 + 0___ ++ + ++++ _ 1 21321 2 97.We solve the inequality136 1025t45228128 or136 1025t452100. Next, 136100 1025t452 , so 13610025t452,3 625t452,t45236 25,a n d t456 5. Solving this last inequality, we have t57a n d t33. Thus, the amount of nitrogen dioxide is greater than or equal to 128 PSI between 10:18 a.m. and 12:42 p.m. 98.False. Take a2,b3, and c4. Thena bc2 342 7,b u ta ba c2 32 486 1214 127 6. 99.False. Take a1,b2, and c3. Then ab,b u t ac132231bc. 100. True.ba1ab1abab. 101. True.abababab. 102. False. Take a3a n d b1. Then a2b291 82 2, butab312. 1.3 The Cartesian Coordinate System Concept Questions page 29 1. a. a0a n d b0 b.a0a n d b0 c.a0a n d b0 2. a. xy 0PÁ(a, b)Pª(_a, b) P¢(a, _b) P£(_a, _b)b.dP1ab00 0a20b2 a2b2, dP2ab00 [0a]20b2 a2b2, dP3ab00 [0a]2[0b]2 a2b2, anddP4ab00 0a2[0b]2 a2b2, so the points P1ab,P2ab,P3ab,a n d P4abare all the same distance from the origin. Exercises page 30 1.The coordinates of Aare33and it is located in Quadrant I. 2.The coordinates of Bare52and it is located in Quadrant II. 3.The coordinates of Care22and it is located in Quadrant IV . 4.The coordinates of Dare25and it is located in Quadrant II.16 1PRELIMINARIES 5.The coordinates of Eare46and it is located in Quadrant III. 6.The coordinates of Fare82and it is located in Quadrant IV . 7.A 8.54 9.E,F,a n d G 10.E 11.F 12.D For Exercises 13 20, refer to the following figure. x1y 013. (_2, 5) 14. (1, 3) 15 (3, _1) 20. (1.2, _3.4)16. (3, _4) 19.(4.5, _4.5)18.3 2()52_, 17.7 2()8, _1 21.Using the distance formula, we find that 412732 3242 255. 22.Using the distance formula, we find that 412402 3242 255. 23.Using the distance formula, we find that [41]2932 52622536 61. 24.Using the distance formula, we find that [102]2612 1225214425 16913. 25.The coordinates of the points have the form x6. Because the points are 10 units away from the origin, we have x02602102,x264, or x8. Therefore, the required points are 86and86. 26.The coordinates of the points have the form 3y. Because the points are 5 units away from the origin, we have 302y0252,y216, or y4. Therefore, the required points are 34and34. 27.The points are shown in the diagram. To show that the four sides are equal, we compute dAB 332742 6232 45, dBC [63]2172 3262 45, dCD [06]2[21]2 6232 45, anddAD 032242 3262 45.xy 0A(3, 4)B(_3, 7) C(_6, 1) D(0, _2) Next, to show that ABC is a right triangle, we show that it satis fies the Pythagorean Theorem. Thus, dAC 632142 9232 903 10 and [dAB]2[dBC]290[dAC]2. Similarly, dBD 903 10, soBAD is a right triangle as well. It follows that BandDare right angles, and we conclude that ADCB is a square.1.3 THE CARTESIAN COORDINATE SYSTEM 17 28.The triangle is shown in the figure. To prove that ABC is a right triangle, we show that [dAC]2[dAB]2[dBC]2and the result will then follow from the Pythagorean Theorem. Now [dAC]2552[22]210016116. Next, we findxy 0 C(5, _2)B(_2, 5) A(_5, 2) [dAB]2[dBC]2[25]2522[52]2252994949116, and the result follows. 29.The equation of the circle with radius 5 and center 23is given byx22 y3252,o r x22y3225. 30.The equation of the circle with radius 3 and center 24is given by [x2]2 y429, or x22y429. 31.The equation of the circle with radius 5 and center 00is given byx02y0252,o rx2y225. 32.The distance between the center of the circle and the point 23on the circumference of the circle is given by d 302202 13. Therefore r 13 and the equation of the circle centered at the origin that passes through 23isx2y213. 33.The distance between the points 52and23is given by d 522[23]2 3252 34. Therefore r 34 and the equation of the circle passing through 52and23is x22 y3234, orx22y3234. 34.The equation of the circle with center aaand radius 2 ais given by [xa]2ya22a2,o r xa2ya24a2. 35. a. The coordinates of the suspect’s car at its final destination are x4a n d y4. b.The distance traveled by the suspect was 5 41, or 10 miles. c.The distance between the original and final positions of the suspect’s car was d 402402 324 2, or approximately 5 66 miles.xy 0 12345612345 36.Referring to the diagram on page 31 of the text, we see that the distance from AtoBis given bydAB 40023002250,000500. The distance from BtoCis given by dBC 80040028003002 1200250021,690,000 1300. The distance from CtoDis given by dCD [800800]280002 08002800. The distance from DtoA is given by dDA [8000]200640,000800. Therefore, the total distance covered on the tour is dABdBCdCDdDA50013008008003400, or 3400 miles.18 1PRELIMINARIES 37.Suppose that the furniture store is located