hw 8 derivatives new

Tec de Monterrey Campus Querétaro Ciencias BásicasMATEMATICS 1 Assignment 8– Derivatives

ITEMS COVERED

DERIVATIVES USING THE DEFINITION

DERIVATIVES WITHOUT CHAIN RULE

DERIVATIVES WITH CHAIN RULE

IMPLICITE DIFFERENTIATION

CRITICAL POINTS, MAXIMUMS/MINIMUMS, CONCAVITY AND INFLEXION POINTS

o MAXIMUMS AND MINIMUMS - CRITERIA OF THE FIRST DERIVATIVEo MAXIMUMS/MINIMUS, CONCAVITY, INFLECTION POINTS - CRITERIA OF THE

SECOND DERIVATIVE

APPLICATIONS OF MAXIMUM AND MINIMUM

1/19


DERIVATIVES WITHOUT CHAIN RULEDifferentiate and simplify your answers. DON’T USE CHAIN RULE:1.

f ( x )= 7 x−5

Ans: f ' ( x )= 7

2.

g( x )=8−3 x

Ans:g '( x )=−3 3.

g( x )=1−2 x−x2

Ans:g '( x )=−2−2 x

4.

f ( x )= 4 x2+x+1

Ans:f ' ( x )= 8x+15.

f ( x )= x3−3 x2+5 x−2

Ans:f ' ( x )= 3 x2−6 x+5

6.

f ( x )= 3 x4−5 x2−2

Ans:f ' ( x )= 12 x3−10x7.

f ( x )= 18x8−x4

Ans:f ' ( x )=x7−4 x3

8.

g( x )= x7−2x5+5 x3−7 x

Ans:g '( x )=7 x6−10x 4+15 x2−7

9.

F ( x )=14t4−1

2t2

Ans:F ' ( x )=t3−t

10.

V (r )= 43πr3

Ans:V '(r )= 4 πr2

11.

G( y )= y10+7 y5− y3+1

Ans:G '( y )= 10 y9+35 y4−3 y2

12.

f ( x )= 4 x4+ 1

4 x4

Ans:

f ' ( x )= 16 x3

− 1

x5

13.

F ( x )=x2+3 x+ 1

x2

Ans:

f ' ( x )=2 x+3− 2

x3

14.

f ( x )= x3

3+ 3x3

Ans:

f ' ( x )=x2− 9

x4

15.

f (s )=√3 (s3−s2 )

Ans:f ' ( x )=√3(3 s2−2 s )

16.

f ( x )= x4−5+x−2−41x−4

Ans:

f ' ( x )= 4 x3− 2

x3+164

x5

17.

f ( x )= 3

x2+ 5

x4

18.

f ( x )= 5

6 x5

2/19


Ans:

f ' ( x )=− 6

x3−20

x5Ans:

f ' ( x )=−25

6 x6

19.

f ( x )=(4 x2−3)2=16 x4−24 x2+9

f ' ( x )=64 x3−48x

20.

G( y )=(7−3 y3 )2=49−42 y3+9 y6

G '( y )=−18 y2(7−3 y3 ) 21.

y=1x

y '=− 1

x2

22.

y=( 2x3 )

2

y '=−24

x7

23.

y= (5 x )−2

y '= −2

25x3

24

y=π 4 x−4

y '=−4 π4

x5

25.

y=6x2+ x−2

y '=12 x−2 x−3

26.

y=5 x4− 1

2 x5

y '=20x3+ 5

2 x6

27.

f ( x )=10

x2+1

f ' ( x )= −20 x

( x2+1 )2

28.

f ( x )=5 (4 x−3 )−1

f ' ( x )= −20

( 4 x−3 )2

29.

G( x )= 3 x+12 x−5

G '( x )= −17

(2x−5 )2

30.

