# 7 - 356 3 CHAPTER4 APPLICATIONS OF DIFFERENTIATION 4.7...

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Unformatted text preview: 356 3 CHAPTER4 APPLICATIONS OF DIFFERENTIATION 4.7 Optimization Problems ——__—___———_~—___——_—— 1. (a) . We needn‘t consider pairs where the ﬁrst number First Number Second Number 22 21 interchange the numbers in such cases. The is larger than the second, since we can just H 20 answer appears to be 11 and 12. but we have 19 considered only integers in the table. 18 17 16 15 14 13 12 2 3 4 5 6 7 8 9 l—|I-—\ l—lo (b) Call the two numbers x and y. Then x + y = 23. so 3/ = 23 — w. Call the product P. Then P : my : x(23 * x) : 2336 ~ m2. so we wish to maximize the function : 23x 7 :52. Since P'(:c) : 23 — 2m. we see that P'(:c) = 0 42> :c 2 § 2 11.5. Thus. the maximum value ofP is P(11.5) : (11.5)2 : 132.25 and it occurs when m : y : 11.5. 0r: Note that P"(ac) : *2 < 0 for all ac. so P is everywhere concave downward and the local maximum at z = 11.5 must be an absolute maximum. 2. The two numbers are at + 100 and x. Minimize = (:17 + 100)z = :02 + 100x. f’(z) = 22: —I— 100 : 0 2 1' : 750. Since f”(ac) : 2 > 0. there is an absolute minimum at as : —50. The two numbers are 50 and —50. . . . 1 2 — 3. The two numbers are :c and —]00. where x > 0. Minimize f(x) : :c + —-100. f’(:c) : — : ——m \$2100. 1 as The critical number is ac = 10. Since f'(a:) < 0 forO < w < 10 and f’(w) > 0 for so > 10. there is an absolute minimum at m = 10. The numbers are 10 and 10. 4. Let as > 0 and let f(w) = a: + 1/m. We wish to minimize f(a:). Now f’(.7c) : 17 ﬁ = 2 i 1) : +1)(;c — 1). so the only critical number in (0.00) is 1. f’(:E) < 0 fort) < x < 1 and f'(;c) > 0forx > 1. so f has an absolute minimum ate: 2 1, and f(1) : 2. 0r: f”(;v) : 2/223 > 0 for all a: > 0. so f is concave upward everywhere and the critical point (1, 2) must correspond to a local minimum for f. 5. If the rectangle has dimensions at and y. then its perimeter is 2m + 2y : 100 m. so y : 50 — it. Thus. the area is A = my : m(50 — :c). We wish to maximize the function A(:c) : 1(50 — cc) : 50w — x2. where 0 < x < 50. Since A'(:c) : 5O — 2:0 : —2(;v — 25). A'(:c) > 0 for O < :c < 25 and A'(;c) < 0 for 25 < ac < 50. Thus. A has an absolute maximum at :c : 25. and A(25) : 252 : 625 m2. The dimensions of the rectangle that maximize its area are :c : y : 25 m. (The rectangle is a square.) SECTION 4.7 OPTIMIZATION PROBLEMS 357 6. If the rectangle has dimensions a: and y. then its area is my : 1000 m2. so y 2 1000/33. The perimeter P : 2x + 2y : 2x + 2000/30. We wish to minimize the function P(:c) : 2m + 2000/10 for :c > 0. P'(m) : 2 — 2000/2;2 = (2/38) (x2 — 1000). so the only critical number in the domain of P is ac : M1000. P"(x) : 4000/:1c3 > 0. so P is concave upward throughout its domain and P(\/1000) = 4 x/ 1000 is an absolute minimum value. The dimensions of the rectangle with minimal perimeter are so = y : x/ 1000 = 10v 10 m. (The rectangle is a square.) 