at the origin Oso that your house is located at A2014. Because dOA 202142 596244, your house is located within a 25-mile radius of the store and you will notincur a delivery charge.xy 1010 0 A(20, _14) 38. x (mi)y (mi) A(0, 0)1000 1000D(1300, 1500) Route 1Route 2C(800, 1500) B(400, 300) Referring to the diagram, we see that the distance the salesman would cover if he took Route 1 is given by dABdBD 40023002 1300400215003002 250,0002,250,000 50015002000 or 2000 miles. On the other hand, the distance he would cover if he took Route 2 is given by dACdCD 800215002 130080022,890,000 250,000 17005002200 or 2200 miles. Comparing these results, we see that he should take Route 1. 39.The cost of shipping by freight train is 0662000100132,000, or $132,000. The cost of shipping by truck is 0622200100136,400, or $136,400. Comparing these results, we see that the automobiles should be shipped by freight train. The net savings are136,400132,0004400, or $4400. 40.T h el e n g t ho fc a b l er e q u i r e do nl a n di s dSQ10,000xand the length of cable required under water is dQM  x20 030002 x230002. The cost of laying cable is thus 310,000x5 x230002. Ifx2500, then the total cost is given by 3 10,00025005 250023000242,02562or $42,02562. Ifx3000, then the total cost is given by 3 10,00030005 300023000242,21320or $42,21320. 41.To determine the VHF requirements, we calculate d 2523526251225 18504301. Models B,C,a n d Dsatisfy this requirement. To determine the UHF requirements, we calculate d 2023224001024 14243774. Models C andDsatisfy this requirement. Therefore, Model Callows him to receive both channels at the least cost.1.3 THE CARTESIAN COORDINATE SYSTEM 19 42. a. Let the positions of ships AandBafter thours be A0yandBx0, respectively. Then x30tandy20t. Therefore, the distance in miles between the two ships is D 30t220t2 900t2400t210 13t. b.The required distance is obtained by letting t2, giving D10 132, or approximately 72 11 miles. 43. a. Let the positions of ships AandBbe0yandx0, respectively. Then y25 t1 2 andx20t. The distance Din miles between the two ships is D x020y2 x2y2 400t2625 t1 22 (1). b.The distance between the ships 2 hours after ship Ahas left port is obtained by letting t3 2in Equation (1), yielding D 400 3 22 625 3 21 22  3400, or approximately 58 31 miles. 44. a. The distance in feet is given by 40002x2 16,000,000 x2. b.Substituting the value x20,000 into the above expression gives 16,000,000 20,000220,396, or 20,396 ft. 45. a. Suppose that Px1y1andQx2y2are endpoints of the line segment and that the point Mx1x2 2y1y2 2 is the midpoint of the line segment PQ. The distance between PandQis x2x12y2y12. The distance between PandMis x1x2 2x12 y1y2 2y12 x2x1 22 y2y1 22 1 2 x2x12y2y12, which is one-half the distance from PtoQ. Similarly, we obtain the same expression for the distance from Mto P. b.The midpoint is given by43 252 2 ,o r1 23 2 . 46. a. x (yd)y (yd) 0 10 20 30 4010203040 A(20, 10)B(10, 40) Mb.The coordinates of the position of the prize are x2010 2and y1040 2,o rx15 yards and y25 yards. c.The distance from the prize to the house is dM152500 15022502 850 2915 (yards). 47.True. Plot the points. 48.True. Plot the points. 49.False. The distance between P1abandP3kckdis d kca2kdb2 kDk ca2db2 k2ca2k2db2 [kca]2[kdb]2.20 1PRELIMINARIES 50.True. kx2ky2a2gives x2y2a2 ka2ifk1. So the radius of the circle with equation kx2ky2a2 is a circle of radius smaller than acentered at the origin if k1. Therefore, it lies inside the circle of radius awith equation x2y2a2. 51.Referring to the figure in the text, we see that the distance between the two points is given by the length of the hypotenuse of the right triangle. That is, d x2x12y2y12. 52.xh2yk2r2x22xhh2y22kyk2r2. This has the form x2y2CxDyE0, where C2h,D2k,a n d Eh2k2r2. 1.4 Straight Lines Concept Questions page 42 1.The slope is my2y1 x2x1,w h e r e Px1y1andPx2y2are any two distinct points on the nonvertical line. The slope of a vertical line is unde fined. 2. a. yy1mxx1 b.ymxb c.axbyc0where aandbare not both zero. 3. a. m1m2 b.m21 m1 4. a. Solving the equation for ygives ByAxC,s oyA BxC B. The slope of Lis the coef ficient of x,A B. b.IfB0, then the equation reduces to AxC0. Solving this equation for x, we obtain xC A.T h i si sa n equation of a vertical line, and we conclude that the slope of Lis unde fined. Exercises page 42 1.(e) 2.(c) 3.(a) 4.(d) 5.(f) 6.(b) 7.Referring to the figure shown in the text, we see that m20 041 2. 