NUMERICALS - easy project | It is a place where people can …… · Web view · 2012-07-11NUMERICALS. MARKET MODEL. QUESTION # 1: QD = 105 - 2p. QS = 35 + 8p. SOLUTION: For p

MANEGERIAL

ECONOMICS

GROUP

ASSIGNMENT

Superior University

Module:

Managerial Economics

Group Assignment:

Numerical of 2 Chapters

Group members:Leader 11338

11315

11353

11321

Prof Zeshan uz Zaman

Session 2010

Class M.com – A 1

NUMERICALS MARKET MODEL

QUESTION # 1: QD = 105 - 2p

QS = 35 + 8p

SOLUTION: For p QD = QS

105 – 2p = 35 + 8p

105 - 35 = 2p + 8p

70 = 10p

70 / 10 = p

7 = P

Putting the value of P in the equation of QD = QS

QD = 105 – 2p , QS = 35 + 8p

QD = 105 – 2(7) , QS = 35 + 8 (7)

QD = 105 – 14 , QS = 35 + 56

QD = 91 , QS = 91

QD = QS 91 = 91

QUESTION # 2:

13p – QS = 27QD + 4p – 24 = 0

SOLUTION: Arrange equation

QD = 24 – 4p

QS = 27 – 13p

For p QD = QS

24 – 4p = 27 – 13p

13p - 4p = 27 – 24

9p = 3

P = 9 / 3

P = 3


QD = 24 – 4p , QS = 27 – 13p

QD = 24 – 4 (3) , QS = 27 – 13 (3)

QD = 24 – 12 , QS = 27 – 39

QD = 12 , QS = 12

QD = QS 12 = 12

QUESTION # 3:

QD = 5 – pQS = 6p – 10


5 – P = 6p – 10

5 + 10 = 6p + p

15 = 7p

15 / 7 = p

2.14 = p


QD = 5 – p , QS = 6p – 10

QD = 5 – (2.14) , QS = 6 (2.14) – 10

QD = 2.85 , QS = 2.85

QD = QS 2.85 = 2.85

QUESTION # 4: QD = 20 – 5pQS = 4 + 3p


20 – 5p = 4 + 3p20 – 4 = 3p + 5p16 = 8p16 / 8 = p2 = p


QD = 20 – 5p , QS = 4 + 3p

QD = 20 – 5 (2) , QS = 4 + 3 (2)

QD = 20 – 10 , QS = 4 + 6

QD = 10 , QS = 10

QD = QS

10 = 10

QUESTION # 5:

QD = 15 – 0.2p

QS = -1 + 0.6p


15 – 0.2p = -1 + 0.6p

15 + 1 = 0.6p + 0.2p

16 = o.8p

16 / 0.8 = p

20 = p


QD = 15 – 0.2p , QS = -1 + 0.6p

QD = 15 – 0.2 (20) , QS = -1 + 0.6 (20)

QD = 15 – 4 , QS = -1 + 12

QD = 11 , QS = 11

QD = QS

11 = 11

QUESTION # 6:

QD = 8 – 2p

QS = 2 + 2p


8 - 2p = 2 + 2p

8 – 2 = 2p + 2p

6 = 4p

6 / 4 = p

1.5 = p


QD = 8 – 2p , QS = 2 + 2p

QD = 8 – 2 (1.5) , QS = 2 + 2 (1.5)

QD = 8 – 3 , QS = 2 + 3

QD = 5 , QS = 5

QD = QS

5 = 5

QUESTION # 7:

Q + P = 5

2p – Q = 5.5


Q = 5 – p

Q = 2p – 5.5

For p QD = QS

5 – P = 2p – 5.5

5 + 5.5 = 2p + p

10.5 = 3p

10.5 / 3 = p

3.5 = p


QD = 5 – p , QS = 2p – 5.5

QD = 5 – 3.5 , QS = 2 (3.5) - 5.5

QD = 1.5 , QS = 7 – 5.5

QD = 1.5 , QS = 1.5

QD = QS

1.5 = 1.5

QUESTION # 8:

Q = 20 + 3p

Q = 160 – 2p


20 + 3p = 160 – 2p

3p + 2p = 160 – 20

5p = 140

P = 140 / 5

P = 28


QD = 20 + 3p , QS = 160 – 2p

QD = 20 + 3 (28) , QS = 160 – 2 (28)

QD = 20 + 84 , QS = 160 – 56

QD = 104 , QS = 104

QD = QS

104 = 104

QUESTION # 9:

P = 7 – x / 2

P = 7 / 6 + x

SOLUTION: For i.e. p QD = QS

7 – X / 2 = 7 / 6 + x

7 – 7 / 6 = x + x / 2

(42 – 7) / 6 = (2x + x)/ 2

35 / 6 = 3x / 2

X = (36 / 6). (2 / 3)

X = 35 / 9

Putting the value of X in the equation of QD = QS

QD = 7 – x / 2 , QS = 7 / 6 + x

QD = 7 – (35 / 9) / 2 , QS = 7 / 6 + (35 / 9)

QD = (7 - 3.89) / 2 , QS = (21 + 70) / 18

QD = 10.111 / 2 , QS = 91 / 18

QD = QS

5.056 = 5.056

QUESTION # 10: QD = 30 – 6p2

QS = 2p2 + 4p + 6


30 – 6p2 = 2p2 + 4p + 6

30 – 6 = 2p2 + 6p2 + 4p

24 = 8p2 + 4p

8p2 + 4p – 24 = 0

4 (2p2 + p - 6) = 0

Equation solve by factorization . . .

