CHAPTER 3 Derivatives and Their Applications

CHAPTER Derivatives and Their Applications Review of Prerequisite Skills, pp. 6 7. a. b. c. d. 0 6 6 0 6 6 6 0 6 6 6 0 6 6 6 6 Calculus and Vectors Solutions Manual e. f. 8 6 0 8 8 6 0 8. a. ( ) ( ) 6 0 6 6 0 b. ( ) ( ) 0( ) ( ) ( ) 0 0 6 7 6 7 7 9 c. t t 0 (t )(t ) 0 t or t d. t t 0 (t )(t ) 0 t or t 6 e. t t t 8t t 8t 0 ( t 6)(t ) 0 6 t or t 6 6 6 -

f. 0 ( ) 0 ( )( ) 0 0 or or g. 8 6 0 ( 8 6) 0 ( ) 0 0 or h. t t t 0 t (t ) (t ) 0 ( t )(t ) 0 ( t )(t )(t ) 0 t or t or t i. t t 9 0 (t 9)(t ) 0 t 6 9 or t 6. a.. 7. 9. b. ( ). 0, 0 or. c.. 0 0 0 ( ), 0 0,,. a. P s 0 s s A s cm b. A lw 8(6) 8 cm c. A pr p(7) 9p cm d. C pr p pr 6 r A pr p(6) 6p cm. a. b. 96p p() h h 6 cm SA prh pr p()(6) p() 8p p 80p cm V pr h c. 6p pr (6) r 6 cm SA prh pr p(6)(6) p(6) 7p 7p p cm SA prh pr d. 0p p()h p() 0p 0ph 0p 70p 0ph h 7 cm V pr h p() (7) 7p cm 6. For a cube, SA 6s and V s, where s is the length of an edge of the cube. a. SA 6() cm V 7 cm b. SA 6(") c. SA 6(") d. SA prh pr p()() p() p p 6p cm V pr h p() () 8p cm V pr h 0 cm V (") " cm 7 cm V (") " cm SA 6(k) k cm V (k) 8k cm - Chapter : Derivatives and Their Applications

7. a. (, `) b. (`, c. d. e. f. 8. a. b. c. d. (`, 0), `) (, 8 (, ) PR0. 6 PR0 #6 PR6 PR0 0 # # 6 PR0,, 6 PR0 #, 06 e. f. 9. a. The function has a maimum value of. f. The function has a minimum value of 7. The function has a minimum value of and no maimum value. b. The function has a maimum value of and no minimum value. c. The function has a minimum value of 7 and no maimum value. d. The function has a maimum value of.. Higher-Order Derivatives, Velocit, and Acceleration, pp. 7 9. v() v() 0 At t, the velocit is positive; this means that the object is moving in whatever is the positive direction for the scenario. At t, the velocit is negative; this means that the object is moving in whatever is the negative direction for the scenario.. a. 0 6 r 0 9 8 s 90 8 90 b. f() # f r() f s() The function has a minimum value of and no maimum value. e. The function has a minimum value of. c. ( ) r ( )() s d. h() hr() 6 hs() 6 6 e. r 6 Calculus and Vectors Solutions Manual -

s f. f() ( )() ()() fr() ( ) ( ) g. r s 6 6 h. g() ( 6) 9 ( 6) i. ( ) r 6( ) s ( ) 8 96 j. h() hr() hs() 0 0 9 9 6 " 6 ( ) ( ) f s() ( ) (0) ()(( )) ( ) gr() ( 6) gs() 9 ( 6). a. s(t) t t v(t) 0t a(t) 0 b. s(t) t 6t 0 v(t) 6t 6 a(t) t c. s(t) t 8 6 t t 8 6t v(t) 6t a(t) t d. s(t) (t ) v(t) (t ) a(t) e. s(t) "t v(t) (t ) a(t) (t ) f. s(t) 9t t 9(t ) 9t v(t) (t ) 7 (t ) a(t) (t ). a. i. t ii., t, iii., t, b. i. t, t 7 ii., t,, 7, t, 9 iii., t, 7. a. s t t t v t t a t b. For v 0, (t )(t ) 0 t or t. 0 The direction of the motion of the object changes at t and t. c. Initial position is s(0) 0. Solving, 0 t t t t 6t 9t t(t 6t 9) t(t ) t 0 or t s 0 or s 0. The object returns to its initial position after s. - Chapter : Derivatives and Their Applications

6. a. s t t v t v() v() () For t, moving in a positive direction. For t, moving in a negative direction. b. s(t) t(t ) v(t) (t ) t(t ) (t )(t t) (t )(t ) (t )(t ) v() 0 v() 9 For t, the object is stationar. t, the object is moving in a positive direction. c. s(t) t 7t 0t v(t) t t 0 v() v() For t, the object is moving in a negative direction. For t, the object is moving in a positive direction. 7. a. s(t) t 6t 8 v(t) t 6 b. t 6 0 t s 8. s(t) 0t t v(t) 0 0t a. When v 0, the object stops rising. t s b. Since s(t) represents a quadratic function that opens down because a, 0, a maimum height is attained. It occurs when v 0. Height is a maimum for s() 60 (6) 80 m. 9. s(t) 8 7t t v(t) 7 t a(t) a. v() 7 0 m>s b. a() m>s Calculus and Vectors Solutions Manual 0. s(t) t (7 t) a. v(t) t (7 t) t t t t t 7 t a(t) 0 t t b. The object stops when its velocit is 0. v(t) t 7 t 7 t ( t) v(t) 0 for t 0 (when it starts moving) and t. So the object stops after s. c. The direction of the motion changes when its velocit changes from a positive to a negative value or visa versa. t 0 " t * t t + v(t) ()() 0 ()() v(t) 7 for t t ( t) v(t) 0 Therefore, the object changes direction at s. d. a(t) 0 for t (6 t) 0. t 0 or t 6 s. t 0 * t * 6 t 6 t + 6 a(t) ()() 0 ()() Therefore, the acceleration is positive for 0, t, 6 s. Note: t 0 ields a 0. e. At t 0, s(0) 0. Therefore, the object s original position is at 0, the origin. When s(t) 0, t (7 t) 0 t 0 or t 7. Therefore, the object is back to its original position after 7 s.. a. h(t) t t v(t) 0t v(0) m>s -

b. The maimum height occurs when v(t) 0. 0t 0 t. s h(.) (.) (.). m c. The ball strikes the ground when h(t) 0. t t 0 t(t ) 0 t 0 or t The ball strikes the ground at t s. v() 0 m>s. s(t) 6t t v(t) t a. Thus, as the dragster crosses the finish line at t 8 s, the velocit is 98 m> s. Its acceleration is constant throughout the run and equals m> s. b. s 60 6 t t 60 0 (t t 0) 0 (t 0)(t ) 0 t 0 or t inadmissible v() 6 0 # t # 8 8 Therefore, the dragster was moving at 8 m> s when it was 60 m down the strip.. a. s 0 6t t v 6 t ( t) a The object moves to the right from its initial position of 0 m from the origin, 0, to the 9 m mark, slowing down at a rate of m> s. It stops at the 9 m mark then moves to the left accelerating at m> s as it goes on its journe into the universe. It passes the origin after ( "9 ) s. t=6 t=0 t= S 0 0 0 0 b. a(t) v(8) 96 98 m>s s t t 9 v t (t ) (t )(t ) a 6t The object begins at 9 m to the left of the origin, 0, and slows down to a stop after s when it is m to the left of the origin. Then, the object moves to the right accelerating at faster rates as time increases. It passes the origin just before s (approimatel.79) and continues to accelerate as time goes b on its journe into space.. t= t=0 0 0 0 0 0 s(t) t 0t v(t) t 0t a(t) 0t 0 a(t) 0, For 0t 0 0 0(t ) 0 t. Therefore, the acceleration will be zero at s. s() 0 9, 0 v() 0, 0 Since the signs of both s and v are the same at t, the object is moving awa from the origin at that time.. a. s(t) kt (6k 0k)t k v(t) kt (6k 0k) a(t) k 0 k Since k 0 and kpr, then a(t) k 0 and an element of the Real numbers. Therefore, the acceleration is constant. b. For v(t) 0 kt 6k 0k 0 kt 0k 6k t k k 0 s( k) k( k) (6k 0k)( k) k k( 0k 9k ) 0k 8k 0k 0k k k 0k 9k 0k 8k 0k 0k k 9k 0k k Therefore, the velocit is 0 at t k, and its position at that time is 9k 0k k. S -6 Chapter : Derivatives and Their Applications