8.Referring to the figure shown in the text, we see that m40 022. 9.This is a vertical line, and hence its slope is unde fined. 10.This is a horizontal line, and hence its slope is 0. 11.my2y1 x2x183 545. 12.my2y1 x2x185 343 13. 13.my2y1 x2x183 425 6. 14.my2y1 x2x142 422 61 3. 15.my2y1 x2x1db ca, provided ac.1.4 STRAIGHT LINES 21 16.my2y1 x2x1bb1 a1a1bb1 a1a112b 2a. 17.Because the equation is already in slope-intercept form, we read off the slope m4. a.Ifxincreases by 1 unit, then yincreases by 4 units. b.Ifxdecreases by 2 units, then ydecreases by 4 28 units. 18.Rewrite the given equation in slope-intercept form: 2 x3y4, 3y42x,a n ds o y4 32 3x. a.Because m2 3, we conclude that the slope is negative. b.Because the slope is negative, ydecreases as xincreases. c.Ifxdecreases by 2 units, then yincreases by 2 3 24 3units. 19.The slope of the line through AandBis102 318 42. The slope of the line through CandDis 15 114 22. Because the slopes of these two lines are equal, the lines are parallel. 20.The slope of the line through AandBis23 22. Because this slope is unde fined, we see that the line is vertical. The slope of the line through CandDis54 22. Because this slope is unde fined, we see that this line is also vertical. Therefore, the lines are parallel. 21.The slope of the line through AandBis25 423 61 2. The slope of the line through CandDis 62 318 42. Because the slopes of these two lines are the negative reciprocals of each other, the lines are perpendicular. 22.The slope of the line through AandBis20 122 12. The slope of the line through CandDis 42 842 121 6. Because the slopes of these two lines are not the negative reciprocals of each other, the lines are not perpendicular. 23.The slope of the line through the point 1aand42ism12a 41and the slope of the line through 28and7a4ism2a48 72. Because these two lines are parallel, m1is equal to m2. Therefore, 2a 3a4 9,92a3a4,1 89a3a12, and 6 a30, so a5. 24.The slope of the line through the point a1and58ism181 5aand the slope of the line through 49and a21ism219 a24. Because these two lines are parallel, m1is equal to m2. Therefore,7 5a8 a2, 7a285a,7a14408a,a n d a26. 25.An equation of a horizontal line is of the form yb. In this case b3, so y3i sa ne q u a t i o no ft h el i n e . 26.An equation of a vertical line is of the form xa.I nt h i sc a s e a0, so x0 is an equation of the line.22 1PRELIMINARIES 27.We use the point-slope form of an equation of a line with the point 34and slope m2. Thus yy1mxx1becomes y42x3. Simplifying, we have y42x6, or y2x10. 28.We use the point-slope form of an equation of a line with the point 24and slope m 1. Thus yy1mxx1,g i v i n g y41x2,y4x2, and finally yx6. 29.Because the slope m0, we know that the line is a horizontal line of the form yb. Because the line passes through32, we see that b2, and an equation of the line is y2. 30.We use the point-slope form of an equation of a line with the point 12and slope m1 2.T h u s yy1mxx1gives y21 2x1,2y4x1, 2yx5, and y1 2x5 2. 31.Wefirst compute the slope of the line joining the points 24and37to be m74 323. Using the point-slope form of an equation of a line with the point 24and slope m3, we findy43x2,o r y3x2. 32.Wefirst compute the slope of the line joining the points 21and25to be m51 22. Because this slope is undefined, we see that the line must be a vertical line of the form xa. Because it passes through 25,w es e e thatx2 is the equation of the line. 33.Wefirst compute the slope of the line joining the points 12and32to be m22 314 41. Using the point-slope form of an equation of a line with the point 12and slope m1, we findy2x1, or yx1. 34.Wefirst compute the slope of the line joining the points 12and34to be m42 312 41 2. Using the point-slope form of an equation of a line with the point 12and slope m1 2,w efind y21 2[x1],y21 2x1,a n d finally y1 2x5 2. 35.We use the slope-intercept form of an equation of a line: ymxb. Because m3a n d b4, the equation is y3x4. 36.We use the slope-intercept form of an equation of a line: ymxb. Because m2a n d b1, the equation isy2x1. 37.We use the slope-intercept form of an equation of a line: ymxb. Because m0a n d b5, the equation is y5. 38.We use the slope-intercept form of an equation of a line: ymxb. Because m1 2,a n d b3 4, the equation is y1 2x3 4. 39.Wefirst write the given equation in the slope-intercept form: x2y0, so2yx,o ry1 2x.F r o mt h i s equation, we see that m1 2andb0. 40.We write the equation in slope-intercept form: y20, so y2. From this equation, we see that m0a n d b2.