Section 3.7 Optimization Problems 1. (a)

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1 Section 3.7 Optimization Problems 303 Section 3.7 Optimization Problems 1. (a) First Number, x Second Number Product, P (110-10)= (110-0)= (110-30)= (110-40) = 800 5O (110-50)= (110-60) = 3000 (b) First Number, x Second Number Product, P (110-10)= (110-0)= (110-30)= (110-40) = (110-50)= (110-60) = (110-70)= (110-80)= (110-90)= ( )= 1000 The maximum is attained near x = 50 and 60. (c) P = x(ll0-x) = ll0x-x (d) asoo o The solution appears to be x = 55. (e)--dp = llo-x = O when x = 55. dx dp --=- <0 dx P is a maximum when x = x = 55. The two numbers are 55 and 55.

2 304 Chapter 3 Applications of Differentiation. (a) Height, x Length & Width Volume 1 4-(1) 114- (1)] = () [4 - ()1 = (3) 3[4 - (3)] = (4) 4[4- (4)1= (5) 5[4 - (5)1 = (6) 6[4 - (6)1 = 864 The maximum is attained near x = 4. (b) V = x(z4-x),0 < x 1 < (c) d~v = x(4 - x)(-) + (4 - x) = (4 - x)(4-6x) = 1(1 - x)(4 - x) = 0 when x = 1, 4 (1 is not in the domain). dv - 1(x- 16) dv -- < 0whenx = 4. dx (d) 100 When x = 4, V = 104 is maximum. o The maximum volume seems to be Let x and y be two positive numbers such that x+y=s. P = xy = x(s-x) = Sx-x dp S -- = S-x = 0whenx =--. dx dp S - < Owhenx =--. dx P is a maximum when x = y = S/. 4. Let x and y be two positive numbers such that xy = 185. S=x+y=x ds when x = 1,,/i-~. dx x ds > 0 when x = ~ dx X 3 S is a minimum when x = y = 1-x/]-~. X 5. Let x and y be two positive, numbers such that xy = S = x+3y =--+3y Y -- ds = wheny = 7. dy y ds 94 > Owheny = 7. dy y3 S is minimum when y = 7 and x = 1. Let x be a positive number. 1 S=x+x ds 1 1 dx 7f 0 when x 1. ds > 0whenx = 1. dx x 3 The sum is a minimum when x = 1 and 1Ix = 1.

3 Section 3.7 Optimization Problems Let x and y be two positive numbers such that x + y = 108. P = xy = y(108- y) = 108y- y dp -- = 108-4y = 0wheny = 7. dp --4 < 0wheny = 7. P is a maximum when x = 54 and y = 7. Let x and y be two positive numbers such that X -F y = 54. P = xy = x(54-x ) = 54x-x 3 dp dx 54-3x = 0 when x 3~/~. dp - dx 6x < 0 when x = 3~r~. The product is a maximum when x = 3~/~ and y = Let x be the length and y the width of the rectangle. x + y = 80 y = 40-x A = xy = x(40-x) = 40x-x da -- = 40-x = 0whenx = 0. da -- =- < 0whenx = Let x be the length and y the width of the rectangle. xy = 3 3 P = x+y = x+ = x+-- x dp 64 dr -~- 0 when x = 4~/~. dp 18 ~ =-- > 0whenx = 4.f~. dr X 3 P is minimum when x = y = 4~/~ ft. 1. Let x be the length and y the width of the rectangle. xy=a A P= x+y=x+(---a~= x+-- A ~,x) dp A 0 when x = ~t-~. dr x dp 4A dr - x 3 > 0 when P is minimum when x = y = ~ cm. (A square!) A is maximum when x = y = 0 m. 10. Let x be the length and y the width of the rectangle. x + y = P P - x P y- - X A = xy = x -x =--x- da P P - x = 0whenx =--. dr 4 da p - - < 0whenx =--. dr 4 A is maximum when x = y = P/4 units. (A square!) = x/(x- = 4x 4-4x * (17/4) Because d is smallest when the expression inside the radical is smallest, you need only find the critical numbers of 17 f(x) = x 4-4x + 7 f (x) = 4x 3-4 = 0 x=l By the First Derivative Test, the point nearest to (, ½)is y