6. a. The acceleration is continuous at t 0 if a(t) a(0). lim ts0 For t $ 0, and and t s(t) t v(t) t (t ) t(t ) (t ) t t (t ) a(t) (t 6t)(t ) (t ) lim a(t) lim ts` ts` (t )(t)(t t ) (t ) (t 6t)(t ) t(t t ) (t ) t 6t t 6t t t (t ) t 6t (t ) 0, if t, 0 Therefore, a(t) t 6t (t ), if t $ 0 0, if t, 0 and n(t) t t (t ), if t $ 0 lim a(t) 0, lim a(t) 0 ts0 ts0 0. Thus, lima(t) 0. ts0 Also, a(0) 0 0. Therefore, lima(t) a(0). ts0 Thus, the acceleration is continuous at t 0. t t b. lim v(t) lim ts` ts` t t t lim ts` t t t 6 t t t t 6 Calculus and Vectors Solutions Manual 7. v #b gs a v? gv a g Since g is a constant, a is a constant, as required. ds Note: dt v dv dt a 8. F m d 0 dt a v! ( v c) b Using the quotient rule, v (b gs) dv dt (b gs) m 0 Since 0 0 dv v ( ) dt c dv v ( ) dt c av b? v dv dt a, m 0( v c ) Sa ( v c ) v a c T v c m av 0 Sac c v a c T ( v c ) m 0 ac c ( v c )? a0 g ds dt b v c m 0 a, as required. c ( v c ). Maimum and Minimum on an Interval (Etreme Values), pp. 8. a. The algorithm can be used; the function is continuous. b. The algorithm cannot be used; the function is discontinuous at. c. The algorithm cannot be used; the function is discontinuous at. c -7

d. The algorithm can be used; the function is continuous on the given domain.. a. ma 8; min b. ma 0; min c. ma 00; min 00 d. ma 0; min 0. a. f(), 0 # # fr() Let 0 for ma or min f(0) f() 8 f() 9 0 ma is at 0 min is at (0, ) ma is at min is at 0 b. f() ( ), 0 # # f r() Let f r() 0 for ma or min 0 f(0) f() 0 6 6 0 6 0 (0, ) (, 0) (, ) (, 0) 6 c. f(), # # f r() 6 Let f r() 0 for ma or min 6 0 ( ) 0 0 or f() f(0) 0 f() 8 f() 7 7 0 min is at, ma is 0 at 0, d. f(), P, f r() 6 Let fr() 0 for ma or min 6 0 ( ) 0 0 or is outside the given interval. f() 0 f(0) 0 f() ma is 0 at 0 min is 0 at 8 0 8 6 0 6 e. f(), P, 0 fr() 6 6 Let fr() 0 for ma or min 6 6 0 0 ( )( ) 0 or -8 Chapter : Derivatives and Their Applications

f() 6 f() 8 f(0) f() not in region ma of 8 at min of at 8 f. f() 6, P0, f r() 6 Let f r() 0 for ma or min 6 0 ( )( ) 0 or f(0) 0 f() f() 9 f() 6 ma is 6 at min is 0 at 0 6 0 0. a. f() f r() 6 Calculus and Vectors Solutions Manual Set f r() 0 to solve for the critical values. 0 0 ( )( ) 0, Now, evaluate the function, f(), at the critical values and the endpoints. Note, however, that is not in the domain of the function. f() f() f(0) 0 0 0 0. So, the minimum value in the interval is when and the maimum value is 0. when 0. b. f()!, # # 9 f r() Let f r() 0 for ma or min! 0! f()! 8.6 f()! f(9)!9 9 min value of when 9 ma value of when c. f() 0 # #, f r() ( ) ( ) ( ) Let f r() 0 for ma or min. ( ) 0 0 f(0) f(), f(), ma value of when min value of when 0, d. f() 6 0 fr() 7 Set f r() 0 to solve for the critical values. 7 0 ( 6) 0-9

( )( ) 0 0,, Now, evaluate the function, f(), at the critical values and the endpoints. f() () () 6() 0 7 f() () () 6() 0 f(0) (0) (0) 6(0) 0 0 f() () () 6() 0 69 f() () () 6() 0 So, the minimum value in the interval is 69 when and the maimum value is 7 when. e. f() # #, f r() ( ) () ( ) Let f r() 0 for ma or min: 0 6 f() 8 f() f() f() 6 7 ma value of when min value of when f. Note that part e. is the same function but restricted to a different domain. So, from e. it is seen that the critical points are and. Now, evaluate the function, f(), at the critical values and the endpoints. Note, however, that and are not in the domain of the function. Therefore, the onl points that need to be checked are the endpoints. f() () () 8.6 f() () () 6 7 8 0.9 So, the minimum value in the interval is 0.9 when and the maimum value is.6 when.. a. v(t) t t $ 0 t, Interval # t # v() v() 6 7 vr(t) ( t )(8t) t (t ) ( t ) 0 t(t 8) 0 t 0, t v() 6 ma velocit is m>s min velocit is m>s b. v(t) t t vr(t) ( t )(8t) (t )(t) ( t ) 8t ( t ) 0 t 8t t 0 8t 8t 8t ( t ) 8t ( t ) 8 t 0 t 0 f(0) 0 is the minimum value that occurs at 0. There is no maimum value on the interval. As approaches infinit, f() approaches the horizontal asmptote. 6. N(t) 0t 0t 00 Nr(t) 60t 0 60t 0 0 t N(0) 00 N() 0(6) 0() 00 0 N(7) 0(9) 0(7) 00 90 The lowest number is 0 bacteria> cm. 7. a. E(v) 600v v 600 0 # v # 00 Er(v) 600(v 600) 600v(v) (v 600) Let Er(N) 0 for ma or min 600v 600 600 00v 0 600v 600 600 v 680-0 Chapter : Derivatives and Their Applications