7. (a) 250 125 75 The areas of the three ﬁgures are 12.500. 12.500. and 9000 ft2. There appears to be a maximum area of at least 12.500 ftz. (b) Let 3: denote the length of each of two sides and three dividers. Let y denote the length of the other two sides. (c) Area A : length >< width : y ~ :c ((1) Length of fencing : 750 => 510 + 2y : 750 (e) 53: + 23; : 750 => y : 375 — 3.2: :» Am) : (375 7 gﬂx : 375:0 — gm? (f) A'(:c) = 375 — 5m : 0 => 2: : 75. Since A"(w) : *5 < 0 there is an absolute maximum when :0 = 75. Then 3; 2 13—5 : 187.5. The largest area is 75(3—35) = 14.0625 ft2. These values ofx and y are between the values in the ﬁrst and second ﬁgures in part (a). Our original estimate was low. 8. (a) : 1 g I (b) Let cc denote the length of the side I ____ _________________ 7 """"""" of the square being cut out. Let y ......................... ._ 3 — z . 1 5‘_l_,; Z L_; 2 2 denote the length of the base. I I I 2 I I I ......................... .. _________________________ __ l l J a The volumes of the resulting boxes are 1. 1.6875. and 2 fts. There appears to be a maximum volume of at least 2 ft3. (c) Volume V : length x width x height => V : y ~ y - ac : sag/2 (d)Lengthofcardboard:3 :> w+y+at=3 => y+2\$23 (e)y+2\$:3 :> 7423722: => V(m):x(3i2at)2 (f) V(a:) : n:(3 — 2n)2 => V'(m) — a: - 2(3 2:6)( 2) + (3 21:)2 - 1 _ (3 i 2m)[—4\$ + (3 i 250)] : (3 e 2x)(—6at + 3). so the critical numbers area: : g and x : Now 0 S a: S g and V(0) = 2 0. so the maximum is V( : : 2 ft3. which is the value found from our third ﬁgure in part (a). 35813 9. 10. 11. 12. 13. 14. CHAPTER 4 APPLICATIONS OF DIFFERENTIATION any 2 1.5 X 106. so y : 1.5 X 106/50 Minimize the amount of fencing. which is 32: + 2y = 3m + 2(1.5 x 10%) : 3x + 3 x 10% = F(a:). F'(:c) : 3 e 3 X 106/1:2 : 3(tc2 — 106) /:c2. The critical number is w = 103 and F/(ac) < 0f0r0 < a: <103 and F’(;c) > 0 ifx > 103. so the absolute minimum occurs when m : 103 and y = 1.5 x 103. The ﬁeld should be 1000 feet by 1500 feet with the middle fence parallel to the short side of the ﬁeld. Let b be the length of the base of the box and h the height. The volume is 32.000 2 bzh :> h : 32.000/b2. The surface area of the open box is S : b2 + 4hb = b2 + 4(32.000/b2)b : b2 + 4(32.000)/b. So S’(b) = 2b — 4(32.000)/h2 : 2(1)3 — 64000) /b2 : 0 <:> b : {7m : 40. This gives an absolute minimum since S’(b) < 0 ifO < b < 40 and S’(b) > 0 ifb > 40. The box should be 40 x 40 x 20. Let b be the length of the base of the box and h the height. The surface area is 1200 = b2 + 4hb : h : (1200 — b2)/(4b). The volume