4 306 Chapter 3 Applications of Differentiation d= (. - 1)~, (-5, 3) + I(x_,)_ = ~(x~+10x+ 5)+(x 4.4x~+8x+4) = ~x 4-4x ~ +x ~ +18x+9 Because dis smallest when the expression inside ~e radical is smallest, you need to find the critical n~bers of g(x) = X X 3 + X + 18x + 9 g (x) = 4x 3-1x a + x + 18 : ~(~ + 1)(~x ~- 8x + 9): 0 x = -1 By the First Derivative Test, x = -1 yields a minimum. so, (-1, 4) is olosest to (-S, 3). = x/x -7x+16 Because d is smallest when the expression inside the radical is smallest, you need only find the critical numbers of f(x) = x ~ -7x+16. f (x) = x-7 = 0 X ~ -- 7 By the First Derivative Test, the point nearest to (4, 0) is (x,,/) 16. f(x) : ~x - 8, (1, 0) = 4X -- 4x x - 8 = x/x ~ - 3x Because d is smallest when the expression inside the radical is smallest, you need to find the critical numbers of g(x) = x ~ - 3x g (x),, : x-3 : 0whenx :-- 3 g"(x),, = > 0atx =-- The point nearest to (1, 0) is 3O 17. xy = 30 ~ y =-- x A = (x + )(3--~ + 3 (see figure) x ~0 / - 0 when x = By the First Derivative Test, the dimensions (x + ) by (y + )are ( + ~)by ( + ~)(approximately by 7.477). These dimensions yield a minimum area. ~x+~ (4, 0)

5 18. xy=36~ y=-- 36 x 108 = x+9 x da -108 dx -- x +3 = 0 ~ 3x = 108 ~ x = 6, y = 6 Dimensions: 9 9 Section 3.7 Optimization Problems x + 3y = 400 is the perimeter. (see figure) 4oo-4x) ~ = loox - x da _ 8_(100 _ x]/ = 0 when x da < Owhenx = 50. dx 3 00 A is a maximum when x = 50 ft and y = -- ft. 3 y dq = kx(qo - x) = kqox - kx ~ dx dzq - kqo - kx = k(qo - x) = 0 when x = -~-. d3q -.k < 0 when x = --. dx 3 dq/dx is maximum when x = Qo/. F= + 0.0v df - O.0v ~ dv ( + 0.0v) = 0 when v = wiilo0 ~ By the First Derivative Test, the flow rate on the road is maximized when v ~ 33 mi/h. 3. (a) A = 4(area ofside) + (area of Top) (1).4 = 4(3)(11) + (3)(3) = 150in. ().4 = 4(5)(5) + (5)(5) = 150 in. (3).4 = 4(3.5)(6) + (6)(6) = 150 in. (b) V = (length)(width)(height) (1) V = (3)(3)(11) = 99in 3. () V = (5)(5)(5) = 15 in 3. (3) V = (6)(6)(3.5) = 117 in a x (c) S = 4xy + x = 150 ~ y - 4x Z = x~y = x~i-150- x~l 75 lx3 ~x ") = T x- V x = 0 ~ x = +5 x 1. xy = 45,000 (see figure) S =x+y where S is the length of fence needed. ds 490, = 0whenx = 700. dx x das 980,000. > 0 when x = 700. dx ~ x 3 S is a minimum when x = 700 m and y = 350 m. x By the First Derivative Test, x = 5 yields the maximum volume. Dimensions: 5 5 x 5 in. (A cube!)

6 308 Chapter 3 Applications of Differentiation 4 S = x + 4xy = = x - -x 3 dv x = 0 =:, x 56.5 => x = 7.5 and y 7.5. dx dv --3x < 0forx = 7.5. dx The maximum value occurs when x = y = 7.5 cm = 4y+x+~rx 3 - x - rex y= 4 A = xy + ~-\-~j.3 - x4 - xx. x + = --8x - l-x - x + x 4 8 da =8-x---x+ =8-xl+ dx 4 dx =- 1+ < 0whenx- 4+~r 3-13/(4 + r)~- ~re3/(4 + n )~ 16 _ y= 4 4+~ 16 3 The area is maximum when y = ~ ft and x - ft. 4+~ 4+~ 6. You can see from the figure that A = xy and y = ~ 8 3 when x = ( r/4] 4 + y da _ 1(6_x), = 0whenx = 3. da --1 < 0whenx = 3. dx A is a maximum when x = 3 and y = 3/.