E(0) 0 E(80) 0 E(00) 9.76 The legal speed that maimizes fuel efficienc is 80 km> h. b. E(v) 600v v 600 0 # v # 0 Er(v) 600(v 600) 600v(v) (v 600) Let Er(N) 0 for ma or min 600v 600 600 00v 0 600v 600 600 v 680 E(0) 0 E(0) 9 The legal speed that maimizes fuel efficienc is 0 km> h. c. The fuel efficienc will be increasing when Er(v). 0. This will show when the slopes of the values of E(v) are positive, and hence increasing. From part a. it is seen that there is one critical value for v. 0. This is v 80. v slope of E(v) 0 # v, 80 80, v # 00 Therefore, within the legal speed limit of 00 km> h, the fuel efficienc E is increasing in the speed interval 0 # v, 80. d. The fuel efficienc will be decreasing when Er(v), 0. This will show when the slopes of the values of E(v) are negative, and hence decreasing. From part a. it is seen that there is one critical value for v. 0. This is v 80. v slope of E(v) 0 # v, 80 80, v # 00 Therefore, within the legal speed limit of 00 km> h, the fuel efficienc E is decreasing in the speed interval 80, v # 00. 8. C(t) 0.t # t # 6 (t ), Cr(t) 0.(t ) 0.t(t ) 0 (t ) (t )(0.t 0. 0.t) 0 t Calculus and Vectors Solutions Manual C() 8 0.006 C() 0.008, C(6) 8 0.007 The min concentration is at t and the ma concentration is at t. 9. P(t) t 0 # t # 6t, Pr(t) (6t ) (6) 0 6 (6t ) 6 t t 80 0 8 t 8t 0 0 (8t )(8t ) 0 8 6 t t t. 0 t 8 0.0 P(0) P(0.0) 0. P().0 Pollution is at its lowest level in 0.0 ears or approimatel 8 das. 0. r() 00 a900 b rr() 00 a900 b 0 Let rr() 0 900, 70,. 0 r(0) 0.8 r(70) 0. r(0) 0.0 A speed of 70 km> h uses fuel at a rate of 0. L> km. Cost of trip is 0. 00 0. $.0.. f() 0.00 0..6 0, 0 # # 7 f r() 0.00 0..6 Set 0 0.00 0..6 0. 6 "(0.) (0.00)(.6) (0.00) 0. 6 0. 0.006 60 or 0 f(0) 0 f(0) f(60) 0 f(7) 6.87 Absolute ma. value at (0, ) and absolute min. value 0 at (0, 0) and (60, 0). -

. a. 6 8 0 8 6 b. D: # # c. increasing: #, 0, # decreasing: 0,,. Absolute ma.: Compare all local maima and values of f(a) and f(b) when domain of f() is a # # b. The one with highest value is the absolute maimum. Absolute min.: We need to consider all local minima and the value of f(a) and f(b) when the domain of f() is a # # b. Compare them and the one with the lowest value is the absolute minimum. You need to check the endpoints because the are not necessaril critical points.. C() 000 9 0.0, # # 00 Unit cost u() C() 000 9 0.0 000 9 0.0 Ur() 000 0.0 For ma or min, let Ur() 0: 0.0 000 60 000 8.9 U() 009.0 U().90 U().98 U(00). Production level of units will minimize the unit cost to $.9.. C() 6000 9 0.0 U() C() 6000 9 0.0 6000 9 0.0 Ur() 6000 0.0 Set Ur() 0 and solve for. 6000 0.0 0 0.0 6000 0 000 8 6. However, 6. is not in the given domain of # # 00. Therefore, the onl points that need to be checked are the endpoints. f() 6009.0 f(00) Therefore, a production level of 00 units will minimize the unit cost to $. Mid-Chapter Review, pp. 9 0. a. h() hr() 6 hs() 6 6 b. f() ( ) f r() 6( ) f s() ( ) 8 0 c. ( ) r ( ) s 0( ) 0.0 6000 0 ( ) d. g() ( ) gr() ( ) gs() ( ) ( ) ( ) ( ). a. s() () () 90() 7 89 70 08 b. v(t) sr(t) t t 90 v() () () 90 - Chapter : Derivatives and Their Applications

7 0 90 c. a(t) vr(t) 6t a() 6() 8. a. v(t) hr(t) 9.8t 6 The initial velocit occurs when time t 0. v(0) 9.8(0) 6 6 So, the initial velocit is 6 m> s. b. The ball reaches its maimum height when v(t) 0. So set v(t) 0 and solve for t. v(t) 0 9.8t 6 9.8t 6 t 8 0.6 Therefore, the ball reaches its maimum height at time t 8 0.6 s. c. The ball hits the ground when the height, h, is 0. h(t) 0.9t 6t 6 6 "6 9. t 9.8 Taking the negative square root because the value t needs to be positive, 6 8.67 t 9.8 t 8.0 So, the ball hits the ground at time t.0 s. d. The question asks for the velocit, v(t), when t.0. v(.0) 9.8(.0) 6 8 8.67 Therefore, when the ball hits the ground, the velocit is 8.67 m> s. e. The acceleration, a(t), is the derivative of the velocit. a(t) vr(t) 9.8 This is a constant function. So, the acceleration of the ball at an point in time is 9.8 m> s.. a. v(t) sr(t) t 6t v() () 6() 8 0 So, the velocit at time t is 0 m> s. a(t) vr(t) t a() () 0 So, the acceleration at time t is 0 m> s. Calculus and Vectors Solutions Manual b. The object is stationar when v(t) 0. v(t) 0 t 6t 0 (6t )(t ) t t, Therefore, the object is stationar at time t s and t s. Before t, v(t) is positive and therefore the object is moving to the right. Between t and t, v(t) is negative and therefore the object is moving to the left. After t, v(t) is positive and therefore the object is moving to the right. c. Set a(t) 0 and solve for t. a(t) 0 t t 7 6 t t 8. So, at time t 8. s the acceleration is equal to 0. At that time, the object is neither accelerating nor decelerating.. a. f() f r() 6 Set f r() 0 to solve for the critical values. 6 0 ( ) 0 0, Now, evaluate the function, f(), at the critical values and the endpoints. f() () () f(0) (0) (0) f() () () So, the minimum value in the interval is when 0 and the maimum value is when. b. f() ( ) f r() ( ) Set f r() 0 to solve for the critical values. 0 Now, evaluate the function, f(), at the critical values and the endpoints. f() ( ) () f() ( ) 0 f() ( ) () -

So, the minimum value in the interval is 0 when and the maimum value is when. c. f() f r() 6 Set f r() 0 to solve for the critical values. 6 0 0 ( ) 0 0 6" Note, however, that " and 0 are not in the given domain of the function. Now, evaluate the function, f(), at the critical values and the endpoints. f() () 0 f(") " (") 8 0.8 f() () So, the minimum value in the interval is 0 when and the maimum value is 0.8 when ". 6. The question asks for the maimum temperature of V. V(t) 0.000 067t 0.008 0 t 0.06 6t 999.87 Vr(t) 0.000 0t 0.07 008 6t 0.06 6 Set Vr(t) 0 to solve for the critical values. 0.000 0t 0.07 008 6t 0.06 6 0 t 8.69 900 t 9.70 9 0 Using the quadratic formula, t 8.96 and t 8 80.66. However, 80.66 is not in the domain of the function. Now, evaluate the function, V(t), at the critical values and the endpoints. V(0) 999.87 V(.96) 8 999.7 V(0) 00.79 So, the minimum value in the interval is 999.7 when temperature t.96. Therefore, at a temperature of t.96 C the volume of water is the greatest in the interval. 7. a. f() b. f r() f r() () 0 f() 8 f r() 6 8 f r() 6() 8() c. f() 7 f r() 6 f s() 6 f s() 6 d. f() 6 f r() 6 f s() 6 f s() () 6 78 e. f() 6 f r() 8 f r(0) 8(0) f. f() f r() f s() 0 6 f s() () 0() 6() 8 g. f() 6 f r() 0 9 f s() 0 8 f sa b 0a b 8a b 0 7 6 0 7 h. f() 7 f r() 9 f ra b 9a b a b 8 6 8 6 - Chapter : Derivatives and Their Applications