F ( x )=2−3 x7−x

F ´ (x )= −19

(7−x )2

31.

y= (6x−1 )2

y '=72 x−12

32.

y=( x4+5 x )2

y '=8x7+50 x4+50 x33.

g( t )= t2

2 t2+t+1

34.

y= x−2

x−3+x−2+1

3/19


g '( x )= t2+2 t(2t2+t+1)2

y '= x−6−2 x−3

(x−3+x−2+1)2

35.

h( v )= 1

v+v2+v3+v4

h ´ ( v )=−(1+2v+3v2+4 v3)

(v+v2+v3+v 4 )2

36.

f (u)=1u+ 1

u2+ 1

u3+ 1

u4

f ´ (u)=− 1

u2− 2

u3− 3

u4− 4

u5

37.

p( x )= 1+ 1x+ 1

x2+ 1

x3

p ´ ( x )= −1

x2− 2

x3− 3

x4

38.

f ( x )= 1

1+x+ x2+x3

f ´ ( x )=−(1−2x+3 x2)(1+x+x2+x3 )

39.

h( z )=8 z32

h ´ ( z )=12 z12

40.

g( t )=6 t53

g ´ ( t )=10 t23

41.

f ( t )=12−3 t4+4 t6

f ´ ( t )=−12 t3+24 t5

42.

f (s )=15−s+4 s2−5 s4

f ´ (s )=−1+8 s−20 s3

43.

f ( x )=3 x2+3√ x4

f ´ ( x )=6 x+ 43

3√ x

44.

g( x )=x4−4√ x3

g ´ ( x )=4 x3− 3

44√ x

45.

f ( x )=x12 (x2+x−4 )

f ' ( x )=52x

32+ 3

2x

12− 2

x12

46.

h( x )=x23 (3 x2−2x+5 )

h '( x )=8 x5

3−103x

23−10

3x

−13

47.

f ( x )= 4 x−53x+2

48.

S( x )=(3 x+1)−2= 1

9x2+6 x+1

4/19


f ' ( x )= −7

(3 x+2)2S '( x )= −6

(3x+1 )3

49.

h( z )= 8−z+3 z2

2−9 z

f ' ( x )=−27 z2+12 z+70(2−9 z )2

50.

f (w )= 2w

w3−7

f ´ (w )= −4w3−14(w3−7 )2

51.

G( v )= v3−1v3+1

G '( x )= 6v2

(v3+1 )2

52.

h( x )= (5 x−4 )2

h '( x )=50 x−40

53.

f ( x )=2ex−xe

f '( x )=2ex−exe−1

54.

f ( x )=2x−3x2

f ´ ( x )=2x ln 2−6 x55.

f ( x )=2−3ex

f ´ ( x )=−3e x

56.

f ( x )=4ex−5f ´ ( x )=4ex

57.

f ( x )= x2+33x

f ´ ( x )=2x−( x2+3 ) ln3

3x

58.

f ( x )= 5x

2x−1

f ´ ( x )=(2 x−1)5x ln 5−5x(2 )

(2x−1 )2

59.

f ( x )= ex

x

f ´ ( x )=x (ex )−ex

x2

60.

f ( x )= ln xx

f ´ ( x )=1−ln x

x2

61.

f ( x )=2 ln x−2x

f ´ ( x )=2x−2x ln2

62.

f ( x )=5x−5 ln x

f ´ ( x )=5x ln 5−5x

63.

f ( x )= 12xe x

64.

f ( x )= 23x ln x

5/19


f ´ ( x )= 12xe x+ 1

2ex f ´ ( x )= 2

3+ 2

3ln x

65.

f ( x )= x2 sen x

f ' ( x )=x2cos x+2 xsenx

66.

f ( x )= 3x2 cos x

f ' ( x )=−3 x2 senx+6 xcos x67.

f ( x )=2 ln x−log3 x

f ' ( x )=2x− 1x ln 3

68.

f ( x )=x log5 x

f ' ( x )= 1ln5

+ log5 x

69.