7 Section 3.7 Optimization Problems (a) y x-1 y=+-- x-1!;=,,/x+y = x+ +~ = x x>l (b) lo ol [ (.587,4.16) o L is minimum when x =.587 and L ~ ~xy= x+ =x+~ X (x-1)-x 1 A (x) = 1 + = (x -1) ~ (x - 1) (X -- 1) = 1 x-1 =_+1 x = 0, (select x = ) Theyy = 4and A = 4. Vertices: (0, 0), (, 0), (0, 4) 8. (a) A= lbasexheight = ~(x/36-h~)(6+ h)= x/36-h(6+ h) da:l(dh,36 - h)-l/(-h)(6 + h) + (36 - h) 1/ = (36 - h)-l/~[-h(6 + h)+ (36 - h~)] = -(h + 3h -18) = -(h + 6)(h - 3) ~36 - h ~ ~36- h da = 0 when h = 3, which is a maximum by the First Derivative Test. So, the sides are ~/36 - h = 6w/ ~, an dh equilateral triangle. Area = 7w/~ sq. units.

8 310 Chapter 3 Applications of Differentiation (b) cos cr = ",J36 - h tan c~ - 6+h Area= (½/(~J36-h )(6+h) = (6+ h) tan~z = 144cos 4atana A (o 0 = 144[cos4 o~ sec~ ~z + 4 cos~(-sin a)tan c~] = 0 :~ COS 40~ sec o~ = 1= 1 4 sin o~ = 4 cos ~ a sin a tan a 4 cos o~ sin a tan c~ sin z c~ 1 - ~ ~z = 30 and A = 7~r~. (c) Equilateral triangle 9. A = xy = x,,/5- x ~ (see figure) g = ~-~--~.) = Owhenx = y =~ da x(l ]( -x 1 + "j5 - x = ( 5-x / By the First Derivative Test, the inscribed rectangle of maximum area has vertices Width: 5"v/~; Length: 5w/-~ , >--x 30. A = xy = x,jr - x (see figure) da ( r - x ) 45r -- = - 0 when x - dx 4r _ x Bythe First Derivative Test, A is maximum when the rectangle has dimensions ~/~r by (45r)/. (-r, O) (r, O)

9 Section 3.7 Optimization Problems (a) P = x+rcr = x+ n-y = 00 (b) Length, x Width, y Area, xy 10 ~(100-10) (10)-~(100-10) = ~(100-0) (0)-~(100-0) ~ ~(100-30) (30)~(100-30) ~ oo - 4ot (40)-~(100-40) ~ 158 5O ~(100-50) (50)~(100-50) ~ 159 6O ~(100-60) (60)(100-60) ~ 158 The maximum area of the rectangle is approximately 159 m. (c) A= xy= x~(100-x)--~(100x-x ) (d) A = --(100 - x). A = 0 when x = Maximum value is approximately 159 when length = 50 m and width - (e) 000 o Maximum area is approximately m (x = 50 m).

10 31 Chapter 3 Applications of Differentiation 3. V = zrh = cubic inches or h = ~ ]/-r (a) Radius, r Height Surface Area 0. z(0.) z(0.)0. + z(0.) z(0.4) z(0.4)0.4 + ~ z(0.4) ~ z(0.6) z(0.6)0.6 + z(0.6) z( ~ 59.0 (b) Radius, r Height Surface Area 0. z(0.) (o.) 0. + z(o.1_ z(0.4) z(0.4)0.4 + z(0.4) ~ z(0.6) z(0.6) z(o~ z(0.8) / (0.8) 7/ (0.8) z(1.o) z(1.0)1.0 + z(1.0) z(1.) z(1.)1. + z(1.) _ z(1.4) z(1.4) 1.4+~ z(1.6) z(1.6) z(1.6) z(1.8) z(1.8) z(1.8) z(.0) z( ~ 47.1 The minimum seems to the about 43.6 for r = 1.6. (c) S = zr + zrh zr(r + h) = rcr + = zr + m r