8. s(t) ta 6 t b ms > 9. s(t) 89t t 7 a. sr(t) 89 7 t b. 6 t t sr(t) t ss(t) 8.7 sr(0) 89 7 (0) 89 ms > sr(t) 0 89 7 t 0 7 t 89 t 8 t (8) t t 7 s c. s(7) 89(7) (7) 7 0 87 96 m d. ss(t) 8 9 t ss(8) 8 9 (8) 6 9 8 6. ms > It is decelerating at 6. m> s. 0. s(t) t t sr(t) t 6t To find when the stone stops, set sr(t) 0: 6t 0 6t t t () s() () () 8 6 m The stone travels 6 m before its stops after s. Calculus and Vectors Solutions Manual. a. h(t).9t t 0. 6 "9.8 t 9.8 t 8. or t 8 0.0 (rejected since t $ 0) Note that h(0) 0.. 0 because the football is punted from that height. The function is onl valid after this point. Domain: 0 # t #. b. h(t).9t t 0. To determine the domain, find when hr(t) 0. hr(t) 9.8t Set hr(t) 0 0 9.8t t 8. For 0, t,., the height is increasing. For., t,., the height is decreasing. The football will reach its maimum height at. s. c. h(.).9(.) (.) 0. h(.) 8..9 0. h(.) 8.9 The football will reach a maimum height of.9 m.. Optimization Problems, pp. 7. W 0.9t t 0. t 6 "() (.9)(0.) (.9) L Let the length be L cm and the width be W cm. (L W) 00 L W 0 L 0 W A L? W (0 W)(W) A(W) W 0W for 0 # W # 0 Ar(W) W 0 Let Ar(W) 0: W 0 0 W A(0) 0 A() 6 A(0) 0. -

The largest area is 6 cm and occurs when W cm and L cm.. If the perimeter is fied, then the figure will be a square.. L W Let the length of L m and the width W m. W L 600 L 600 W A L? W W(600 W) A(W) w 600W, 0 # W # 00 Ar(W) w 600 Length of the bo is 00. Width of the bo is 0. V (00 )(0 )() for domain 0 # # 0 Using Algorithm for Etreme Value, dv d (00 )(0 ) (0 )() 000 80 8 80 60 000 dv Set d 0 0 000 0 W da For ma or min, let dw 0: W 0 A(0) 0 A(0) (0) 600 0 000 A(00) 0 The largest area of 000 m occurs when W 0 m and L 00 m.. Let dimensions of cut be cm b cm. Therefore, the height is cm. 00 00 0 0 0 6 "7600 6 0 6 8.8 6 8.8 or 7.9 Reject 7.9 since 0 # # 0 When 0, V 0 8.8, V 8 80 cm 0, V 0. Therefore, the bo has a height of 8.8 cm, a length of 00 8.8 8. cm, and a width of 0 8.8. cm.. 0 A() (0 ) A() 0 Ar() 0 Set Ar() 0. 0 0 0 0 0 0 Ar(0) 0, 0 Ar(0) 0. 0 maimum: The dimensions that will maimize the rectangles area are 0 cm b 0 cm. 6. a b 6 ab 6 P a b P a a 6 a b P a 8 a P a 8a Pr 8 Set Pr 0 0 8 a 8 a a a 6 a 8 ( 8 is inadmissible) b 6 8 b 8 Pr() 6, 0 Pr(9) 8.6. 0 maimum: The rectangle should have dimensions 8 m b 8 m. -6 Chapter : Derivatives and Their Applications

7. Given: 000 000 A 000 A a b A 000 8 Ar 000 6 Set Ar 0 0 000 6 6 000 000 () 8 66.67 Ar(0) 000, 0 Ar(0) 000. 0 maimum: The ranger should build the corrals with the dimensions m b 66.67 m to maimize the enclosed area. 8. Netting refers to the area of the rectangular prism. Minimize area while holding the volume constant. V lwh V A Total A Side A Top A Side A End A A A a b A A Ar Calculus and Vectors Solutions Manual Set Ar 0 0 6 6 Ar() 9, 0 Ar(8) 9.. 0 minimum: The enclosure should have dimensions m 6 m 6 m. 9. h Let the base be b and the height be h h 000 h 000 6 Surface area h A h a 000 b 000 for domain 0 # # 0" Using the ma min Algorithm, da 000 0 d 0, 000 000 0 A 00 00 600 cm Step : At S 0, A S ` Step : At 0"0, A 000 000 0!0!0!0 000 0"0 Minimum area is 600 cm when the base of the bo is 0 cm b 0 cm and height is 0 cm. -7

0. 0 0 Let the length be and the height be. We know 00. 6"00 Omit negative area "00 for domain 0 # # 0 Using the ma min Algorithm, da d "00? (00 ) (). da Let d 0. "00 "00 0 (00 ) 0 00 0 ",. 0. Thus, ", L 0" Part : If 0, A 0 Part : If 0, A 0 The largest area occurs when " and the area is 0""00 0 0""0 00 square units.. a. Let the radius be r cm and the height be h cm. Then pr h 000 h 000 pr Surface Area: A pr prh pr pra 000 pr b da dr L = For ma or min, let 0 pr 000, pr 000 r da pr 000 dr 0. 0 r 0 # r #` When r 0, A S ` r. A 8 660.8 r S `, A S ` The minimum surface area is approimatel 66 cm when r.. b. r., h 000 p(.) 8 0.8 h d 0.8. Yes, the can has dimensions that are larger than the smallest that the market will accept.. a. cm L W ( L) cm Let the rectangle have length L cm on the cm leg and width W cm on the cm leg. A LW L B similar triangles, W 60 L W 60 W L (60 W)W A for domain 0 # W # Using the ma min Algorithm, da W 60. cm. dw 60 W 0, When W. cm, A Step : If W 0, A 0 Step : If W, A 0 The largest possible area is cm and occurs when W. cm and L 6 cm. b. 8 cm ( L) cm Let the rectangle have length L cm on the cm leg and width W cm on the 8 cm leg. A LW B similar triangles, L W L (60 0). cm. W 8 r 00 p 0 8L W 00 0 W r L Å p 8. 8-8 Chapter : Derivatives and Their Applications

(0 W)W A for domain 0 # W # 8 8 Using the ma min Algorithm, da W 0 dw 0 0W 0, 8 0 cm. When W cm, A 0 cm. 8 Step : If W 0, A 0 Step : If W 8, A 0 The largest possible area is 0 cm and occurs when W cm and L 7. cm. c. The largest area occurs when the length and width are each equal to one-half of the sides adjacent to the right angle.. a. Let the base be cm, each side cm and the height h cm. 60 60 w A h (wh) h wh B w (0 60) C Divide b!: 0 0 0. To find the largest area, substitute 0, 0, and 0. A(0) 0 A(0) 0!(0)!(0) " (0) 0 A(0) 0!(0)!(0) " (0) 8 90 The maimum area is 0 cm when the base is 0 cm and each side is 0 cm. b. Multipl the cross-sectional area b the length of the gutter, 00 cm. The maimum volume that can be held b this gutter is approimatel 00(0) or 60 000 cm.. a. A 0 h 0 A B C h From ^ ABC h cos 0 h cos 0 " w sin 0 w sin 0 Therefore, A (60 )a " b " A() 0!! " 0 # # 0, Appl the Algorithm for Etreme Values, Ar() 0!! " Now, set Ar() 0 0!! " 0. Calculus and Vectors Solutions Manual h 6 h or h Area h " ( ) " A() " Ar() " 0 # #., For ma or min, let Ar() 0, 8.0. A(0) 0, A(.0) 8., A(.) 8. The maimum area is approimatel. cm and occurs when 0.96 cm and h.09 cm. b. Yes. All the wood would be used for the outer frame. -9