is V 2 b2}; : 02(1200 i 02)/4b : 300b i 173/4 :> V’(b) = 300 — gs? V’(b) 0 300 gb2 r b2 400 s b m 20. Since V/(b) > 0 for 0 < b < 20 and V'(b) < 0 for b > 20. there is an absolute maximum when b : 20 by the First Derivative Test for Absolute Extreme Values (see page 334). If b = 20. then It : (1200 — 202)/(4 - 20) = 10. so the largest possible volume is b2h : (20)2(10) : 4000 cm3. - V : lwh => 10 : (2w)(w)h : 2111211. So it : 5/1122. The COSt is 10(2w2) + 6[2(2wh) + 2(hw)] : 20w2 + 36wh. so it C(w) : 20w2 + 36w(5/w2) : 20w2 + 180/10. W 2w C'(w) : 4010 7180/03 2 40 (w3 7 gym? :> w = ﬁ/g is the critical number. There is an absolute minimum for C when w : i/g since C’(w) < 0 for 0 < w < {/g and C'(w)>0forw> {/g. C(g/g) :20(\3/g)2+ 180 W m \$163.54. - 10 = (2w)(w)h : 2w2h. so h = 5/7122. The cost is h C(w) = 10(2w2) + 6[2(2wh) + 2th + 6(2w2) w : 32w2 + 36wh : 32w2 + 180/10 2w C’(w) : 64w — 180/w2 : 4(16w3 i 45)/w2 => 111 z 3% is the critical number. C’(w) < 0 for 0 < w < 3 £5; and C’('w) > 0 form > 3 %. The minimum cost is C( 3 (a) Let the rectangle have sides :0 and y and area A. so A : any or y : A/x. The problem is to minimize the perimeter = 250 + 2y : 256 + 214/27 : Now P’(:z:) : 2 — 2A/zc2 : 2(x2 — A) /x2. So the critical number is m : Since P'(.7:) < 0 for 0 < ac < x/Z and P’(w) > 0 fora: > there is an absolute minimum at a: : The sides of the rectangle are x/Z and A/x/Z : so the rectangle is a square. 45 16) : 32(2.8125)2/3 +180/x/2.8125 :e \$191.28. (b) Let p be the perimeter and :0 and y the lengths of the sides. so 1) : 250 + 23/ :> 2y : p — 22: => y:ép—sc.TheareaisA(;c):pr—w):%px—:c2.NowA’(x):0 => %p—2:c:O => SECTION 4.7 OPTIMlZATION PROBLEMS ‘3 359 25c : —:—p :> cc : ip. Since A"(:c) : #2 < 0. there is an absolute maximum for A when as 2 ip by the Second Derivative Test. The sides of the rectangle are ﬁp and %p — i p : ip. so the rectangle is a square. 15. The distance from a point g) on the line 3/ : 4m + 7 to the origin is t/ (m — 0)2 + (y — 0)2 : «\$2 + 312. However, it is easier to work with the square of the distance; that is. 2 ‘ ’ . . v . . . . . D(:c) : («:62 + y2 ) : :02 + y2 : \$2 + (4m + 7)2. Because the distance is posttlve, its minimum value Will occur at the same point as the minimum value of D. D'(a:):2\$+2(4\$+7)(4)=34m+56.soD'(at):O <:> m:—E 17- D”(w) : 34 > 0. so D is concave upward for all :3. Thus. D has an absolute minimum at a: = — g. The point closesttothe origin is — (—f§,4( l 7) =( 177). 