f ( x )=e x senx

f ' ( x )=e x cos x+ex senx

70.

f ( x )=3 sen x

f ´ ( x )=3 cos x

73.

f ( x )=2x cos x

f ' ( x )=−2x sin x+2x ln2 cos x

74.

f ( x )=2x senx

f ' ( x )=2x cos x+2x ln 2 senx75.

f ( x )=log10 x

x

f ' ( x )= 1x2 ln 10

−log10 x

x2

79.

f ( x )= sen x2− cos x

f ' ( x )= 2cos x−1

(2−cos x )2

81.

f ( x )= x cos x + sen x

f ´ ( x )=−x sen x+2 cos x

82.

f ( x )= 2 sen x cos x

f ´ ( x )=−2 sen2 x+ 2 cos2 x83.

f ( x )= sen xx

f ´ ( x )=x (cos x )−sen x

x2

84. f ( x )= x sen x − cos x

f ´ ( x )= xcos x+2 sen x

6/19


DERIVATIVES WITH CHAIN RULE

Differentiate the following functions using the Chain rule and others.1) f (x) = (2x + 1)3

f’(x) = 3(2x + 1)2(2) = 6(2x + 1)2

2) F(x) = (x2 + 4x – 5) 4

F’(x)=4(2x+4)(x2+4x-5)3

3) f (t) = ( 2t4 – 7t3 + 2t –1)2

f’ (t) = 2(2t4 – 7t3 + 2t –1) (8t3 – 21t2 + 2)

4) f (x) = (x2 + 4 ) –2

f' (x )=-4x

( x2+ 4 )3

5) f (x) = 4 cos3x – 3 sen4x

f' (x )= -12 sin3x-12cos4x

6)

7)

H ( x )= (x2−1 )(3 2)

(x2−4 )(1

2)

H' ( x )=x (x2−1 )

3

2

( x2−4 )12

+3x (x2−1)12 ( x2−4 )

12

8)

f (x) = ( x−7x+2 )

2

f' (x )=18( x-7 )( x+2 )3

9)

g( x )=√ x4−x2

g' ( x )= 4+x2

2 √ x √( 4+x2 )3

10) f ( t )=sin2 ( 3t2−1 )

f' ( t )= 12tsin(3t2−1 )cos (3t2−1)

15)

h( x )=ln √4+5x

h' (x )= 52(4+5x )

16)

g ( t )= ln2 (3t + 1)

g' ( x )=6 ln (3t+1)

(3 t+1 )17)

f ( x )= cos ( ln x)

f' (x )=−sin( ln x )

x

18)

f ( x )= ln [(5x - 3 )4(2x2+7 )3 ]

f' (x )=205x−3

+12 x

2 x2+7 39)

Dx(√9+√9−x )

= −1

4 √9−x √9+√9−x

40)

f ( x )= 3√1+√ x

f ' ( x )= 1

6 √x 3√(1+ √x )2

7/19


43)

f ( x )= x

√7−3 x

f ' ( x )= 14−3 x

2 √(7−3 x )3

44)

y=( x2+1)

3√x2+2

y '=2 x (4 x2+7 )3( x2+2)2/3

45)

g( x )=√1+4 x2

f ' ( x )= 4 x

√( 4 x2+1)

46)

f (s )=√2−3 s2

f ' ( x )=− 3 s

√2−3 s2

47)

f(x) =(5−3 x )

23

f ' ( x )=− 2

(5−3x )1

3

48)

g( x )=3√4 x2−1

f ' ( x )= 8 x

33√( 4 x2−1 )2

49)

g( y )= 1

√25− y2

f ' ( x )= y

√(25− y2 )3

50)

f ( x )=(5−2 x2 )−1

3

f ' ( x )= 4 x

3 (5−2 x2)4

3

51)

h( t )=2cos√ t

f ' ( x )= −sen√ t√t

52)