. W N S E z There is a critical number at t.0 hours v t,.0.0 t..0 d9(t) 0 Graph Dec. Local Min Inc. Let z represent the distance between the two trains. After t hours, 60t, ( t) z 600t ( t), 0 # t # z dz dt dz dt 700t 00( t) 700t 00( t) z dz For ma or min, let dt 0. 700t 00( t) 0 t 0.6 When t 0, z, z t 0.6, z 600(0.6) ( 0.6) z 9 z 6 t, z!600 60 The closest distance between the trains is 6 km and occurs at 0.6 h after the first train left the station. 6. Vehicle d P Vehicle At an time after :00 p.m., the distance between the first vehicle and the second vehicle is the hpotenuse of a right triangle, where one side of the triangle is the distance from the first vehicle to P and the other side is the distance from the second vehicle to P. The distance between them is therefore d "(60t) ( 80t) where t is the time in hours after :00. To find the time when the are closest together, d must be minimized. d "(60t) ( 80t) d "600t 800t 600t d "0 000t 800t 0 000t 800 dr "0 000t 800t Let dr 0: 0 000t 800 "0 000t 800t 0 Therefore 0 000t 800 0 0 000t 800 t.0 hours There is a local minimum at t.0, so the two vehicles are closest together.0 hours after :00, or :0. The distance between them at that time is km. 7. L a + b ab w a b L a b a b L a b W ab W ab a b (a b L) A LW da Let dl a b L 0, L a b and W ab a b ca b a b d ab. The hpothesis is proven. 8. Let the height be h and the radius r. Then, pr h k, h k. pr Let M represent the amount of material, M pr prh pr prha k pr b pr k 0 # r #` r, Using the ma min Algorithm, dm k pr dr r dm Let r k r 0 or dr 0, p, When r S 0, M S ` r S `, M S ` r a k p b ab a b a L b L L r a k p b. -0 Chapter : Derivatives and Their Applications

d a k p b h Min amount of material is Ratio k pa k k p b p? (p) k? k M pa p b a k h d p b? a k p b A P 00 B 9. Cut the wire at P and label diagram as shown. Let AP form the circle and PB the square. Then, pr r p 00 And the length of each side of the square is. Area of circle pa p b p Area of square a 00 b The total area is A() where 0 # # 00. p a00 b, Ar() p a00 ba b p 00 8 For ma or min, let Ar() 0. p 00 0 8 00p p8 r A(0) 6 A() p ka p k b. a k p b? a k p b a00 b 8 0 a. The maimum area is 796 cm and occurs when all of the wire is used to form a circle. b. The minimum area is 0 cm when a piece of wire of approimatel cm is bent into a circle. 0. 0 8 6 (, ) (a, (a ) ) 8 6 0 6 8 6 An point on the curve can be represented b (a, (a ) ). The distance from (, ) to a point on the curve is d "(a ) ((a ) ). To minimize the distance, we consider the function d(a) (a ) (a 6a 6). in minimizing d(a), we minimize d since d. alwas. For critical points, set dr(a) 0. dr(a) (a ) (a 6a 6)(a 6) if dr(a) 0, a (a 6a 6)(a 6) 0 a 8a 9a 0 (a )(a 6a ) 0 6 6 "8 a, or a There is onl one critical value, a. To determine whether a gives a minimal value, we use the second derivative test: dr(a) 6a 6a 9 ds() 6 6 9 $ 0. Then, d() 7. The minimal distance is d "7, and the point on the curve giving this result is (, ). A(00) 00 p 8 796 Calculus and Vectors Solutions Manual -

. 8 6 A(a, a) D C 0 6 8 B(b, b) 6 8 Let the point A have coordinates (a, a). (Note that the -coordinate of an point on the curve is positive, but that the -coordinate can be positive or negative. B letting the -coordinate be a, we eliminate this concern.) Similarl, let B have coordinates (b, b). The slope of AB is a b a b a b. Using the mid-point propert, C has coordinates a a b, a bb. Since CD is parallel to the -ais, the -coordinate of D is also a b. The slope of the tangent at D is d given b for the epression. d Differentiating. d d d d And since at point D, a b, d d a b. But this is the same as the slope of AB. Then, the tangent at D is parallel to the chord AB.. B P(, ) 0 A 0 Let the point P(, ) be on the line 0 0. Area of ^APB 0 or 0 A() (0 ) 0, 0 # # Ar() 0. For ma or min, let Ar() 0 or 0 0,., A(0) 0 A(.) (0 )(.). A() 0. The largest area is. units squared and occurs when P is at the point (,.).. (0, k) A is (, ) and B(, ) Area where k A() (k ) k, k # # k Ar() k 6 For ma or min, let Ar() 0, 6 k 6 k! When k a k 6 k! b!, k Ma area is k 9 A D A k! k k!!! square units. (k, 0). Optimization Problems in Economics and Science, pp.. a. C(6) 7(!6 0) Average cost is 6 $.80. b. C() 7(! 0) 7! 70 Cr() 7! Cr() 7! $.07 c. For a marginal cost of $0.0/L, 7! 0. 7! 6 The amount of product is 6 L. - Chapter : Derivatives and Their Applications C B

. a. a. 6t 6 (t t ) Let Lr(t) 0, then b. L() c. d. The level will be a maimum. e. The level is decreasing.. C 000 h dc dh Set N(t) 0t t N() 60 9 N() 0 6 6 terms b. Nr(t) 0 t Nr() 0 6 terms> h c. t. 0, so the maimum rate (maimum value of Nr(t) ) is 0. 0 terms> h 6t. L(t) t t Lr(t) 6(t t ) 6t(t ) (t t ) dc dh 0, t 6. 6 6. 000 000 h therefore, 6t 6 0, t 000 000, 000 # h # 0 000 h 000 000 0, h h 000 000 h 000, h. 0. Using the ma min Algorithm, 000 # h # 0 000. When C 000 000 000 000 h 000,, 000 8 9 067. 000 000 000 When h 000, C 000 000 6000. When h 0 000, C 8 608. The minimum operating cost of $6000> h occurs when the plane is fling at 000 m. Calculus and Vectors Solutions Manual. Label diagram as shown and let the side of length cost $6> m and the side of length be $9> m. Therefore, ()(6) ()(9) 9000 00. Area A 00 But. 00 A() a b 00 for domain 0 # # 00 Ar() 00 Let Ar() 0, 7. Using ma min Algorithm, 0 # # 00, A(0) 0, A(7) 00(7) (7) 9 70 A(00) 0. The largest area is 9 70 m when the width is 0 m b 7 m. 6. Let be the number of $ increases in rent. P() (900 )(0 ) (0 )(7) P() (0 )(8 ) P() 0 0 8 P() 0 Pr() 0 Set Pr() 0 0 0 0 8. 8 or 9 Pr(0). 0 Pr(0) 7, 0 maimum: The real estate office should charge $900 $(8) $00 or $900 $(9) $ rent to maimize profits. Both prices ield the same profit margin. 7. Let the number of fare changes be. Now, ticket price is $0 $0.. The number of passengers is 0 000 00. The revenue R() (0 000 00)(0 0.), R() 00(0 0.) 0.(000 00) 000 00 000 00. -

Let Rr() 0: 00 000. The new fare is $0 $0.() $.0 and the maimum revenue is $0 00. 8. Cost C a v 6b t Where or C(v) a v 6ba00 v b 0v 08 000, where v $ 0. v 08 000 Cr(v) 00v v Let Cr(v) 0, then 00v 08 000 v 00 v 6 v 6. The most economical speed is 6 nautical miles> h. 9. h vt 00 h t 00 v. Label diagram as shown. We know that ()(h)(h) 0 000 or h 0 000 0 000 08 000 v Cost C 0(h) h(00) 00()(h h) 80h 00h 00h 00h 680h 00h 0 000 Since h, 0 000 C(h) 680ha b 00h, 0 # h # 00 h 6 800 000 C(h) 00h h 6 800 000 Cr(h) h 800h. Let Cr(h) 0, 800h 6 800 000 h 800 h 8 0.. h Appl ma min Algorithm, as h S 0 C(0) S ` 6 800 000 C(0.) 00(0.) 0. 99 800 C(00) 06 000. Therefore, the dimensions that will keep the cost to a minimum are 0. m b 0.8 m b.0 m. 0. Let the height of the clinder be h cm, the radius r cm. Let the cost for the walls be $k and for the top $k. V 000 pr or h 000 h pr The cost C (pr )(k) (prh)k or C pkr pkr a 000 pr b C(r) pkr 000k, r $ 0 r Cr(r) 8pkr 000k r Let Cr(r) 0, then 8pkr 000k r or r 000 8p r 8. h 000 p(.) 7.. Since r $ 0, minimum cost occurs when r. cm and h 7. cm.. a. Let the number of $0.0 increase be n. New price 0 0.n. Number sold 00 7n. Revenue R(n) (0 0.n)(00 7n) 000 0n.n Profit P(n) R(n) C(n) 000 0n.n 6(00 7n) 800 7n.n Pr(n) 7 7n Pr(n) 0, Let 7 7n 0, n 8 0. Price per cake 0 $ Number sold 00 70 0 b. Since 00 6, it takes price increases to reduce sales to 6 cakes. New price is 0 0. $.0. The profit is 6 $8. - Chapter : Derivatives and Their Applications

