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

Info iconThis preview shows pages 1–18. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 18
This is the end of the preview. Sign up to access the rest of the document.

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 (where m < 0) through (3. 5) has equation y — 5 = m(ar * 3) or y : mm + (5 — 3m). The y—intercept is 5 ~ 3m and the w—intercept is iS/m + 3. So the triangle has area A(m) : %(5 — 3m)(~5/m + 3) : 15 — 25/(2m) 259 2 2m220(>m9 _ 2 2m. Now A’(m) ' m —§ (since m < 0). _25 m3 A"(m) > 0. so there is an absolute minimum when m = ~g. Thus. an equation of the line is y — 5 = ‘ man (memory: —§m+10. Every line segment in the first quadrant passing through (a, b) with endpoints on the sc— and y-axes satisfies an equation of the form y — b : m(:c — a). where m < 0. By % 7 0)- [(a — a) — 012 + [0 — (b — amnz. setting a: = 0 and then y : 0. we find its endpoints, A(0, b — am) and B(a — The distance d from A to B is given by d : b 2 2 2 2ab b2 2 2 2 S(m): a—— +(am—b) =a ——+—2+am 72abm+b.Thus. m m m , 2b 22 2 S (m) : -i2 7 is: +2a2m * 2ab : —3(abm — b2 —I—a2m‘1 — abmg) m m m 2 — m[b(am — b) + am3(am * b)] : $(am — b)(b+ am3) Thus, S’(m) : O <=> m : b/a orm = — f/g Since b/a > 0 and m < 0, m must equal — g/g. Since 2 W < 0. we see that S’(m) < 0 form < 7 fig and S’(m) > 0 form > — f/g. Thus. 5' has its absolute minimum value when m 2 — fi/g. That value is s(—3 g): (a+bg/§)2+(—a{/§—b>2 (a+W)2+(€/¢%+b)2 :a2+2a4/3b2/3+a2/3b4/3+a4/3b2/3+2a2/3b4/3+b2 2a2+3a4/3b2/3+3a2/3b4/3+b2 The last expression is of the form x3 + 3122/ + 33107;2 + y3 I: (m —I— y)3] with x : (12/3 and y : b2/3. so we can write it as ((12/3 + b2/3)3 and the shortest such line segment has length \/§ 2 (612/3 + b2/3)3/2. => 3/ = 120302 7 15204, so the tangent line to the curve at a: : a has slope m(a) : 120a2 — 156‘. Now m'(a) : 240cc 7 600,3 : —E~0a(a2 v 4) : —60a(a + 2)(a ~ 2). so m'(a) > 0 fora < *2. and 0 < a < 2. and m’(a) < 0 for —2 < a < 0 and a > 2. Thus. m is increasing on (A00, —2). decreasing on (—2, 0). increasing on (0. 2). and decreasing on (2. oo) . Clearly. m(a) —> —00 as a —> :I:oo. so the maximum value of m(a) must be one of the two local maxima. m(—2) or m(2). But both m(i2) and m(2) equal 120 I 22 — 15 - 24 : 480 7 240 = 240. So 240 is the largest slope. and it occurs at the points (—23 —223) and (2A 225). Note: a : 0 corresponds to a local minimum of m. SECTION 4.7 OPTIMIZATION PROBLEMS 367 47. Here 32 = h2 + (22/4. so 17,2 : s2 — (22/4. The area is A = %b t/s2 # (22/4. Let the perimeter be p. so 25 + b = p or s 2 (p — b)/2 => A(b) : gb ‘ / (p — 192/4 — b2/4 : b y/p2 — 2pb/4. Now /——2 _ _ 2 A’(b) — p 2pb bp/4 * 3pb + p . Therefore. A’(b) : 0 => 4 x/p2—2pb—4 p2—2pb 73p?) + p2 = O :> b 2 10/3. Since A’(b) > 0 for b < 17/3 and A/(b) < O for b > 17/3, there is an absolute maximum when b : p/3. But then 25 + p/3 p. so 5 12/3 > s b the triangle is equilateral. 