11 Section 3.7 Optimization Problems 313 (d) loo (1.5, 43.46) -10 The minimum seems to be for r ~ 1.5. ds = 4n-r (e) -~r r = 0 when r = 3~l/n- ~, 1.5 in. h = ~ = 3.04 in. n-r Note: Notice that h - (111/3"~ n-r n-(llln-)/3 - ~-~-~-)= r. 33. Let x be the sides of the square ends and y the length of the package. P = 4x+y = 108 ~ y = 108-4x V = xy = x(108-4x) = 108x - 4x 3 dv -- = 16x-1x = 1x(18 -x) = 0 when x = 18. dv x = -16 < 0 when x = 18. The volume is maximum when x = 18 in. and Y = 108-4(18) = 36 in. 34. V = n-rx x + n-r = 108 => x = n-r (see figure) V = n-r(108- n-r)= n-(lo8r - n-r 3) dv _ a (16r - 6n-r ) = 6n-r(36 - n-r) dr 36 = Owhenr =--andx = 36. dv 36 - n-(16-1n-r) < 0 when r = -- dr Volume is maximum when x = 36 in. and r = 36/n" ~ in. 35. V = ln-xh = ln-x(r + ~/r - x)(see figure) n x r + ~/r - x dx 3.,Jr _ x n-x (r+rx/r_x_3x)= _ 0 3.,Jr _ x ~ r + rx/tgr ~ - x - 3x = 0 (0, r) rx/-~r ~ _ x = 3x _ r 4r(r - x ) ~- 9x 4 _ 1xr + 4r 4 0 = 9x 4-8xr = x(9x - 8r ) xi~r 3 By the First Derivative Test, the volume is a maximum when ~/~r x - and h = r + x/r _ x = 4r Thus, the maximum volume is V= ln-(8r1 4rl = 3n-r3 cubicunits

12 314 Chapter 3 Applications of Differentiation 36. V = rcxh = n x(~)= rcx~r ~ -x (see figure) dx = x/ r~ - x ~ -r - 3x- = O when x = O and x 3 ~ -f~r and h = r By the First Derivative Test, the volume is a maximum when x = -- /-;. 3 Thus, the maximum volume is V = 37. No. The volume will change because the shape of the container changes when squeezed. 38. No, there is no minimum area. If the sides arex andy, then x + y = 0 ~ y = 10 - x. The area is A(x) = x(10 - x) = 10x - x. This can be made arbitrarily small by selecting x ~ V = 14 = 4~r3 + ~crh (4/3)~cr h ~ -- r n-r ;rr 3 S = 4/rr + urh = 4~r + ur(14~r 4~) as a /al ~r-~ = Owhenr =r~ ~ 1.495cm. dr 3 r~ ~~ d~s 8 56 /1 - ~+ > Owhenr = 8 8 8~r = 4~r +-- r 3 3 r The surface area is minimum when r =,3/~ cm and h = 0. The resulting solid is a sphere of radius r = cm. h

13 Section 3.7 Optimization Problems 315 A 40. V = 4000 = Z~rr 3 + rcrh r ~r 3 Let k = cost per square foot of the surface area of the sides, then k = cost per square foot of the hemispherical ends. By the Second Derivative Test, you have d~c = k n" + [~ > 0 when r = 3~7~ -. 0 The cost is minimum when r = 3~ ff and h ~ ft. 41. Let x be the length of a side of the square and y the length of a side of the triangle. 4x + 3y = 10 (10-3y) ~ ~ ~ y + 4~/-~y=0 3O y- 9 +4x/~ daa 9 + 4Vr~ -- dy ~ 8 (10-3y)(-3) + --~-y = 0 >0 30 lo~r~ A is minimum when y = and x ~"~ (a) Let x be the side of the triangle andy the side of the square. A = ~(c t-~)x: + ~(c t-~)y: where 3x + 4y = 0 When x = 0, A= 5, when x = 60/(4-,/~ + 9),A, , andwhen x = 0/3, A = Area is maximum when all 0 feet are used on the square.

14 316 Chapter 3 Applications of Differentiation (b) Let x be the side of the square andy the side of the pentagon. A = ~lcot-~lx +~(cot-~)ywhere4x + 5y= 0 = x x,0 < x_< 5. A = x / ~x = 0 x ~.6 When x = 0, A ~ 7.58, when x ~.6, A ~ 13.10, and when x = 5, A ~ 5.Area is maximum when all0 feet are used on the pentagon. (c) Let x be the side of the pentagon andy the side of the hexagon. A = ~(cot-~)x + ~(cot-~ly where 5x + 6y = 0 = cot x + 6 0_<x_<4. A = cot x = 0 x ~ When x = 0, A ~ 8.868, when x ~.0475, A ~ , and when x = 4, A ~ Area is maximum when all 0 feet are used on the hexagon (d) Let x be the side of the hexagon and r the radius of the circle. A =-~@ot~)x +~rrwhere6x+~rr =0 10 = x+~ - 0NxN~. 3 x ~ When x = 0, A ~ , when x ~ 1.748, A ~ , andwhen x = 10/3, A ~ ~ea is maximum when all 0 feet are used on the circle. In general, using all of the wire for the figure with more sides will enclose the most area. 43. Let Sbe the strength and kthe constant of proportionality. Given h + w --- 0, h = 0 - w, S = kwh S = kw(400 - w ) m k(400w - w 3) ds - k[400" - 3w " = 0 when w 0./-~ - in. and h = ~ dw \ j 0-~/-~. 3 3 ds -- =-6kw < 0whenwdw 0x/-~ 3 These values yield a maximum. in.