c. If ou increase the price, the number sold will decrease. Profit in situation like this will increase for several price increases and then it will decrease because too man customers stop buing.. Let be the base length and be the height. Top/bottom: $0> m Sides: $0> m 000 cm a m 00 cm b 0.00 m 0.00 0.00 A Top A Bottom A Side C 0( ) 0() C 0 0a 0.00 b C 0 0.8 Cr 80 0.8 Set Cr 0 0 80 0.8 80 0.8 0.006 8 0.8 0.00 0.8 8 0. Cr() 79.. 0 Cr() 80.8, 0 maimum The jeweller bo should be. cm 8. cm 8. cm to minimize the cost of materials.. Let be the number of price changes and R be the revenue. R (90 )(0 ) Rr (90 ) (0 ) Set Rr 0 0 (90 ) (0 ) 0 0 0 0 00 0 0 00 0 Price $90 $0 Price $0 Rr(0) 00. 0 Rr(00) 600, 0 maimum: The price of the CD plaer should be $0.. Let be the number of price changes and R be the revenue. R (7 )( 000 800), # 7. Rr 800(7 ) ()( 000 800) Set Rr 0 0 60 000 000 70 000 000 0 000 8000. Price $7 $(.) Price $8. Rr() 6000. 0 Rr() 6 000, 0 maimum: The price of a ticket should be $8... P() (000 )(000) ( 000 000 800 000 7 ) P() 000 000 000 000 000 800 000 7 P() 07 00 000 000 000 Pr() 0 0 00 000 Set Pr() 0 0 0 0 00 000 0 0 00 000 8 9.70 Pr(0) 00 000. 0 Pr(0) 000, 0 maimum: The computer manufacturer should sell 9 70 units to maimize profit. 6. P() R() C() Marginal Revenue Rr(). Marginal Cost Cr(). Now Pr() Rr() Cr(). The critical point occurs when Pr() 0. If Rr() Cr(), then Pr() Rr() Rr() 0. Therefore, the instantaneous rate of change in profit is 0 when the marginal revenue equals the marginal cost. 7. h r Calculus and Vectors Solutions Manual -

Label diagram as shown, Let cost of clinder be $k> m. V 00 pr h pr Note: Surface Area Total cost C Cost C (prh)k (pr )k But, 00 pr h or 600 pr h pr 8 pr 600 pr Therefore, h pr. 600 pr C(r) kpra pr b 8kpr ka Since h # 6, C(r) 00k r 00k r 600 pr r r # a 600 p b 8kpr 6kpr b 8kpr or 0 # r #.6 kpr Cr(r) 00k kpr r Let C r(r) 0 00k r kpr 0 r pr pr 0 r 0 p r.9 h 8 8.97 m Note: C(0) S ` C(.) 8 6.k C(.6) 8 0.6k The minimum cost occurs when r 0 cm and h is about 900 cm. 0 8. C. (.)0 8.(s 0) s C 7. 7.s 9 s 7.s 7.s 77 C 9s.s 77 7s C 9s.s To find the value of s that minimizes C, we need to calculate the derivative of C. Cr 7.s 86 s 8 0 7 (9s.s ) Let Cr 0: 7.s 86 s 8 0 7 0 (9s.s ) s 8. There is a critical number at s 8. km> h There is a local minimum for s 8., so the cost is minimized for a speed of 8. km> h. 9. v(r) Ar (r 0 r), 0 # r # r 0 v(r) Ar 0 r Ar vr(r) Ar 0 r Ar Let vr(r) 0: Ar 0 r Ar 0 r 0 r r 0 r(r 0 r) 0 r 0 or va r 0 b Aa 9 r 0 bar 0 r 0 b 7 r 0A A(r 0 ) 0 The maimum velocit of air occurs when radius is Review Eercise, pp. 6 9. Cr 7(9s.s ) (9s.s ) Cr ( s 7.s ) (9s.s ) v(0) 0 r 0. (8 0 7 9 700s s ) (9s.s ) r r 0. f() ( 77 7s)(9.s) (9s.s ) s s, 8. 8. s. 8. C9(s) 0 Graph Dec. Local Min Inc. f r() f s() 0 6-6 Chapter : Derivatives and Their Applications

. 9 7 d d 98 d d 77. s(t) t (t ). s(t) t 7 t t 7 t v(t) t a(t) 0t. s(t) t t v(t) 0t For v(t) 0, t.. Therefore, the upward velocit is positive for 0 # t,. s, zero for t. s, negative for t.. s. v(t) metres/second v sr(t) t (t ) () a ss(t) (t ) () 0 0 0 t (t ) (t ) t 0 " t *.. t +. v(t) 0 6 8 0 t (seconds) 6. a. f() 9 f r() 6 8 For ma min, fr() 0: 6( ) 0 0 or. f() 9 min 0 0 ma 7 6 The minimum value is. The maimum value is 0. b. f() f r() For ma min, f r() 0: ( ) 0 or f() 9 6 6 ma 6 min c. f() 8 f r() 8 For ma min, f r() 0: 8 9 6. f() 8 0 0 8.6 The minimum value is. The maimum value is 0. 7. a. s(t) 6 6t t v(t) 6 t s(0) 6 Therefore, the front of the car was 6 m from the stop sign. b. When v 0, t 8, s(8) 6 6(8) (8) 6 8 6 Yes, the car goes m beond the stop sign before stopping. Calculus and Vectors Solutions Manual -7

c. Stop signs are located two are more metres from an intersection. Since the car onl went m beond the stop sign, it is unlikel the car would hit another vehicle travelling perpendicular. 8. s(t) t 8 t v(t) 8(t ) (t) a(t) 6(t ) 6t()(t ) t 6(t ) 6t (t ) 6(t ) t t For ma min velocities, a(t) 0: t t 6 "6. t v(t) 6t (t ) 0 min! 6! ( ).8 The minimum value is. The maimum value is!. 9. u() 6 0.0 ur() 6 0.0 For a minimum, ur() 0 6 00 0 u() 6 0.0 6.0 0.. min 00 6 00 6. 6! 6 9! ma 6t (t ) Therefore, 0 items should be manufactured to ensure unit waste is minimized. 0. a. C() 000 i. C(00) 00 000 00 00 ii. 00 $.0 iii. Cr() The marginal cost when 00 and the cost of producing the 0st item are $.00. b. C() 0.00 0 8000 i. C(00) 60 6 000 8000 60 60 ii. 00 $6.60 iii. Cr() 0.008 0 Cr(00) 0.008(00) 0.0 Cr(0) 0.008(0) 0 $. The marginal cost when 00 is $.0, and the cost of producing the 0st item is $.. c. i. ii. iii. C()! 000 C(00) 0 000 $00 C(00) 00 00 $. C r()! Cr(00) 0 $0.0 8 $0.0 Cr(0)!0 $0.0 8 $0.0 The cost to produce the 0st item is $0.0. d. C() 00 700 i. C(00) 00 000 700 0 $70 ii. C(00) 70 00 $6.87 $6.88 iii. Cr() 0 Cr(00) 0 (0).006 $.0 Cr(0) $.0 The cost to produce the 0st item is $.0.. C() 0.00 0 6 000 Average cost of producing items is -8 Chapter : Derivatives and Their Applications