48. See the figure. The area is given by 14(35): wmmgpmnm) : m(x+ x2+b2 —a2) for 0§m§a.Now/l'(m) : a2—w2(1+ Except for the trivial case where :c = O. a = b and A(;v) = O. we have as + Vac? + b2 — a2 > 0. Hence. cancelling this factor gives 2 2 x a —w _ => m\/$2+b2fia2:a27$2 :> a2_$2 $2+b2_a2 2:2 (x2 + b2 7 a2) : a4 i 2a2x2 —I— :64 => 332(1)? — a2) : a4 i 2a2m2 a2 Now we must check the value of A at this point as well as at the endpoints of the domain to see which gives the => x2(b2+a2):a4 :> 1-: maximum value A(0) : ax/b2 — a2. A(a) : 0 and a2 2 a2 2 a2 a2 2 A 2 2 i a 2 2 2 2 + / + b2 ‘ a2 x/a +b \/(1 +12 x/a +b a2+b2 i ab [ a2 I b2 ab(a2 + ()2) b x/a2+b2 x/a?+b2 I x/a2+b2 (12+!)2 7a 2 Since b 2 Vb2 — a2. A(a2 Va2 + b2) 2 A (0). So there is an absolute maximum when as : ——a—. In this Va? + b2 i . 2ab l . a2 + b2 case the horizontal lece should be —— and the vertical tece should be — : x/ 2 ()2. p 1/a2+b2 p /a2+b2 a + 368 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 49. Note that [ADI : |AP| + IPDI :> 5 : :c + IPDI => |PD| = 5 — 3:. Using the Pythagorean Theorem for APDB and APDC gives us L(x) = IAPI + |BP| + IC’PI : w-I— \/(5 — :02 +22 + \/('5 — 95)? +32 =m+\/a:2—10$+29+ w2—1013+34 2:75 x—5 —— + ——. V532 ~10x + 29 V202 ~10$ + 34 12 0.3 - 3 I 4 0 5 9 —0.3 From the graphs of L and L’. it seems that the minimum value of L is about L(3.59) : 935 m. :> L/(SC) : 1+ 50. We note that since c is the consumption in gallons per hour. and ’U is the velocity in miles per hour. then c _ gallons/hour # gallons 1) miles/hour mile gives us the consumption in gallons per mile‘ that is. the quantity G. 'To find the UE —cd—v 11E *0 minimum. we calculate E : : M * —dv— do do 112 ’1] 02 This is 0 when 1) E — c = 0 <=> E : E. This implies that the dv dz; 2) tangent line of C(v) passes through the origin. and this occurs when 1) m 53 mi/h. Note that the slope of the secant line through the origin and a point (1;. C(11)) on the graph is equal to 00;) and it is intuitively clear that G is minimized in the case where the secant is in fact a tangent. 51. A The total time is a T(a:) : (time from A to C) + (time from C to B) x Cd-x m+ b2+<d—a:)2 U2 ‘ : 0<$<d ’Ul d__.| viva2+w2 v2 b2+(d—$)2— v1 v2 . . sin 91 sin 02 The minimum occurs when T’(1:) = 0 :> z . ’U1 U2 [Notes T"(x) > 0| SECTION 4.7 OPTIMIZATION PROBLEMS 369 52. P Ifd = |QT|. we minimize f(61) : [PRI + [RSI : acsc 61 + bcsc 62. S . . . . _ df Differentiating With respect to 61. and setting W equal to 0. we get 1 a 1) d6 5le = 0 : *acsc61 cot 61 — bcsc62 cot 62 Q R T . d62 . . So we need to find an expresston for We can do this by observmg that IQTl : constant : acot 61 + bcot 62. 1 . , . . . . . . . . 2 2 d62 Differentiating this equation implicnly With respect to 61. we get —a csc 61 — bcsc 62 E : 0 => 2 :43: : 7%. We substitute this into the expression for dfol to get 2 2 6 , 6 t 6 —aesc61 cot61— bcsc62 cot62 —w : 0 ¢> —acsc61 cot 61 + am 2 0 ¢> b csc2 62 csc 62 cot 6 cot 6 . cot 61 csc 62 : csc 61 cot 62 <=> 1 = 2 <=> cos 61 : cos 62. Since 61 and 62 are both acute. we csc 61 csc 62 have 61 Z 02. 