55) f (s )=(2 s3−3 s+7)4

f ' ( x )= 4(2 s3−3 s+7 )3

(6 s2−3 )

56)

F ( x )=( 4 x4−4 x2+1 )−1

3

f ' ( x )=−13(4 x4−4 x2+1 )

−43 (16 x3−8 x )

59)

Dx [( x+1) senx−x cos x ]= xcos x+xsenx+senx

60)

Dt ( sen23 t )= 6 sen3 t cos3 t

63) 64)

8/19


f ( x )=( 2xx2+1 )

2

f ' ( x )=−8 x ( x2−1 )

( x2+1)3

65) 66)

f ( x )=1+x2

senx

f ' ( x )= 2 xsenx−cos x−x2 cos xsen2 x

67)

y=e5 x

y '= 5e5 x

68)

y=e−7 x

y '=−7 e−7 x

69)

y=e−3 x2

y '=−6xe−3 x 2

70)

y=ex2−3

y '= 2xe x2−3

71)

y=ecos x

y '=−ecos x senx

72)

y=e2 sen 3 x

y '= 6e2sen 3 x cos3 x73)

y=ex sen (e x)f ' ( x )= e2 x cos (ex )+ex sen( ex )

74)

y= e x

x

f ' ( x )= xex−ex

x2

77)

y= e x−e− x

ex+e− x

y '= 4 e−2 x

(ex+e− x )2

78)

y=lne4 x−1e4 x+1

y '= 8e4 x

e8 x−1

79) 80)

y=ln(ex+e−x )

y '= e x−e− x

ex+e− x

87)

y = 35 x

y '=5 ( 35 x ) ln 3

88)

f ( t )=43 t2

9/19


f ' ( t )=6 t (43 t2) ln 491) y = ln( ln x )

y '= 1x ln x

92) y = ln (sen5x )

y '=5 cos5 xsen5 x

10/19


IMPLICITE DIFFERENTIATIONDifferentiate implicitly:

1.

2 y2+xy=x2+3 , Ans:

y '=2 x− y4 y+x

2.

xe y= y−1 , Ans:

y '= e y

1−xe y3.

yx− y

=x2+1, Ans:

y '=2 x (x− y )2+ y

x

4. y=ln (x2+ y2)

Ans:

y '= 2x

x2+ y2−2 y

5. e y=ln (x3+3 y ) Ans:

y '= 3 x2

e y ( x3+3 y )−3

6.

ln ( x+ y )−ln (x− y )=4 , Ans:

y '= yx

7.

y2e2 x+xy 3=1 , Ans:

y '=− y2+2e2 x y2e2 x+3xy

16.

cos (x− y )= ysenx , Ans:

y '=y cos x+sen ( x− y )−senx+sen ( x− y )

17.xseny+cos 2 y=cos y ,

Ans:

y '= −senyx cos y−2 sen2 y+seny

18.x cos y+ y cos x=1 , Ans:

y '= ysenx−cos ycos x−xseny

19.y2=ln xy , Ans:

y '= y

2 xy2−x 20. y=ln ( x+ y ) , Ans:

y '= 1x+ y−1

21.

x+ y2=lnxy , Ans:

y '=y2(1− x )x (2 y2+1) 22. y=e

x+ y, Ans:

y '= ex+ y

1−ex+ y

23. ln y=x+ey

, Ans:

y '= y

1− ye y 24. y = cos (exy ) Ans:

y '= − ye xy senexy

1+xexy senexy

Critical Points, Maximums/Minimums, Concavity and Inflexion Points

MAXIMUMS AND MINIMUMS - CRITERIA OF THE FIRST DERIVATIVEIn the following exercises :

a) Determine the interval where the function is increasing/decreasing.b) Determine the critical points.c) Determine if the critical points corresponds to a relative maximum/minimum using the criteria of the first

derivative.