C() C() 6 000 C() 0.00 0 To find the minimum average cost, we solve C r() 0 6 000 0.00 0 6 000 000 0 000 000 000,. 0 From the graph, it can be seen that 000 is a minimum. Therefore, a production level of 000 items minimizes the average cost.. a. s(t) t 0 v(t) 6t v() 8 v(). 0, so the object is moving to the right. s() 7 0 7. The object is to the right of the starting point and moving to the right, so it is moving awa from its starting point. b. s(t) t t 0 s(0) 0 Therefore, its starting position is at 0. s() 7 6 0 v(t) t 8t v() 7 Since s() and v() are both negative, the object is moving awa from the origin and towards its starting position.. s 7t 6 t a. v 8t 6 t 8t 6 t 0 8t 6 t 6 8 t 6 t. 0 Therefore, t. 0, t. 0 Calculus and Vectors Solutions Manual b. a dv dt At t a 6, 6 Since a. 0, the particle is accelerating.. Let the base be cm b cm and the height h cm. Therefore, h 0 000. A h 0 000 But h, A() 0 000 a b 00 000, for $ 00 000 Ar(), Let Ar() 0, then 00 000 7.. 6t t Using the ma min Algorithm, A() 80 000 80 0 A(7.) 8 7 The dimensions of a bo of minimum area is 7. cm for the base and height.7 cm.. Let the length be and the width. 000 P 6 and 000 or 000 P() 6 7 000 P(), 0 # # 00( 0) 7 000 Ar() 00 000 Let Ar() 0, 7 000 6 000 8 90. Using ma min Algorithm, A(0) 0 700 70 m A(90) 8 79 m A(00) 0 060-9

The dimensions for the minimum amount of fencing is a length of 90 m b a width of approimatel 6 m. 6. 0 Let the width be w and the length w. Then, w 800 w 00 w 0, w. 0. Let the corner cuts be cm b cm. The dimensions of the bo are shown. The volume is V() (0 )(0 ) 0 800, 0 # # 0 Vr() 0 800 Let Vr() 0: 0 800 0 60 00 0 60 6 "600 00 6 8.8 or., but # 0. Using ma min Algorithm, V(0) 0 V(.) 0 cm V(0) 0. Therefore, the base is 0..6 b 0..6 The dimensions are.6 cm b.6 cm b. cm. 7. Let the radius be r cm and the height h cm. V pr h 00 A pr prh Since h 00 6 # h # pr, A(r) pr pra 00 pr b pr 000 r 0 for # r # Ar(r) pr 000. r Let Ar(r) 0, then pr 000, r 000 p r 8.. Using ma min Algorithm, A() 8 0 A(.) 8 9 A() 8 7 For a minimum amount of material, the can should be constructed with a radius of. cm and a height of 8.6 cm. 8. R km Let be the distance CB, and 8 the distance AC. Let the cost on land be $k and under water $.6k. The cost C() k(8 ).6k", 0 # # 8. Cr() k.6k ( ) () k.6k " Let C r() 0, k.6k " k 0.6 " k.6!.6.6 8 0.6 0.8,. 0 Using ma min Algorithm, A(0) 9.6k A(0.8) k(8 0.8).6k" (0.8) 9.k A(8).9k The best wa to cross the river is to run the pipe 8 0.8 or 7. km along the river shore and then cross diagonall to the refiner. 9. S A A z 8 + C 8 B Let represent the distance the westbound train is from the station and the distance of the B -0 Chapter : Derivatives and Their Applications

northbound train from the station S. Let t represent time after 0:00. Then 00t, (0 0t) Let the distance AB be z. z "(00t) (0 0t), 0 # t # dz dt (00t) (0 0t) 00 00t 0 (0( t)) dz Let that is dt 0, 00 00t 0 0( t) 0 "(00t) (0 0t) or 0 000t 8 800( t) 8 800t 88 000 t 88 or. min. 88 8 0.9 h When t 0, z 0. t 0.9 z "(00 0.9) (0 0 0.9) 76.8 km t, z 00 The closest distance between trains is 76.8 km and occurs at 0:. 0. Let the number of price increases be n. New selling price 00 n. Number sold 0 n. Profit Revenue Cost P(n) (00 n)(0 n) 70(0 n), 0 # n # 0 600 0n n Pr(n) 0 n Let Pr(n) 0 0 n 0 n.. Therefore, n or. Using ma min Algorithm, P(0) 600 P() 9 P() 9 P(0) 0 The maimum profit occurs when the portable MP are sold at $0 for 68 and at $06 for 67 portable MP.. p km 0 R C A 0 km Let represent the distance AC. Then, RC 0 and. PC " The cost: C() 00 000" 7 000(0 ), 0 # # 0 Cr() 00 000 ( ) () 7 000. Let C r() 0, 00 000 7 000 0!! 6 9( ) 7 8 8.7. Using ma min Algorithm, A(0) 00 000! 7 000(0) 000 000 A(.7) 00 000!.7 7 000(0.7) 80 7.60 A(0) 06.8. The minimum cost is $ 80 7 and occurs when the pipeline meets the shore at a point C,.7 km from point A, directl across from P.. cm h cm w A hw 8 (h 6)(w ) 8 h 6 w 8 h 6 w 8 (h 6) w h 6 h 7 h 6 w Substitute for w in the area equation and differentiate: Calculus and Vectors Solutions Manual -

h 7 A (h) h 6 A h 7h h 6 Ar (8h 7)(h 6) (h 7h) (h 6) Ar 8h 9h h 7h (h 6) Ar h 8h (h 6) Let Ar 0: h 8h (h 6) 0 Therefore, h 8h 0 Using the quadratic formula, h 7.0 cm h t, 7.0 7.0 t. 7.0 A9(h) 0 Graph Dec. Local Min Inc. There is a local minimum at h 7.0 cm, so that is the minimizing height. 8 (h 6)(w ) 8.0(w ) 7. w w. cm The dimensions of the page should be. cm 7.0 cm.. = Brick = Fence C (9 8) 9() C 0 8 000 000 000 Substitute for in the cost equation and differentiate to find the minimizing value for : 000 C 0 8 0 000 C 8 Cr 0 000 8 Cr 8 0 000 Let Cr 0: 8 0 000 0 Therefore 8 0 000 0 8 0 000 9. m, 9. 9.. 9. C9() 0 Graph Dec. Local Min Inc. There is a local minimum at 9. m, so that is the minimizing value. To find, use the equation 000 000 9.. m The fence and the side opposite it should be. m, and the other two sides should be 9. m.. Boat d Boat Dock The distance between the boats is the hpotenuse of a right triangle. One side of the triangle is the distance from the first boat to the dock and the other side is the distance from the second boat to the dock. The distance is given b the equation d(t) "(t) ( t) where t is hours after :00 d(t) "69t 88t To find the time that minimizes the distance, calculate the derivative and find the critical numbers: 78t 88 dr(t) "8t 8t Let dr(t) 0: 78t 88 "8t 8t 0 Therefore, 78t 88 0 78t 88 t.9 hours - Chapter : Derivatives and Their Applications