15 Section 3.7 Optimization Problems LetA be the amount of the power line. A = h-y+x/x +y da y x -- =-1+ = 0wheny - dy ~/x ~ + y.~-~ d~a x ~ x - > 0fory = ~-. dy ~ (x + y) 3/ The amount of power line is minimum when y = x/~f~. (0, h) h - y /(-x, o) (x, O) 45. R = Vo sin 0 g dr Vo ~ x 3n" cos 0 = 0when0- do g 4 4 dr 4v ~ sin 0 < 0 when 0 = --. do ~ g 4 By the Second Derivative Test, R is maximum when 0 = ~r/4. : 1 1 [0, 4] 46. f(x) -~x g(x) = 1-!gx 4 - -~x on (a) -1 4 (c) +/ -- z( t x 3 if(x) = x--~x = O ~ gx = ~ x = 0, xi~(in [0, 4]) The maximum distance is d = 4 when x = f (x) = x, Tangent line at (~/~, 4)is y - 4 = x/~(x - ~,/~) y = ~/~x - 4. g (x) : ¼x 3 - x, Tangent line at ("xi ~, O)is y- 0 = (¼("xi~) 3 - "x/~)(x - -xi~) y = ~,J~x - 8. The tangent lines are parallel and 4 vertical units apart. (d) The tangent lines will be parallel. Ifd(x) = f(x) - g(x), then d (x) = 0 = f (x) - g (x) implies that f (x) = g (x) at the point x where the distance is maximum.

16 318 Chapter 3 Applications of Differentiation h h ~r 47. sina =- ~ s -,0 < a <-- s sin a h tan a tana =- ~ h = tana ~ s - sin a - seca I - ksina _ ksina _ ksinacos: s: 4 sec a 4 di = _~Esin a(- sin a cos a)+ cos: a(cos a)~ da k aecos a sin a ] = ~ COS -- 4 k ae1 3 sin: a~ = -- COS -- 4 zr 3z 1 = 0 when - a, or when sin a = +_-~. 1 Because a is acute, you have sina -~ ~ h = tana = = ~/~ ft. Because (di)/(da ) = (k/4)sin a(9 sina - 7) < 0when sin a = 1/x/~, this yields a maximum. 48. Let F be the illumination at point P which is x units from source 1. ki1 ki F =-~-+ (d_ x) df -ki~ ki + dx x3 (d- x)~ X - 0 when k/~ x 3 -- k/ (d - df 6kI1 --._~ -- 6ki dx x4 (d- x)4 This is the minimum point. > 0 when x =

17 Section 3.7 Optimization Problems s -- 4yx~ + ~, ~ = ~/~ + (~ _ x/~ ~x +4 ~x -6x+10 Time = T dt x x-3 + dx ~x + 4 4~/x =0-6x + 10 x You need to find the roots of this equation in the interval [0, 3]. By using a computer or graphing utility you can determine that this equation has only one root in this interval (x = 1). Testing at this value and at the endpoints, you see that x = 1 yields the minimum time. So, the man should row to a point 1 mile from the nearest point on the coast. 50. (a) ~~~,~ I ~ 3-tanot h tanet ""~ \ Q sec a 3 - tan a 3 - tan a T - + = sec a dt 1 (b) -- = seea tan a --see a = 0 da 1 tana =-seca 1 sin a = - 7 r(o) = ~ Endpoint /arot n( Endpoint l) Minimum time: xi~ ~ hours 4