53. X B A y2 : x2 + 22. but triangles CDE and BOA are similar. so ‘ 4m 2/8 : 33/(4x/w — 4) :> z : 2IE/\/$ — 4. Thus, we minimize . C 2 2 2 3 f(:c)=y =3: +4.73 /(:c—4)=:c Mac—4). 4<m§8. f/(m) _ W “gm”: " 4) ‘35] 7 25520” _ 6) (ac — 4)2 (a: — 4)2 (x _ 4)2 D E when a: = 6. f’(;c) < 0 when 9: < 6. f/(m) > 0 whensc > 6. so the minimum occurs when so = 6 in. 54. Paradoxically. we solve this maximum problem by solving a minimum problem. Let L be the length of the line ACB going from wall to wall touching the inner corner C. As 6 —> 0 or 6 a we have L —> 00 and there will be an angle that makes L a minimum. A pipe of this length will just fit around the comer. Fromthediagram.L:L1+L2=9csc6+6sec6 => dL/d6=—9csc6cot6+6sec6tan6=0when 2/3 65ec6 tan6 : 90806 cot6 <=> tan36 : 9 = 1.5 <=> tan6 = {71.5. Then sec26 : 1+ and 6 csc26: 1+ (%)_2/3 NIW _ 1/2 1 2 . so the longest pipe has length L : 9[1 + 2/3] + 6[1 +( )2/3] /_% 21.07 ft. Or, use 6 : tan’1(\3/1.5) % 0852 => L : 9csc6 + 63ec6 m 21.07 ft. 370 I3 CHAPTER4 APPLICATIONS OF DIFFERENTIATION 55. It suffices to maximize tan 0. Now 3t _ ¥tanwj : 6) _ tanw+ltani9 i t—I—tanO' 1 1—tan1ptanI9 1—ttan0 $03t(1~ttan6):t+tan0 => 2t:(1+3t2)tane :> 2t 2t tanO: .Let t:t 0: I 3, 1+3t2 f() a" 1+3t2 Ht) 2(1 + 3252) e 2t(6t) 7 2(1 — 3t?) ' _ (1+3t2)2 ’ (1+3t2)2 ’ ‘ 1—3t2:0 <=> t: fisincetZO. Now f’(t) > 0 for 0 S t < f and f’(t) < 0 fort > so f has an absolute maximum whent : % 2(W3> and tan0 — e 1 0 i ’Y. Substituting fort and 6 in 3t : tan(1/1 + 0) Gives us 1 +3(1/\/§)2 “3 6 c x/g:tan(1/1+%) => 2/»: g. 55. 10 10 10 We maximize the cross—sectional area A (0) : 10h + 2(gdh) : 10h + dh : 10(105i11 6) + (10 cos 6)(105i110) =100(sin0 + Sin0 C080). 0 g 6 S % 14,09) : 100(c056‘ + cos2 0 ~ sin2 (9) :100(c0s0 + 2cos2 0 i 1) :10()(2c0s0 ~1)(c086 + 1) :Owhencosfizé <=> 6:". (c0s6#—1since0§0§§.) E Now A(0) 2 0. : 100 and : 75 \/§ x 129.9. so the maximum occurs when 6 : 5 2 . 57. From the figure. tana : -— and tanfi = . Since .73 3 — 11) O 71 '5 —1 2 a+fl+6v180 i7r,6—7r tan tan :> 5 z: 3 — :0 d6 1 < 5 1 2 2 dz. ’ 5 2 $2 2 2 (3 _ $)2 1 — 1 A P 8 £232 5 (3 i h)? 2 Nowd—e 0 —> 5 i 2 dan— x2+25_m2—6w+13 3x2 7 30:16 + 15 : 0 :> $2 —10;r, + 5 : 0 => :1: : 5 i 2 x/S. We reject the root with the + sign. since it is larger than 3. dQ/dx > 0 for so < 5 A 2 \/5 and dO/dac < 0 for a: > 5 — 2 so I9 is maximized when |API = :c : 5 e 2V5 m 0.53. _> 291? : 50—5x2730m+65 :> SECTION 4.7 OPTIMIZATION PROBLEMS 371 58. Let a: be the distance from the observer to the wall Then. from the given figure. 6 : tan71<h+d> —tan’1<g>.x > 0 => 3: :0 d0 1 [ h+d] 1 [ d]i h+d d dz _ 1+ [(h+d)/m]2 2:2 1+ (Ci/w)2 3:2 $2 + (h +d)2 + $2 +d2 dire? + (h +002] — (h + d)(w2 + d2) i h2d+ Ide — hm2 : 0 Q _ [x2 + (h + (1)2] ($2 + d?) [x2 + (h + dfl (x2 + d2) hmz = h2d+ hat2 <2 :02 : hd+ d2 ¢> x : t/d(h + d). Since d6/dar > 0 for all cc < t/d(h +d) and d6/dac < 0 for all a: > t /d(h + d). the absolute maximum occurs when ac = d(h + d). 