2p2 + 4p – 3p – 6 = 0

2p (p+2) -3 (p-2) = 0

(P + 2) (2p – 3) = 0

P + 2 = 0 , 2p – 3 = 0

P = -2 , 2p = 3

P = -2 , p = 3 / 2


QD = 30 – 6p2 , QS = 2p2 + 4p + 6

QD = 30 – 6 (3 / 2)2 , QS = 2 (3 / 2)2 + 4 (3 / 2) +

6

QD = 30 – 6 (9 / 2) , QS = 2 (9 / 4) + 6 + 6

QD = 30 – 27 / 2 , QS = 9 / 2 + 12

QD = (60 – 27) / 2 , QS = (9 + 24) /2

QD = 33 / 2 , QS = 33 / 2

QD = 16.5 , QS = 16.5

QD = QS

16.5 = 16.

QUESTION # 11:

P + Q2 + 3Q – 20 = 0

P – 3Q2 + 10Q = 5


P = - Q2 – 3Q +20

P = 3Q2 – 10Q + 5

For p QD = QS

- Q2 – 3Q + 20 = 3Q2 – 10Q + 5

- Q2 – 3Q2 -3Q +10Q +20 – 20 = 0

- 4Q2 +7Q + 15 = 0

- (4Q2 – 7Q – 15) = 0

Equation solve by factorization . . .

4Q2 - 12Q + 5Q - 15 = 0

4Q (Q-3) + 5 (Q – 3) = 0

(4Q + 5) (Q – 3) = 0

4Q + 5 = 0 , Q – 3 = 0

4Q = - 5 , Q = 3

Q = -5 / 4 , Q = 3

Putting the value of Q in the equation of QD = QS

QD = - Q2 – 3Q + 20 , QS = 3Q2 - 10Q + 5

QD = - (3)2 – 3 (3) + 20 , QS = 3(3)2 – 10 (3) + 5

QD = -9 – 9 + 20 , QS = 27 – 30 + 5

QD = 2 , QS = 2

QD = QS

2 = 2

QUESTION # 12: QD = 10 – 3P2

QS = 4 + P

SOLUTION: Taking derivative of QD & QS

QD = 10 – 3P2 ,

QS = 4 + P

dQD = - 6P

As we know that equilibrium price is settled where QD = QS

- 6P = 4 + P

- 6P - P = 4

7P = 4

P = - 4/7

P = - 0.57

Putting the value of p in the equation of QD = QS

QD = - 6P , QS = 4 + p

QD = - 6 (-0.57), QS = 4 + (- 0.57)