16. The square of the distance from a point y) on the line y : ~6z + 9 to the point (—3, 1) is D(m) : (ac + 3)2 + (y — 1)2 : (a: + 3)2 + (—6w + 8)2 : 37352 i 903: + 73. D’(:c) 2 74st — 90. so D'(x) : 0 <=> :c : %. D"(x) : 74 > 0, so D is concave upward for all x. Thus, D has an absolute minimum at :1: : %. §@ The point on the line closest to (—3‘ 1) is (373 37). From the ﬁgure. we see that there are two points that are farthest away from A(1., 0). The distance d from A to an arbitrary point P(w, y) on the ellipse is d : (ac — 1)2 + (y — 0)2 and the square of the distance is S=d2=\$2—2x+1+y2=x2 2w:1+(4 4952): 3952 23515. S/:~6w—2and5’:0 => m:7%.NowS":76<0,sowe know that Shas amaximum at as : 7%. Since ~13 a: S 1.S(—1)= 4. The distance d from (1. 1) to an arbitrary point P(:c, g) on the curve : taunt is al : (an — 1)? + (y — 1)2 and the square of the distance is S : d2 : (a: — 1)2 + (tanx —1)2. S’ = 2(x i 1) + 2(tanar —1)sec2 cc. Graphing S" on (—g, gives us a zero at :c z 0.82. and so tanm % 1.081 The point on y : tanm that is closest to (1. 1) is approximately (0.82.108). The area of the rectangle is (2\$)(2y) : 4503;. Also r2 : 1:2 + y2 so y = \/1"2 — 302. so the area is A(:c) = 4:1: \/7"2 — 122‘ Now 2 2 2 — 293 A’ an — 4( r2 m2 m > 7 4 T . The critical ( ) /T2 _ \$2 /T2 _ 9132 number is :3 2 ﬁr. Clearly this gives a maximum. 2 y : t/T2 — : t/éﬂ : ﬁr : at. which tells us that the rectangle is a square The dimensions are 25c : ﬁr and 23/ : ﬁr. 360 :I CHAPTEFM APPLICATIONS OF DIFFERENTIATION 2 2 20. The area of the rectangle is (2x)(2y) : 4mg. Now % + (3—2 = 1 gives b y = g\/a2 — at? so we maximize A(:c) : 493mm? — 322. a I _ 4_b 1 2 2 -1/2 2 2 1/2 A(m)ia 35-501 —9:) (—2\$)+(a —\$) -1] _ 4b 2 2 71/2 2 2 2 g 4b 2 2 _a(a SE) [m+a—£C]—a\/a2—_—\$2[CL—2IE] So the critical number is a: 2 fat, and this clearly gives a maximum. Then y : «Lib, so the maximum area is 4(\/i§a) (ﬁb) = 2ab. 21. The height h of the equilateral triangle with sides of length L is g L. sinceh2—I—(L/2)2:L2 :> h2:L2,:L2:%L2 : ﬁLVy ﬁll h:gL.Usingsimilartriangles. 2 : 5/2 =\/§ => ﬁm:§L—y :> ysz—x/gx :> y:§(Li2x)i The area of the inscribed rectangle is A(:c) : (2:5)3/ : x/EML 7 2m) 2 x/ng — 2 x/gsc2. where 0 g a: g L/2. Now 0 : A’(a:) : ﬁL — 4\/§w => x = : L/4. Since [1(0): A(L/2) : 0, the maximum occurs when a: : L/4. and y 2 39L 7 gL : lgL. so the dimensions are L/2 and 34éL. 22. The rectangle has area A(:c) = 2m; : 2x(8 7 \$2) : 16:0 7 2203. where ng§2\/§.NOWA’(:II):1676\$2 :0 :> x=2\/§.smce A (0) : A(2\/§ ) : 0. there is a maximum when at = 2 Then y = 1%. so the rectangle has dimensions 4 g and g. 23. T The area of the triangle is l A(:c) = %(2t)(r+m) :t(r+\$) : x/r2 —a:2(7"+ar).Then _2x 729; O—AI — + 1'2 x2 I :z A l (m) V 2 :1: +rzc 2 2 _— —T2—\$2+\/7" :L' —> 2 m2+rx2=\/7T——m2 => x2+rwzr2im2 => 0:2\$2+Tx*T2=(2\$—T)i\$+rl :> r —:c SECTION 4.7 OPTIMIZATION PROBLEMS 361 %r or a: : —’I‘. Now A0") : 0 : A(ir) => the maximum occurs where m : £7". so the triangle has height r+%r: grandbaseZHT’2 ~ (%r)2 :2,/%r2 : x/gr. m: 24. The rectangle has area my. By similar triangles 3 ; y = Z :> —4y+ 12 = 3:0 org 2 —%w+3. So the areais 14(33): \$(igﬂi + 3) : 732:2 + 3x whereO S x S 4. Now O=A’(;c) : 7%364—3 => \$22andy= Since A(0) : A(4) : 0‘ the maximum area is A(2) 2 : 3 cm2. 