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1. Exercise 1.1, #61 a.

C ( q )=0.01 q2 +0.9 q +2 C ( 10 )=0.01 ( 10 )2 +0.9 ( 10 ) +2 C(10)=12 2

b.

9=[0.01 (10 ) + 0.9 ( 10 )+2] 0.01 ( 9 )2+ 0.9 ( 9 ) +2 ¿ -[ C ( 10 )−C ¿ C ( 10 ) – C (9)=1.09

2. Exercise 1.2, #41

cost=15 ∙ qty

price=x

qty=5 ( 27−x )=135−5 x

Let C = cost, R = revenue, P = profit then,

C ( x )=15(135−5 x)

C ( x )=2025−75 x 2

P ( x )=R ( x )−C (x)

P ( x )=135−5 x −(2025−75 x)

Maximum profit is when

P' ( x )=−10 x+210

R ( x ) =x( 135−5 x)

0=−10 x +210

10 x=210

x=21 optimal price is $21 per

qty=135−5 (21)

qty=30 30 sets will be sold each week. 3. Exercise 1.3, #37

Let

C=total cost

and

y=C ( x)=60 x +5,000

P ( x )=−5 x 2+ 210 x−2025

P' ( x )=0

game

qty=135−5 x

R ( x ) =135 x−5 x 2

x= productioncost per unit .

4. Exercise 1.3, #46 a.

{

50 N if 1,000≤ N ≤10,000 F( N ) 40 N if 10,000< N ≤ 20,000 35 N if 20,000< N ≤ 50,000

b. N

F(N)

1,000

50,000

10,000

500,000

10,000

400,000

20,000

800,000

20,000

700,000

50,000

1,750,00 0

5. Exercise 1.4, #3 a.

P ( x )=3 x−17,000

b.

P (20,000 )=3 ( 20,000 )−17,000

P (20,000 )=60,000−17,000

P (20,000 )=43,000

Producing 20,000 units gives a profit of $43,000.

P (5,000 )=3 ( 5,000 )−17,000 P

( 5,000 )=15,000−17,000

P (5,000 )=−2,000 Producing 5,000 units gives a loss

of $2,000. 6. Exercise 1.4, #32 x

y

0

20

30

41

72

62

100

55

a.

b. Split equally between the 2

x= y=50 ; use

26+0.5 x since

30 ≤5070 P( y )= 51− y for 70 ≥ y >28 55+ y for 28≥ y ≥ 0 7. Exercise 1.4, #42 a.

x=being thenumber of trees

Average yield: The total yield:

y =400−4 x 60+ x y=( 400−4 x )( 60+ x )

y=−4 x2 +160 x+ 24,000 b.

Vertex=

c.

−b 2a

Vertex=

−160 2(−4)

Vertex=20 60+20=80 The grower should grow 80 trees to maximize yield. 8. Exercise 1.4, #52

x=no . of dining room tables ¿ sell

R ( x ) =500 x

C ( x )=30,000+350 x a.

R ( x ) =C( x ) 500 x=30,000+350 x

150 x 30,000 = 150 150 b.

x=200 ; 200 tables must be sold to breakeven

P ( x )=R ( x )−C (x) 6,000=500 x−350 x−30,000

6,000=150 x−30,000 36,000 150 x = 150 150 c.

x=240 ; 240 tables must be sold to make a profit of $6,000

P ( x )=150 x−30,000 P (150 )=150 ( 150 )−30,000 150 tables are sold.

d.

R ( 200 )=500(200)

P (150 )=−7,500 There will be a loss of $7,500 if

R ( 200 )=100,000

9. Exercise 1.6, #48

t=6

The graph is discontinuous at

and t=12 . The company bought more

inventories during those months. 10.Exercise 2.1, #50 a.

E ( x )=x ∙ D ( x )

b.

Eave =

E ( x )=x (−35 x +200)

E ( x )=−35 2+ 200 x

E ( 5 )−E(4) 5−4 2

2

Eave =−35 ( 5 ) + 200 (5 )−[−35 ( 4 ) +200 ( 4 ) ]

Eave =125−240

Eave =−115

The average change in consumer expenditures is -$115 per unit. c.

E' ( 4 ) =lim

h→0

E ( 4 +h )−E (4 ) h

−35 ( 4+ h )2+ 200 ( 4+ h )−[−35 ( 4 )2+ 200 ( 4 ) ] h h→0

'

E ( 4 ) =lim (−35 h−80) h→0

¿4

per unit when

2

−35 h −80 h h h→0

E' ( 4 ) =lim

E' ( 4 ) =lim

E' ( 4 ) =−80 The instantaneous rate of change is -$80

. The expenditure is decreasing when

x=4 .

11.Exercise 2.2, #64 Let

G(t)

be the GDP in billions of dollars where

t

is years and

represents 1995. Since the GDP is growing at a constant rate,

G(t)

t=0 is a linear

function passing through the points (0,125) and (8,155).