1. f ( x )=x3−3 x2

Ans:a)Function is increasing in: (−∞ , 0 )∪(2 , ∞ ) Decreasing in ( 0 , 2 )b) The critical points are: x = 0 , x = 2.c) x = 0 is a local maximum; x = 2 is a local minimum.

2.f ( x )=1

3x3−x2−8 x+1

Ans:a) Function is increasing in:: (−∞ ,-2) u(4,∞ ) Decreasing in : (-2,4)b) The critical points are: x =-2 , x =4, c) f(-2) is a local maximum y f(4) is a local minimum

3. f ( x )=x 4−4 x3+9

Ans:a) Function is increasing in:: [0,+) Decreasing in (-,0]b) The critical points are: x = 3, 0 c) x = 3 is a local minimum

11/19


4. f ( x )=4 x5−10 x4+2Ans:

a)Function is increasing in: (−∞ ,0 )∪(2 ,∞)

Decreasing in (0,2)b) The critical points are: x = 0 , x =2, c) x=0, máx local

x=2, min. local

5.f ( x )=x+ 1

xAns:a)Function is increasing in:

(−∞ ,−1 ) U (1 ,∞ ) Decreasing in (−1 , 0 ) U ( 0,1 )b) The critical points are: x =-1 , x = 1

d) x = -1 is a local maximum x = 1 is a local maximum

6.f ( x )= 5

x2+1Ans:

a)Function is increasing in: (−∞ , 0 )

Decreasing in: (0 , ∞ )b) Puntos es : x = 0 c) x = 0 is a local maximum

7. f ( x )=x3−3 x

Ans:a)Function is increasing in:

(-∞ ,−1 )∪(1 ,∞) Decreasing in (-1,1)b) The critical points are: x =-1 , x =1, c) f(-1) is a local maximum, f(1) is a local minimum

8.

f ( x )=13x3−1

2x2+1

Ans:a)Function is increasing in: (-,0] U [1, ) Decreasing in [0,1]b) The critical points are: x = 0, x = 1, c) f(0)=1 es máximo local y f(1)= = 5/6 es mínimo local

9. f ( x )=−x3+3 x2+9 x−1

Ans:a)Function is increasing in: (-1,3)

Decreasing in (−∞ ,−1)∪(3 ,∞) b) The critical points are: x = -1 , x = 3, c) x=-1, min.

X=3, máx.

10. f ( x )=( x2−1 )3

Ans:

a)Function is increasing in: (0 , ∞ ) Decreasing in : (−∞ , 0)b) The critical points are: x = 1 ,x =0 ,x = 1 c) x = 0 is a local minimum, x = 1 neither min, nor max loc

graph

11. f ( x )=x3+3 x2+3 x−3

Ans:a)Function is increasing in: R Decreciente nuncab) The critical points are: x =-1 c) Doesn´t have local extrems

MAXIMUMS/MINIMUS, CONCAVITY, INFLECTION POINTS - CRITERIA OF THE SECOND DERIVATIVE In the following exercises :

a) Determine the interval where the function is increasing/decreasing.b) Determine the critical pointsc) Determine if the critical points corresponds to a relative maximum/minimum using the criteria of the

SECOND derivatived) Determine the points of inflection , intervals where the function is concave upwards , downwards and

points of inflection.e) Sketch the graph of the function based on the above information.

1. f ( x )=x3+x−3Ans:a)Function is increasing in: Reals Decreasing in : Doesn’t existb) The critical points are: Doesn’t exist

2.

f ( x )= 14x4+ 4

3x3+2 x2

Ans:

3. f ( x )=2x4−16 x2+3Ans:a)Function is increasing in: Decreasing in b) The critical points are: x =-2, x

12/19


c) Points of inflection: f(0) = -3

d) The interval where the function is concave upwards: (0, )

The interval where the function is concave downwards:

(-,0)

a)Function is increasing in: (0 ,∞)

Decreasing in (−∞ ,−2 )∪(−2,0 )b) The critical points are: x = 0 ( Mín ) , x = -2, c) Points of inflection:

x=-2, x=-2/3

d) The interval where the function is concave upwards:

(−∞ ,−2 )∪(−23,∞)


(-2,-2/3)

=0, x =-2c) mínimo en x =-2 , x =-2

máximo en x =0 .

d) Points of inflection:

± 2

√3

The interval where the function is concave

upwards:(−∞ ,− 2

√3 ), ( 2

√3, ∞)


downwards: (− 2

√3,

2

√3 )

4. f ( x )=( x2−1 )2

Ans:

a)Function is increasing in: (−1 , 0 )∪(1 , ∞ )

Decreasing in: (−∞ , −1 )∪(0 , 1 )b) The critical points are: x = 0 (Máx. loc.) , x = 1 (mín loc), x

c) Points of inflection:

x=± 1

√3

d) The interval where the function is concave upwards: (−∞ , −1 )∪(1 , ∞ )

The interval where the function is concave downwards: (−1 , 1 )

5. f ( x )=x 4+4 xAns:a)Function is increasing in:

(−1,∞) Decreasing in (-∞ ,−1 )b) The critical points are: x =-1 c) Points of inflection: no tiene

d) The interval where the function is concave upwards: R

6. f ( x )=−3x 4+8 x3−6 x2−2Ans:a)Function is increasing in: (-,0] Decreasing in [0, )b) The critical points are: x = 0 , x = 1, c) Points of inflection: x = 1/3, 1

d) The interval where the function is concave upwards: (1/3,1)

The interval where the function is concave downwards: (-,1/3) U (1, )

7. f ( x )=4 x5−5 x4

Ans:

a)Function is increasing in: (−∞ ,0 )∪(1 ,∞)

Decreasing in (0,1)b) The critical points are: x =0 ( max ), x =1 ( mín )c) Points of inflection:

x=3/4


8. f ( x )=−x2 (x−3 )2Ans:a)Function is increasing in: Decreasing in b) The critical points are: x = 0 , x =3/2, x =3c) Points of inflection:

x=3+√32

, x=3−√32

9. f ( x )= (x−2 )2 ( x+3 )3Ans:a)Function is increasing in: (−∞ , 0 )∪(2 , ∞ )

Decreasing in (−∞ , −1 )∪(0 , 2 )b) The critical points are: x = 3, x = 0,x = 2 x = 0 máx. loc.; x = 2 mín rel.

13/19


( 34,∞)


(−∞ ,0 )∪(0 ,34)



c) Points of inflection: X = 3 ,

x=±√ 32

d) The interval where the function is concave

upwards: (−3 ,−√ 3

2 )∪(√ 32, ∞)


upwards: (−∞ , −3 )∪(−√ 3

2, √ 3

2 )Graph

10.

f ( x )=6x 4+2 x 3 −12x2+3Ans:a)Function is increasing in:

(−1 .132 ,0 )∪ (0 .882 ,∞ ) Decreasing in

(−∞ ,−1. 132 )∪(0 ,0 . 882 )b) The critical points are: x = 0 ( máx ) , x = 0.882 ( mín), x = -1.132 ( mín),c) Points of inflection:

x=−23

x=12


(−∞ ,−23 )∪(1

2,∞)


(−23,12 )

11.f ( x )=1

3x3−2 x2−12 x

Ans:12.f ( x )= 1

4x4−2x2

Ans:

13. f ( x )=6x5−10 x3

Ans:a)Function is increasing in: x < -1

14/19


a)Function is increasing in: (−∞ , −2 )∪(6 , ∞ )

Decreasing in (−2 , 6 )b) The critical points are: x =-2 , x = 6, c) Points of inflection: x = 2

d) The interval where the function is concave upwards: (2 , ∞ )


(−∞ , 2 )

a)Function is increasing in: (−2 , 0 )∪(2 , ∞ )