18 30 Chapter 3 Applications of Differentiation sec a 3 - tan o~ (c) r dt - sec a tan a - -- sec o~ = 0 da vl v tan o~ V 1 -- seca V depends on (d) Cost = ( sec o~)c 1 + (3 - tan a)c - sec a (3 - tan a) From above part (c), minimum cost when sin 1/Cl a = _ C ~_ ~ - /c Cl 51. x/x +4 qx T= + -6x+10 V1 11 dt -,~ x + x-3 =0 dx hx/x + 4 va/x - 6x + 10 Because x x-3 - sin 01 and = -sin ~x +4 ~x -6x+10 you have sin 01 sin 0 sin0~ sin0 _ 0 ~ V Because dt >0 dx v~(x + 4) 3/ 11(X - 6X + 10) 3/ this condition yields a minimum time. 5. r _,,Jx + d1 + x/d~_~- + (a- x) 11 dt x x-a dx 1114 X + dl v4d + (a - x) =0 Because x - sin 01 and x - a = -sin 0 ~/x + dl 4d + (a - x) you have sino~ sino _ 0 ~--- sin O~ sin VZ Because dt d~ d + dx 111(X + d1) 3/ 11[d + (a- x)] 3/ this condition yields a minimum time. >0

19 Section 3.7 Optimization Problems 31 g3. f(x) = -sinx 55. Fcos0 = k(w-fsino) y ~x I 4 (a) Distance from origin to y-intercept is. Distance from origin to x-intercept is a / ~ (b) d = ~/x +y = 4x +(_sinx) 3 (0.7967, ) Minimum distance = at x = (c) Let f(x) = d(x) = x q- ( - sin X). f (x) = X + (- sin X)(-- COS X) Setting f (x) = 0, you obtain x ~ , which corresponds to d = C(x) = k~x +4 +k(4-x) xk C (x)- ~x+ 4 k=o x = ~x +4 4X = X + 4 3X = 4 x-x/~ Or, use Exercise 50(d): -sin 0 - C 1~ 0 = 30. So, x - W/~ ~ 4-x 56. kw F= cos 0 + k sin 0 df = -kw(k cos 0 - sin 0) = 0 do (cos 0 + k sin 0) kcos0 = sin0 ~ k = tan0 ~ 0 = arctank Because cos 0 + k sin 0 1 k v/~5 + 1, the minimum force is kw kw F = - cos 0 + k sin 0 %/k + 1 V = l~rrh = 1~rr-,/144 - r ~r r r ~) + - dr 3 _ ln.i88r 7_.3r3 3 L.]144-rl F r(96 - r ) x/144 - r 0 when r O, 4x/-~. = ~ k... = = By the First Derivative Test, V is maximum when r = 4x/-~ and h = 4x/-~. Area of circle: A = at(1) = 144n" Lateral surface area of cone: S = ~(4-,,/-~)~(4~~) + 4W/-~ ( ) = 48"~n" Area of sector: 144n ,j-~a = lot = 70 0= 144a ,f~a" 7 = -~-~ ~r (3 - w/-~) ~ radians or 66

20 3 Chapter 3 Applications of Differentiation 57. (a) (b) Base 1 Base Altitude Area cos 10 8 sin cos 0 8 sin cos 30 8 sin cos 40 8 sin cos 50 8 sin cos 60 8 sin Base 1 Base Altitude Area cos 10 8 sin cos 0 8 sin cos 30 8 sin cos 40 8 sin cos 50 8 sin cos 60 8 sin cos 70 8 sin cos 80 8 sin cos 90 8 sin The maximum cross-sectional area is approximately 83.1 ft. (c) A = (a+b)~ = [-8 + ( cos 0)~ 8 sin 0 = 64(1+ cos0) sin0,0 < 0 < 90 da = 64(1 + cos 0)cos 0 + (-64 sin 0)sin 0 (d) d--~ (e) loo = 64(cos (9 + cos (9 - sin O) = 64(cos :0+ coso-1) = 64( cos 0-1)(cos 0 + 1) = 0 when 0 = 60, 180, 300. The maximum occurs when 0 = 60. (60, 83.1) Let d be the amount deposited in the bank, i be the interest rate paid by the bank, and P be the profit. P = (0.1)d -id d = ki(because d is proportional to fl) p =,(0.1)(ki)_ i(ki ) = k(0.1i _ i3) dp k(0.4i - 3i z ) = 0 when i di ~ " 3 d~p- = k(0.4-6i) < 0 when i= 0.08 (Note: k > 0). dfl The profit is a maximum when i = 8%. C= ~,--~-- 100(00 +) l<xx x -~ 30" - C =100(-400~ x3 +-~-~(x 30 Approximation: x ~ hundred units, or 4045 units P = -- ~--~-S 3 q- 6S ~ dp 3 (a) - s ~ + 1s ds 10 (b) = --~os(s - 40) = 0whenx = 0, s = 40. dp 3 -,s + 1 ds 5 ~ {0) > 0 ~ s = 0 yields a minimum. dp-~s(40) < 0 ~ s = 40 yields a maximum. The maximum profit occurs when s = 40, which corresponds to $40,000 (P = $3,600,000). d~p 3 s+1 = 0whens = 0. ds 5 The point of diminishing returns occurs when s = 0, which corresponds to $0,000 being spent on advertising. o