There is a local minimum at t.9 hours, so the ships were closest together at :.. Dundas Ancaster rest 6 stop 8 C 8 D Let the distance from C to the rest stop be and so the distance from the rest stop to D is 8, as shown. The distance from Ancaster to the rest stop is therefore " "6, and the distance from the rest stop to Dundas is "6 (8 ) "6 6 6 "00 6 So the total length of the trails is L "6 "00 6 The minimum cost can be found b epressing L as a function of and eamining its derivative to find critical points. L() "6 "00 6, which is defined for 0 # # 8 Lr() t t,.9.9 t..9 d9(t) 0 Graph Dec. Local Min Inc. "6 6 "00 6 "00 6 ( 8)"6 "(6 )(00 6 ) The critical points of A(r) can be found b setting Lr() 0: "00 6 ( 8)"6 0 (00 6 ) ( 6 6)(6 ) 00 6 6 6 6 6 0 0 6 0 0 ( 6)( 6) 0 So. and 6 are the critical points of the function. Onl the positive root is within the interval of interest, however. The minimum total length therefore occurs at this point or at one of the endpoints of the interval: L(0) "6 0 "00 6(0) 0 L(.) "6. "00 6(.). 8.8 L(8) "6 8 "00 6(8) 8 8.9 So the rest stop should be built. km from point C. 6. a. f() 6, # # 7 f r() Set f r() 0 0 f() () () 6 f() 6 f() 9 f(7) (7) (7) 6 f(7) 9 6 f(7) f() () 6 f() 6 f() Absolute Maimum: f(7) Absolute Minimum: f() b. f(), # # f r() Set f r() 0 0 0 ( ) or 0 f() () () f() 7 9 f() 8 f a b a b a b f a b 8 7 9 f a b 7 f(0) (0) (0) f(0) 0 f() () () f() 7 9 f() 6 Absolute Maimum: f() 6 Absolute Minimum: f() 8 c. f(), # # f r() Calculus and Vectors Solutions Manual -

Set f r() 0 0 or f() () () f() 60 f() 6 f() () () f() 8 f() f() () () f() 8 f() 8 f() () () f() 60 f() 67 Absolute Maimum: f() 67 Absolute Minimum: f() 6 d. f(), # # f r() Set f r() 0 0 0 ( ) 0 ( )( ) or 0 or f() () () f() 96 0 f() 6 f(0) (0) (0) f(0) 0 Note: (0, 0) is not a maimum or a minimum f() () () f() 07 0 f() 7 f() () () f() f() f() () () f() f() Absolute Maimum: f() 7 Absolute Minimum: f() 6 7. a. s(t) 0t 0.t sr(t) 0 0.9t The car stops when sr(t) 0. 0 0.9t 0 0.9t 0 0 t Å 0.9 t 8.7 (.7 is inadmissible) s(.7) 0(.7) 0.(.7) 8 6.9 m b. From the solution to a., the stopping time is about.7 s. c. ss(t).8t ss().8().6 ms > The deceleration is.6 m> s. 8. a. f r() d d ( ) So b. f r() d d ( ) So c. So d. f s () d d ( ) 0 f s() 0() 60 6 f s () d d (6 ) f s() () 6 f r() d ( ) d ( ) () 6( ) f s() d d (6( ) ) 6()( ) () 9( ) f s(0) 9((0) ) 9 f r() d d a b ( )() ()() ( ) 0 ( ) f s() d d a 0 ( ) b ( ) (0) (0)(( )) ( ) - Chapter : Derivatives and Their Applications

0 ( ) 0 So f s() ( ) 6 e. can be rewritten as f() ( ) f(). Then f r() d d (( ) ) ( ) f s() d d a ( ) b (t ) s() () 6 6 6 sr() ( ) 6 6 a ba b ( ) ( ) So f s() ( ) 08 f. f() can be rewritten as f(). Then f r() d d ( ) a b f s() d d aa b b a ba b a 9 b So f s(8) a 9 b (8) 7 9. a. s(t) t t (t )() t() sr(t) (t ) ss(t) (t ) (0) 6((t ) ) (t ) t 6 t (t ) 6 (t ) 6(t 6) (t ) (t ) (t ) ss() ( ) 8 At t, position is, velocit is 6, acceleration is 8, and speed is 6. b. s(t) t t (t )(0) () sr(t) (t ) ss(t) 0 (t ) (0) (t )() (t ) 0 (t ) s() 8 sr() 9 9 6 6 (t ) 0(t ) (t ) ( ) Calculus and Vectors Solutions Manual -

ss() 0 7 8 At t, position is, velocit is 9, acceleration is 0 7, and speed is 9. 0. a. s(t) (t t), t $ 0 a(t) b. v avg 0 ( ) v(t) (t t) (t ) c (t t) (t )(t ) (t ) d a b (t t) (t ) 6(t ) 9 (t t) (t t 6t 6t) 9 (t t) (t t ) s() s(0) 0 ( ) (0 0) 0 0 8.9 The average velocit is approimatel.9 m> s. c. v() ( ) (() ) (0) () 8.60 The velocit at s is approimatel.6 m> s. v() v(0) d. Average acceleration which is 0 undefined because v(0) is undefined. e. a() 9 ( ) (() () 6 ) 9 (0 )(9) 8 0. The acceleration at s is approimatel 0. m> s. Chapter Test, p. 60. a. 7 9 r 9 s b. f() 9 6 f r() 6 f s() 80 c. 0 r 0 s 60 60 d. f() ( 8) f r() ( 8) () ( 8) f s() ( 8)() 96( 8). a. s(t) t t 6t v(t) 9t 0t 6 v() 9(9) 0 6 7 a(t) 8t 0 a() 8() 0 b. s(t) (t ) v(t) (t ) (). a. b. 6(t ) v() 6( ) 6 a(t) (t )() (t ) a() ( ) s(t) t t v(t) t a(t) t 0 t. s s(.). (.) 0. c. t t 0 (t )(t ) 0 t or t 0 v()0 0 0 0 v()0 0 0 The speed is m> s when the position is 0. d. The object moves to the left when v(t), 0. t, 0 t,. The object moves to the left between t 0 s and t. s. -6 Chapter : Derivatives and Their Applications

e. v() 0 7 m>s v() m>s average velocit 7 m>s. a. f() f r() 0 ( ) 0 0 or Test the endpoints and the values that make the derivative 0. f() 60 6 min f(0) f() 6 8 8 f() 60 67 ma b. f() 9 9 f r() 9 9 0 9 0 9 0 9 0 6 is not in the given interval. f() 9 0 ma f() 6 min f(6) 6. 7.. a. h(t).9t t 0. hr(t) 9.8t Set hr(t) 0 and solve for t. 9.8t 0 9.8t t 8. s The graph has a ma or min at t. s. Since the equation represents a parabola, and the lead coefficient is negative, the value must be a maimum. b. h(.).9(.) (.) 0. 8.9 The maimum height is about.9 m. 6. Let represent the width of the field in m,. 0. Let represent the length of the field in m. 000 A From : 000. Restriction 0,, 00 Substitute into : A() (000 ) 000 Ar() 000. For a ma min, Ar() 0, 0 A() (000 ) 0 lim A() 0 S0 0 A(0) 000 ma 000 lim A() 0 S000 0 and 00. Therefore, each paddock is 0 m in width and 00 m in length. 7. Let represent the height. Let represent the width. Let represent the length. Volume 0 000 Cost: C 0.0() (0.0)( ) (0.0)() 0.() 0.0 0. 0. 0. 0. 0. But 0 000 000. Therefore, C() 0.a 000 b 0. 700 0., $ 0 C r() 700 0.. Calculus and Vectors Solutions Manual -7

Let C r() 0: 700 0. 0 0. 700 0 8 6.. Using ma min Algorithm, C(0) S ` C(6.) 700 6. 0.(6.) 7.. Minimum when 6.,. and 9.0. The required dimensions are 6 mm b mm b 90 mm. 8. Let = the number of $00 increases, $ 0. The number of units rented will be 0 0. The rent per unit will be 80 00. R() (80 00)(0 0) Rr() (80 00)(0) (0 0)(00) 800 000 000 000 000 00 Set Rr() 0 0 00 000 000 00.7 but $ 0 To maimize revenue the landlord should not increase rent. The residents should continue to pa $80> month. -8 Chapter : Derivatives and Their Applications