Decreasing in: (−∞ , −2 )∪( 0 , 2 )b) The critical points are: x = 2 ( mín. loc.), x = 0 ( máx. loc.)


x=± 2

√3


upwards: (−∞ ,− 2

√3 )∪( 2

√3, ∞)


downwards: (− 2

√3,

2

√3 )

y x > 1, Decreasing in -1 < x < 0 y 0 < x < 1, b) The critical points are: x =-1 , x =0, x=1

c) Points of inflection: x=−√2

2 x=0,

x=√22

d) The interval where the function is concave upwards: (0,∞ )

(−√22,0)∪(√2

2,∞)


downwards:(−∞ ,−√2

2 )∪(0 , √22 )

14. f ( x )=2x3−9 x2+2Ans:a)Function is increasing in: (-,0]U[3, ) Decreasing in [0,3]b) The critical points are: x = 0 , x =3, c) Points of inflection: 3/2

d) The interval where the function is concave upwards: (3/2,+ )


(-,3/2)

15. f ( x )=x3−3 x2−9 xAns:a)Function is increasing in:

(−∞ ,−1 )∪(3 ,∞ ) Decreasing in (−1,3 )b) The critical points are: x = - 1 ( máx ), x = 3 (mín), c) Points of inflection: x = 1


upwards: (1 ,∞ )


downwards: (−∞ ,1 )

16.f ( x )=1

5x5−5

3x3+4 x+1

Ans:a)Function is increasing in:

(−∞ , −2 ) U (−1 , 1 ) U (2 ,∞) Decreasing in (−2 ,−1 ) U (1, 2 )b) The critical points are:

x=±2 , x=±1


x=0 , x=±√ 52


downwards:(−∞ , −√ 5

2 ) U (0 , √ 52 )


downwards: (−√ 5

2, 0) U ( √ 5

2, ∞)

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17. f ( x )=x5−5 x3−20x−2Ans:

a)Function is increasing in: (−∞ , −2 )∪(2 , ∞ )

Decreasing in: (−2 , 2 ) b) The critical points are: x = 2 (máx. loc.) , x = 2 (mín loc.),


x=0 , x=±√ 32


(−√ 32, 0)∪(√ 3

2, ∞)


(−∞ , √ 32 )∪(0 , √ 3

2 )

APPLICATIONSAPPLICATIONS OF MAXIMUM AND MINIMUM

15. The factory M elaborates two products: A and B. If C is the total cost of production in a turn of 8 hours, then C = 3x2 + 42 y , where x is the number of machines used for the elaboration of product A and y is the number of machines used for the elaboration of product B. In a given turn of 8 hours, a total of 15 machines are working. Determine how many machines have to be used to elaborate product A and how may for product B having a minimum cost. ANS : 7 for A; 8 for B.

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Solve:1.. Lynbrook West, an apartment Complex, has 100 two-bedroom units. The monthly profit (in dollars) realized from renting out x

apartments is given by P( x )=−10x2+1760 x−50 ,000 . To maximize the monthly rental profit, how many units should be rented out? (ans x = 88) What is the maximum monthly profit realizable? P(88) = $27440

2.. The estimated monthly profit (in dollars) realizable by Cannon Precision Instruments for manufacturing and selling x units of its model MI camera is

P( x )=−0 .04 x2+240 x−10 ,000To maximize its profits, how many cameras should Cannon produce each month?

x=3000 Cameras P (3000 )=$350000

3.