59. In the small triangle with sides a and c and hypotenuse W, sin 6 = % and cos 6 2 In the triangle with sides b and d and hypotenuse L. sin6 : g: and c0s6 : Thus. a : Wsin6. c : Wcos6, d = Lsin6. and b = L cos 6. so the area of the circumscribed rectangle is A(6) : (a + b)(c + d) : (W sin6 + Lcos6)(W cos6 + Lsin6) = W2 sin6 cos6 + WLsin2 6 + LWc0s26 + L2 sin6 cos6 : LW sin2 6 + LW cos2 6 + (L2 + W2)sin6 c0s6 : LW(sin2 6 + C082 6) + (L2 + W2) - % ~25in6 cos6 :LW+§(L2+W2)sin20. ogegg This expression shows. without calculus, that the maximum value of A(6) occurs when sin 26 : 1 ¢> 26 : g :> 6 : So the maximum area is AGE) : LW + %(L2 + W2) : %(L2 + 2LW + W2) 2 %(L + W)2. 60. (a) Let D be the point such that a = |AD|. From the figure, sin6 : lB—bCl => IBCI 2 bcsc6 and cos6 : Egg—g" = Lig? => |BCI : (a — |AB|)seC6. Eliminating |BC| gives (a 71AB|)sec6 = bcsc6 => bcot6 = a — |AB| => |AB| : a — bcot 6. The total resistance is |ABI |BC| u i bc0t6 bcsc6 R 6 : C C : __ ( ) r? + T; C Til + T3 , , b 2 t , , (b) R (6) : C< csc1 6 V bcsc64c0t6> : bCCSC0<csi6 _ 0026) r1 r2 r1 r2 6 cot6 r4 C0t6 R’ 6 :0 <:> CSC : <=> —2 : : , . ( ) 1"? r3 r31 csc6 C080 , c306 cot6 4 , 4 R (6) > 0 42> T4 > T4 => cos6 < % and R (6) < 0 when cos6 > :—i. so there is an absolute 1 2 1 1 minimum when cos 6 : r4/rf. (c) When n : in. we have c036 : ($4. $06 = cos’1(§)4 z 790. 372 3 CHAPTERlI APPLICATIONS OF DIFFERENTIATION If k : energy/km over land, then energy/ km over water = 1.4k. So the total energy is E: 1.4kx/25+m2+k(13727),0 S a: S 13. dE 1.4192: and so — : _— _ k‘ dw (25+m2)1/2 dE ‘ 2 1/2 2 2 2 5 Set — : 0.1.4ka: : k(25+a:) :> 1.9635 2 :1: +25 => 0.962: = 25 => 3: : m 5.1. dz 0.96 Testing against the value ofE at the endpoints: E(0) : 1.419(5) + 1319 : 20k. E(5.1) z 17.9k. E(13) % 19.519. Thus, to minimize energy. the bird should fly to a point about 5.1 km from B. (b) If W/L is large. the bird would fly to a point C that is closer to B than to D to minimize the energy used flying over water. If W/L is small. the bird would fly to a point C that is closer to D than to B to minimize the . 1 dB Wat distanceofthefliht.E:W 25+x2+L13—:c :> ————:———L:Owhen g x/ ( ) dm r——25 +12 . By the same sort of argument as in part (a). this ratio will give the minimal expenditure of K _ V25 -I- m2 L _ 2: energy if the bird heads for the point as km from B. (C) For flight direct to D. an : 13. so from part (b), W/L : g m 1.07. There is no value of W/L for which the bird should fly directly to B. But note that lim+(W/L) : 00. so if the point at which E is a minimum is 3H0 close to B. then W/L is large. (d) Assuming that the birds instinctively choose the path that minimizes the energy expenditure. we can use the equation for dE/dsc = 0 from part (a) with 1.4k : c. a: : 4. and k z 1: (c)(4) : 1 ~ (25 + 42)1/2 c : MELT/4 m 1.6. 