25. A The c linder has volume V : 7r 2 2a: . Also 3:2 + y2 : r2 => - y 23/ ( ) y2 : r2 7 :22. so V(;1:) = 7T(7’ — m2)(2m) : 27r(7"2:c — :63). where S x g 7‘, V'(a:) : 27r(r2 — 3x2) : 0 => m : Now V V(O) : V(7") : 0. so there is a maximum when :0 : r/\/§ and V(7" = 7r(7"2 — r2/3)(2r/\/§) : 47rr3/(3x/g). 26. By similar triangles. y/a: : h/r, so 3; : hat/r. The volume of the AA cylinder is 7rm2(h i y) : 7rhac2 i (wh/r)m3 : Now “ ' h V'(\$) : 2m: — (37rh/r)9:2 : ahx(2 * 323/7"). J So V'(:z:) : O :> an 2 0 or as 2 gr. The maximum clearly occurs when :1; : gr and then the volume is r 7rh:c2 — (Wh/T)£E3 = 7rhcc2(1— :c/T) : 7r(§r)2h(1 — = %7rr2hi 27. - The cylinder has surface area 2(area of the base) + (lateral surface area) = 27r(radius)2 + 27r(radius)(height) : 27ry2 + 27ry(23:)i k Nowm2+yg:r2 :> y2=r27x2 => yzx/rz—x2.sothe ‘ , surface area is 8(1) :271'(7"2 — 3:2) +47m: V73 — \$2. 0 S I g 7" : 27rr2 — 27rch + 47r<m V 7'2 — 172) Thus. S’(m) : 0 — 477x + 47r[m- %(r2 — x2)_1/2(—2:c) + (r2 — m2)1/2-1] 2 2 2 2 \$ imm—x +7“ 7w :471' imi—‘+\/T2—\$2 :47r-——————-— T2_\$2 T2_x2 S’(;c):O => xx/rg—m22r2i2932 (*) 2 (:c\/r2—;c2)2:(r2—2:v2)2 : 362 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 9:2 (7‘2 ~ x2) = r4 — 472.102 + 4:64 => r2562 — 3:4 : r4 — 4722/:2 + 455’ => 52:4 — 5r2m2 + r4 : 0. This is a quadratic equation in .102. By the quadratic formula. x2 = i-iﬁérz. but we reject the root with the + sign since it doesn’t satisfy (H. [The right side is negative and the left side is positive] So 2: = V 5:105 7". Since 8(0) : 5(7") : 0. the maximum surface area occurs at the critical number and x2 : miﬁ :> 10 2 _ ‘ . 3/2 : r — \$13 : #W :> the surface area IS 10 4—5 18/572 : 7(7’2 [2 - is 5 -I- 4 ﬁﬂ/gxsﬂ/g] : 71'1"2 [5‘35 + 2m] = 10 10 _ Perimeter=30 => 2y+w+7r(£):30 : 2 1 y — 2 (30 :1: — 15 —; The areais the area ofthe 2 rectangle plus the area of the semicircle. or my + . so A(:c) : 20(15 7 g i + émﬁ 215:6 — éxz ~ €362. 1 60 . . A'(;v)—15 (1:1)23—0 —> w—1+:/4—4+7T. A//(IIJ):*(1+E)<0.SOthnglVCSa 60 30 157T 60 + 157r — 30 — 157r 30 I . I I ‘ : :1 — — : —-——-—— : ft maxrmum The dimenswns area: 4+” ft andy 5 4+” 4+” 4+” 4+” so the height of the rectangle is half the base. 29. my : 384 :> y : 384/!B. Total area is A(:c) : (8 + x)(12 + 384/13) 2 12(40 + at + 256/:c). so A'(:z) : 12(1 7 256/552) 2 0 => 3: : 16. There is an absolute minimum when m : 16 since A'(x) < 0 for0 < a: < 16 and A’(w) > 0 for a: > 16. When :3 = 16. y : 384/16 : 24. so the dimensions are 24 cm and 36 cm. 30. my : 180. so y = 180/1 The printed area is (a: i 2)(y i 3) : (a: — 2)(180/a: ~ 3): 186 — 3.7: — 360/55 : A'(a:) : i3 + 360/sc2 : 0 when x2 : 120 => :2 = 2%. This gives an absolute maximum since A’(x) > 0 for 0 < a: < 2 v30 and A’(z) < 0 forac > 2 x/iﬁi When a: : 2 «30. y : 180/(2 x/30 ). so the dimensions are 2 v30 in. and 90/ x/ 30 in. SECTION4] OPTIMlZATIONPRDBLEMS 3' 363 31. 