PIB de un país en desarrollo. The GDP of a developing country during the period 1988 to 1996 is given by

G( t )=−0.2 t3+2 .4 t 2+60 0≤t≤8Where G(t) is measured in millions of dollars and t 0 0 corresponds to 1988. Show the máximum GDP corresponds to 1992.3. The management of Trappee and Sons producers of the famous TexaPep hot sauce, estimate that their profit (in dollars) from the daily production and sales of x cases (each case consisting of 24 bottles) of the hot sauce is given by

P( x )=−0 .000002 x3+6 x−400

What is the largest possible profit Trappee can make in 1 day? x=1000 cases and P (1000 )=$3600

4.- The quantity demanded each month of the Walter Serkin recording of Beethoven's Moonlight Sonata manufactured by Phonola

Record Industries, is related to the price/compact disc. The equation p=−0. 00042 x+6 0≤ x≤12 ,000 where p denotes the unit price in dollars and x is the number of discs demanded, relates the demand to the price. The total monthly cost (in dollars) for

pressing and packaging x copies of this classical recording is given by C ( x )=600+2 x−0 . 00002 x2 0≤ x≤20 ,000To maximize its profits, how many copies should Phonola produce each month?Hint: The revenue is R(x) = px, and the profit is P(x) = R(x) - C(x). (Ans 5000 discs)

5.- A manufacturer of tennis rackets finds that the total cost C(x) (in dollars) of manufacturing x rackets/day is given by

C ( x )=400+4 x−0 .0001 x2. Each racket can be sold at a price of p dollars, where p is related to x by the demand equation

p = 10 - 0 . 0004 x .If all rackets that are manufactured can be sold, find the daily level of production that will yield a maximum profit for the

manufacturer. Ans: x=10000 rackets.

6. The weekly demand for the Pulsar 25-in. color console television is given by the demand equation

p=−0 .05 x+600 0≤x≤12 ,000where p denotes the wholesale unit price in dollars and x denotes the quantity demanded. The weekly total cost function associated with manufacturing these sets is given by

C ( x )= 0 . 000002x3−0 .03x2+400+80 ,000where C(x) denotes the total cost incurred in producing x sets. Find the level of production that will yield a maximum profit for the manufacturer. Hint: Use the quadratic formula.Ans: x = 3334 TVs

7. Suppose the total cost function for manufacturing a certain product is C ( x )= 0 .2(0 . 01x2+ 120 ) dollars, where x represents the number of units produced. Find the level of production that will minimize the average cost. Ans: x = 110 units.

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9. The daily total cost (in dollars) incurred by Trappee and Sons for producing x cases of TexaPep hot sauce is given by the function

C (X )=0 .000002 x3+5 x+400 Using this function:

a. Find the average cost function C.b. Find the level of production that results in the smallest average production cost. c. Find the level of production for which the average cost is equal to the marginal cost. d. Compare the result of part (c) with that of part (b).

(a) C (X )=0 .000002 x2+5+400

x b) x=464 c) x=464

12. Suppose the quantity demanded per week of a certain dress is related to the unit price p by the demand equation

p=√800−x where p is in dollars and x is the number of dresses made. To maximize the revenue, how many dresses should be made and sold each week?

Hint: R(x) = px Ans: x=534 dresses.

13. The quantity demanded each month of the Sicard wristwatch is related to the unit price by the equation

p=50

0 . 01x2+10≤x≤20

where p is measured in dollars and x is measured in units of a thousand.

To yield maximum revenue, how many watches must be sold? Ans: x=10000

17.Nivel de producción Un fabricante estima que el costo unitario de producir x (millares) de artículos por mes es Cu( x )=4 x ln x+10 para x>0. Halle el nivel de producción que minimiza esta función de costo.

x=368 artículos18.Empleados El departamento de personal de una industria encuentra que el costo en dólares para procesar una solicitud de trabajo se modela por

C ( x )=0 . 01x2−18 ln x+70

cuando x es el número de empleado de en personal ¿Cuál es el número de empleados que minimiza el costo?

x=30 empleados19.Restaurante En un pequeño restaurante se observa que el número promedio N(x) de clientes para el almuerzo depende del precio x(en dólares) del “especial plato azul”, según la función

N ( x )=1000

x2 (1+ lnx4 )

Para x$1.5. Encontrar el precio que maximiza N(x) y el número máximo correspondiente de clientes

x=$2 .42 V (2. 42 )=85 Clientes

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