155−125 t+125 8−0

Then

G ( t )=

Then

G(t)=

In 2010,

t=15

15 t+125 4 and the model predicts a GDP of

G ( 15 )=181.25

billion dollars.

12.Exercise 2.4, #66

C ( q )=0.2 q2 +q +900 dC =2 ( 0.2 q ) +1 d

q ( t )=t 2+ 100t

dC =0.4 q+1 d

t=1

dC d = (0.2 q 2+ q+900) d dq

dq d 2 = (t + 100t ) dt dt

dC dC dq = ∙ =(0.4 q+1)(2 t+100) dt dq dt

q ( 1 )=( 12) +100 (1)

dC ∨¿ q=101= [ 0.4 ( 101 )+1 ] [2 ( 1 ) +100] dt t =1 ¿

q ( 1 )=101 units dC =4222.8 dt 13.Exercise 2.5, #1 a.

1 C( x )= x 2 +4 x +57 5 1 C' ( x ) = ( 2 x ) +4 5

2 C' ( x ) = x+ 4 5

R ( x ) =x ( P ( x ) )

( x4 )

¿ x 9− ¿ 9 x−

b.

x2 4

R' ( x )=9−

x 2

2 ' C ( x ) = x+ 4 5 2 ' C ( 3 )= (3)+ 4 5 C' ( 3 )=$ 5.20

c.

C ( 4 )−C ( 3 )=

[

C ( 4 )−C ( 3 )=$ 5.40

d.

R ( x )=9−

x 2

R' ( 3 )=9−

3 2

'

R' ( 3 )=$ 7.50

][

1 1 (4 )2+ 4( 4)+57 − (3)2+ 4(3)+57 5 5

]

dq =2 t+100 dt q ( 1 )=1+100 dC ∨¿ q=101=(41.4 )(102) dt t =1 ¿

e.

4 ¿2 ¿ 3 ¿2 ¿ ¿ 9 ( 3 )−¿ ¿−¿ 9 ( 4 ) −¿ R ( 4 )−R ( 3 )=¿ ' ' R ( 4 ) −R ( 3 ) =$ 7.25

14.Exercise 2.5, #13 a. Marginal cost = Derivative of total cost function

MC=(3 q2 +q+ 500) MC=6 q +1

M C ' =6 ( 40 )+1 MC ’=$ 241 b.

Actual=MC (q+ 1)−MC (q) Actual=3 ( 41 )2 +41+500−[ 3 ( 40 )2 + 40+500 ] Actual=5584−( 5340 ) Actual=244 $244

15.Exercise 2.5, #24

Q ( L ) =300 L2 /3

'

Q ( L )=

200 L1 /3

' 12.5=∆ Q≈ Q (512 ) ∆ L=25 ∆ L

Q ' ( 512 )=25 12.5 25 ∆ L = 25 25

0.5=∆ L

∆ L=0.5 more worker-hours are needed

16.Exercise 2.6, #55

Q=2 x 3+ 3 x 2 y 2+ ( 1+ y )3 d dy dy 2 x 3 +3 x 2 y 2 + ( 1+ y )3 ]=6 x 2 +3 x 2 ∙ 2 y + y 2 ∙ 2 x + 3 (1+ y )2 ∙ [ dx dx dx

(

)

d 3 2 2 3 2 2 dy 2 2 dy 2 x +3 x y + ( 1+ y ) ]=6 x +6 x y +6 x y +3 ( 1+ y ) [ dx dx dx dy dy + 3 (1+ y )2 dx dx −6 x 2−6 x y 2 = 6 x 2 y+ 3 (1+ y )2 6 x 2 y +3 ( 1+ y )2

6 x2 y

−6 ( 30 )2−6 ( 30 ) ( 20 )2 dy ∨¿ x=30 = dx y=20 6 ( 30 )2 ( 20 ) +3 ( 1+20 )2 ¿

dy −6 x 2−6 x y 2 = dx 6 x 2 y +3 ( 1+ y )2

dy −77,400 ∨¿ x=30 = dx y=20 109,323 ¿

Δy =Q' Δx ∆ y=Q ' (∆ x) ∆ y=

−77,400 ( 0.8) 109,323

∆ y ≈ 0.566 units 17.Exercise 3.1, #59 a.

b.

S ( x ) =−2 x 3+ 27 x2 +132 x+207 3

2

S ( 0 )=−2 ( 0 ) +27 ( 0 ) +132 ( 0 )+ 207 207 units will be sold

S ( 0 )=207

c.

S ( x ) =−2 x 3+ 27 x2 +132 x+207 S ' =−6 x2 +54 x +132

S ' =−6( x 2−9 x−22)

S ' =−6( x−11)( x +2)

4 (−6)(132) ¿ 54 2−¿ −54 ± √ ¿ x=¿ x=−2

0< x 1

Elastic for p > 25; or Inelastic for p < 25; or

|E ( p )|=−0.04 p