10 Let a: be the length of the wire used for the square The total area is x lO—x ,_ m 2 1 10—:c \/§ 10—7; «3 — : — - _ Di LEW?) A“) (4) Hi 3 >2( 3 l O-x '3 =ﬁm2+gaoimﬁ 092310 _ _ 40\/§ A’(m):lm—§(10im):0 <=> %m+%m—%§#O <=> mim.Now A(O) = (@100 m 4.81. A(lO) : 11%? = 6.25 and A(9—4%) z 2.72. so (a) The maximum area occurs when a: : 10 m. and all the wire is used for the square (b) The minimum area occurs when so = M m 4.35 m. 9+4\/§ 2 10— 2 2 10— 2 32- 10 Total area is A(at) : (E) + 7r< w) 2 \$— + x 104x 4 27r 16 47r 10—2: 1 1 5 i < <1.A’ :57 : _ _ __:0 D4 G 0—95—0 <9”) 8 2W (afﬁx W e r: ‘02:," z : 40/(4 + 7r). A(O) : 25/7r z 7.96.A(10) : 6.25. and A(40/(4 + 7r)) z 3.5. so the maximum occurs when a: : 0 m and the minimum occurs when .2: = 40/(4 + 7r) m. 33. - The volume is V = 7r7‘2h and the surface area is S(T):7TT‘2+27FTh:7TT2-l-27TT L :7r7'2—l—2K. h 71.72 T , 2 S(r):27rr——‘2/=O => 27rr3:2V : r: 3 Zcm. 1' 7r . . . - ~ / 3 V I 3 V This gives an absolute minimum smce S (7") < 0 for O < r < — and S (7’) > 0 forr > —. When 7r 7r T—3Kh—L———V —3Kcm _ 7r~ _ 7r1°2 _ 7r(V/7r)2/3 — 7r ' 34. L:8csc€+4sec9.0<t9< dL E — —8csc€ c0t9+4sec0 tanG : Owhen secGtan0=2cs06cot€ 42> tan3622 <=> tanﬁ: 3/2— <=> 6:tan’1 dL/de < 0 when 0 < 0 < tarr1 6/2 dL/d0 > 0 when tan‘1 \3/5 < 0 < so L has an absolute minimum when x a (9 : tan‘1 and the shortest ladder has length 1 /1 2 2/3 L : 8—;T +4x/1+22/3 m 16.65 ft. 4 y Another method: Minimize L2 = x2 + (4 + y)2. where 35. 36. 37. 38. 39. 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION e h2 +r2 a R2 _> V _ grzh _ §(R2 112)}; 31th If). V’ (h) = §(R2 — 3h2) : 0 when It : \$312. This gives an absolute maximum. since V’(h) > 0 forO < h < %R and V’(h) < O for h > \$11 The maximum volume is V051?) : Mam 3H3) _ 2 3 mﬁ'R. The volume and surface area ofa cone with radius 1" and height h are given by V : lwrzh and S : 7rrx/ r2 + h2. 3 We‘ll minimize A : 52 subject to V = 27. V : 27 :> gmzh : 27 => r2 = 8—; (1). 7T 81 81 812 i2 - 812 A Z 2 2 2 2 : 2 _ _ 2 : / Z : 7rr(r +h) 7r 7Tb Wh—l—h hz +817rh,soA 0 :> —h3 +817r O :> 2-812 1 2 817r : :s h3 = ~6— => h : 5/1—62 = 33/g m 3.722 From (1). h3 71' 7r 7r 81 1 2 r2*%— 8 7 r~ 3% @2632.A":6-812/h4>0.soAandhenceShasan 7r~3€/6/7r : \3/671'2 absolute minimum at these values ofr and h. v6 6W2 _ By similar triangles. % : H _ h (l). The volume of the inner cone is 7" i 2 « H7" V: §7rr h.sowe ll solve(1)forh,r ? :H—h => H _ h:H_ﬂ:M_£(R,T) (2), ' R R R ,3 2E _ _ﬂ 2* 3 Thus. V(r)_ 3r R(R 7")? 