62. (a) I 0( Strength Of source 2 . Adding the intensities from the left and right Iightbulbs. (distance from source) I: + k _ k + k m2+d2 (10_$)2+d2“x2+d2 x2—20x+100+d‘2' I(.’E) : (b) The magnitude of the constant I: won’t affect the location of the point of maximum intensity. so for convenience wetakekZL W) -—m)‘2 Substituting d : 5 into the equations for I(x)and I’(m). we get 1 1 2% 2(22 7 10) 1 : — d 1’ : __ — —————- 5m x2 + 25 + m2 — 20m + 125 a" 5m (x2 + 25)2 (x2 7 201: +125)2 0.06 0-005 From the graphs. it appears that I 5(w) k APPLIED PROJECT THE SHAPE OF A CAN 373 1 1 ___ _._____ d $2+100+$2-201B+200an (c) Substituting d : 10 into the equations for I($) and I’($) gives 110(x) : HOW) _ 295 2 (x — 10) ~ (2:2 + 100)2 (m2 — 20x + 200)2 0"” 000% From the graphs. it seems that for I d : 10. the intensity is minimized at the o lo endpoints. that is. :c : 0 and :0 : 10. a The midpoint is now the most brightly lit 8014 10 —0.0006 point! (d) From the first figures in parts (b) and (c). we see that the minimal illumination changes from the midpoint (:0 2 5 with d = 5) to the endpoints (a: : 0 and :c = 10 with d : 10). 0.0365 0.023 0.0] 0 —_ l0 0 l0 0 l0 0.0325 x 0.0215 x —0.0| (1' So we try (1 : 6 (see the first figure) and we see that the minimum value still occurs at :c = 5. Next. we let d : 8 (see the second figure) and we see that the minimum value occurs at the endpoints. It appears that for some value of d between 6 and 8. we must have minima at both the midpoint and the endpoints. that is. 1(5) must equal 1(0). To find this value of d. we solve 1(0) : [(5) (with k : 1): i+_1_i 1 + 1 _ 2 d2 100+d2_25+d2 25+d2725+d2 (25+d2)(100+d2) +d2(25+d2) = 2d2(100+d2) :5 2500 + 125112 + d4 + 25d2 + d4 : 200d2 + 2d4 _> 2500 7 50012 d2 _ 50 9 d = 5 \/§ m 7.071 (for 0 g d g 10). The third figure. a graph of [(0) — [(5) with d independent. confirms that [(0) ~ 1(5) 2 0, that is. 1(0) 2 [(5), when d : 5 Thus. the point of minimal illumination changes abruptly from the midpoint to the endpoints when d z 5 APPLIED PROJECT The Shape of a Can 1. In this case. the amount of metal used in the making of each top or bottom is (2r)2 : 4T2. So the quantity we want to minimize is A : 2777/2 + 2(4r2). But V : nrzh <=> h : V/7rr2. Substituting this expression for h in A gives A : 2V/r + 8T2. Differentiating A with respect to r. we get dA/d?" = —2V/7‘2 + 161" : 0 => 167“3 : 2V : 27rr2h <=> 2 S z 2.55. This gives a minimum because d A = 16 + g > O. T dr2 7‘3 ...
View Full Document

This note was uploaded on 12/08/2009 for the course MATH 101 taught by Professor Dr.tahir during the Fall '08 term at King Fahd University of Petroleum & Minerals.

Page1 / 18

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

This preview shows document pages 1 - 18. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online