21 Section 3.7 Optimization Problems S~ = (4m-1) +(5m-6) +(10m-3) ds1 - (4m - 1)(4) + (5m - 6)(5) + (10m - 3)(10) dm 64 = 8m-18 = 0whenm Line: y =--x $ = 4m m m - 3 Using a graphing utility, you can see that the minimum occurs when m = 0.3. Line y = 0.3x $ = 4(0.3) (0.3) (0.3) - 31 = 4.7 mi. S, 63. $ m m I10m - 3 x/~m~ + 1 xi-~m~ + 1 x/~m~ + 1 Using a graphing utility, you can see that the minimum occurs when x ~ 0.3. Line: y ~ 0.3x $3 = 14(0 3) (0.3) (0.3) - 3 = 4.5 mi. N/(0.3) mi = 1-~ i~-- = 14--i- 0- lo-~ (0. / 4.7) (a) Label the figure so that r = x + h. Then, the areaa is 8 times the area of the region given by OPQR: h h 8x r -- X A (x) 8.,/r x 8x = - +8x=0 = 8x+8~r -x~ r _ X x - r = X4F -- X 4x 4-4xr + r 4 = x(r - x ) 5x 4-5xr + r 4 = 0 Quadratic in x. X = 5r ± 45r4 -- 0r4 = F5 - wi~ L Take positive value. x = r + 10 ~ r Critical number

22 34 Chapter 3 Applications of Differentiation [9 x (b) Note that sin--0 = _h and cos -- = -. The area A of the cross equals the sum of two large rectangles minus the common r r square in the middle. A= (x)(h) - 4h = 8xh - 4h = 8rZsin--0 cos 0 A (O) =4r( c s O - sin -- COS O = 0 ~) cos 0 = sin -- cos -- = - sin 0 tan 0 = 0 = arctan() ~ or 63.4 (c) Note that x = --[5 + ~-~], and r - x = ---[5 - w/-~). 10 x 10 x A~x)" " =~ /~x,,/ r ~ - x ~ + 4x -4r :8 (5+ ~/-~)~0(5- ~/~) + 4~0(5 + ~/~) - 4r LlO ] 5 =~r~_r + ~r 5 5 : r=[~-~ ] : r(~/-~ - 1) Using the angle approach, note that tan 0 =, sin 0 = ~5-5 and sin (~) = ~(1 - cos 0) = ~(1 - ~5]. So, A(0) = 4r sin0-sin ~ = 4r - 1--~ = 65. f(x) = x 3-3x; X < 13x X x + 36 = (x - 9)(x - 4) = (x-3)(x-)(x+)(x+3) < 0 So,-3 _< x_<-or _< x_< 3. f (x) = 3x - 3 = 3(x + 1)(x- 1) fis increasing on (-m,-1) and (1, m). So, f is increasing on [-3,-] and [, 3]. f(-) = -, f(3) = 18. The maximum value off is 18.

23 Section 3.8 Newton s Method Let a = x + and b = X 3 -- X > O. X 3 Let f(x) : / (x l/x) 6- (x 6 1Ix 6 ) /x + v4~+ (~ + ~/~) a _ b -a-b a+b = x 3 +3x x 3 + x =3x+--=3x+. x 1 1 Let g(x) : x +-,x g (x) : 1--~ = 0 :=> x : 1. g"(x) = =and g"(1): > 0. So g is a minimum at x = 1: g(1)=. x J Finally, f is a minimum of 3() = 6. Section 3.8 Newton s Method The following solutions may vary depending on the software or calculator used, and on rounding. 1. y(x) : x- 5 f (x) : x n Xl =. f(x,,) f (x.) x. f (x.) f(x) = x 3-3 f (x) : 3x Xl = 1.4 f(x,) f(x.) f(x.) n X n x. f (x,)