3R(Rr r) :> H mm = g—gQRr i 372) = g—RTQR 7 37"). V’( 0 0 2R 3 2R df (2)h—£(R—ER)*£(1R)—1H 7") >7" or r>r3anrom.—R 3—R3 —3. V'(r) changes from positive to negative at r = §R. so the inner cone has a maximum volume of V : érrrﬁh : §7r(gR)2(lH) : i - énR2H. which is approximately 15% of the volume of the larger cone. 3 3 27 aLv3 (v i u)3v2 — v3 : E’ : L———— : 0 b (a) 19(1)) _ => (1)) a (U _ “)2 ( ) when 2113 3th 21) 3n 1; gu. The First Derivative Test shows that this value of 11 gives the minimum value of E. S : 65h 7 \$82 c0t0 + 3523? csc6 (a) — 332 CSC26 i 332\$ 2 csc0 c0t00r gSZCSCQ (cscﬂ i ﬁcotel SECTIONllJ OPTIMIZATIONPROBLEMS C 365 d8 1 cos6 1 . . . —- : ‘ — : a 3 = O :‘> cos6 : —. The First Derivative (b) d6 0 when csc6 x/gcot 6 0 2) Sing \/_ Sing «3 Test shows that the minimum surface area occurs when 6 = cos_1 % 55°. (c) If c056 : then cot 6 : i and csc 6 = so the surface area is .. a 3 2 1 2 \/§ x/5 L 2 | ..9_ 2 J3 J2 SiGsh 53ﬁ+33 2‘5763h 2&3 .2ﬁs = 6811+ %s = 63(h + 2—355) 40. 15 km}, N Let t be the time. in hours. after 2:00 RM. The position of the boat heading W E south at time t is (0. 72025). The position of the boat heading east at time t is (A15 + 15t. 0). If D(t) is the distance between the boats at time t, we minimize f(t) : [1)(t)]2 : 202t2 + 152(2: — 1)? 20 km/h 1 I f’(t) = 8002: + 450(t i 1) : 12502: — 450 = 0 when t : {12—55% : 0.36 h. S 0.36 h x 60am" 2 21.6 min = 21 min 36 5. Since f"(t) > 0. this gives a minimum, so the boats are closest together at 2:21:36 PM. 41. HereT(m): —V\$26+25+ 5;\$.0£:c35 => T’(:c) (“MEWS El; 0 (> 8m 6W <=> 163:2 : 9(552 —l— 25) <:> a: : But % > 5. so T has no critical number. Since T(O) % 1.46 and T(5) % 1.18. he should row directly to B. 42. In isosceles triangle AOB. £0 2 180° 7 6 i 6. so ABOC : 26. The B '; distance rowed is 4 cos 6 while the distance walked is the length of arc BC : 2(26) : 46. The time taken is given by A C 4 T(6): C356+%:2ms6+0. 0363 g. T'(6):725in6+1=0 <=> sin6:1 :> 6: 5 E 6 . Check the value of T at 6 : g and at the endpoints of the domain of T; that is, 6 : 0 and 6 = T(O) 2, 2 \/§ + g m 2.26. and : g z 1.57. Therefore. the minimum value ofT is g when 6 : g; that is. the woman should walk all the way. Note that T”(6) = —2 cos 6 < 0 for 0 S 6 < so 6 : % gives a maximum time. . . . . 3k k: 43. The total illumination is 1(x) = —2 + —. 0 < 2: < 10. Then 3k :1: (10 — x)? k ﬁ—x—dhloixﬂ- 1’(x):_x—?“+ﬁ:o :> 6k(10—x)3:2km3 :> 3(10—w)3:173 :> \3/§(107;1:):\$ : 10x3/_—\?/§ac:m =>10\3/§=w+\3/§m :>10\3/§:(1+\3/3):c :> g6:106:13 1+x7§ 0<x<10. % 5.9 ft. This gives a minimum since I"(x) > 0 for 366 44. 45. .y:1+40\$3—3335 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION y The line with slope m (wh...
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