QD = 3.42 , QS = 3.43

QD = QS

3.42 = 3.42

QUESTION # 13: QD = 5 – p2

QS = 2p – 3

SOLUTION: Taking derivative of QD

dQD = 0 – 2p

QD = 0 – 2p

QS = 2p – 3


0 – 2p = 2p – 3

- 2p – 2p = - 3

- 4p = - 3

P = -3/-4

P = 0.75

Putting the value of p in the equation of QD = QS

QD = 0 – 2p , QS = 2p – 3

QD = - 2(0.75) , QS = 2(0.75) – 3

QD = - 1.5 , QS = - 1.5

QD = QS

- 1.5 = - 1.5

QUESTION # 14: P = 48 – 3Q2

P = Q2 + 4Q + 16


48 – 3Q2 = Q2 + 4Q + 16

- 3Q2 – Q2 – 4Q = - 48 + 16

- 4Q2 – 4Q = - 32

- 4 (Q2 + Q – 8) = 0

Equation solve by quadric equation

[-b + √ b2 - (4ac)] / 2a = 0

[-1 +√ (1)2 – 4 (1) (- 8)] / 2(1) = 0

[-1 + √ 1 + 32] / 2 = 0

(-1 + √ 33) / 2 = 0 , (-1 – √ 33) / 2 = 0

(-1 + 5.74) / 2 = 0 , (-1 – 5.74) / 2 = 0

2.37 , - 3.37


QD = 48 – 3Q2 , QS = Q2 + 4Q + 16

QD = 48 – 3(2.37)2 , QS = (2.37)2 + 4(2.37) +

16

QD = 48 – 3 (5.61) , QS = 5.61 + 9.48 + 16

QD = 31.1 , QS = 31.1

QD = QS

31.1 = 31.1

QUESTION#15: Q = 10p + 5p2

Q = 64 – 8p – 2p2

SOLUTION: For i.e. QD = QS

10p + 5p2 = 64 – 8p – 2p2

5p2 + 2p2 + 8p + 10p – 64 = 0

7p2 + 18p – 64 = 0

Equation solve by quadric equation

[- b + √ (b) 2 – 4 ac] / 2a

[- 18 + √ (18)2 – 4 (7) (-64)] / 2 (7) = 0

[- 18 + √ (324) – 1792)] / 14 = 0

[- 18 + √ (2116)] / 14 = 0

(-18 + 46) / 14 = 0 , (-18 – 46) / 14 = 0

2 = 0 , - 4.5 = 0


QD = 10p + 5p2 , QS = 64 – 8p – 2p2

QD = 10(2) + 5 (2)2 , QS = 64 – 8(2) – 2(2)2

QD = 20 + 5(4) , QS = 64 – 16 – 2(4)

QD = 20 + 20 , QS = 64 – 16 – 8

QD = 40 , QS = 40

QD = QS

40 = 40

QUESTION # 16: P = (Q + 2)2

P = 39 – 3Q2

SOLUTION: (a + b) 2 = (a2 + b2 + 2ab)

(Q + 2)2 = (Q)2 + (2)2 + 2(Q) (2)

(Q + 2)2 = Q2 + 4 + 4Q

P = Q2 + 4 + 4Q , P = 39 – 3Q2

Taking derivative of Q

dQ = 2Q + 4 , dQ = - 6Q

P = 2Q + 4 , P = - 6Q

Q = (P – 4) / 2 , Q = - P/6


- P/6 = (P – 4) / 2

- 2P = 6P – 24

- 2P – 6P = - 24

- 8P = - 24

P = - 24/- 8

P = 3

Putting the value of p in the equation of QD =QS

Q = - P/6 , Q = (P – 4) / 2

Q = - 3/6 , Q = (3 – 4) / 2

Q = - 0.5 , Q = - 0.5

QD = QS

- 0.5 = - 0.5

QUESTION # 17:

P = 6 + Q2 / 4Q = √ 36 – p

SOLUTION:

P = 6 + Q2 / 4P = (24 + Q2) / 44p = 24 + Q2

Taking by under root√ 4p – 24 = √ Q2

√ 4p – 24 = Q

Q = Q

Q = √ 36 – p . . . . . . (i)

Putting the value of Q . . . (i) in the equation

√ 4p – 24 = √ 36 – p

Taking by square

√ (4p – 24)2 = √ (36 – p) 2

4p – 24 = 36 – p

4p + p = 36 + 24

5p = 60

P = 60 / 5

P = 12

Putting the value of p in this equation

Q = √ 36 – p

Q = √ 36 – 12

Q = √ 24

Q = 12

Putting the value of Q in the equation . . . . .

P = 6 + Q2 / 4

P = 6 + (12)2 / 4

P = 6 + 24 / 4

P = (24 + 24) / 4

P = 48 / 4

P = 12

QUESTION # 18: P = 16 – Q2

P = 4 + Q

SOLUTION: Taking derivative of P

P = 16 – Q2 , P = 4 + Q

P = - 2Q , Q = P – 4

2Q = - P

Q = - P/2


- P/2 = P – 4

- P = 2P – 8

-P – 2P = - 8

- 3P = - 8

P = - 8/ - 3

P = 2.66


Q = - P/2 , Q = P – 4

Q = - 2.66 / 2 , Q = 2.66 – 4

Q = - 1.33 , Q = - 1.33

QD = QS

- 1.33 = - 1.33

QUESTION # 19: (Q + 12) (p + 6) = 169

Q-p + 6 = 0

SOLUTION : For i.e. QD = QS

Q = P – 6 (1)

Putting the value of Q in this equation

(Q + 12) (P + 6) = 169

[(p – 6) + 12] (p + 6) = 169

(P – 6 + 12) (p + 6) = 169

(p + 6) (p + 6) = 169

(p + 6)2 = 169

By taking under root

√ (p + 6)2 = √ 169

P + 6 = 13

P = 13 – 6P = 7Putting the value of p in the equation of qQ – P + 6 = 0

Q – (7) + 6 = 0

Q – 1 = 0

Q = 1

Prove QD = QS

(Q + 12) (p + 6) = 169 , Q - p + 6 = 0

(1 + 12) (7 + 6) = 169 , 1 – 7 + 6 = 0(13) (13) = 169 , - 7 + 7 = 0 169 = 169 , 0 = 0

OPTIMIZATION ONE VARIABLE

QUESTION # 20: Y = 5x2 – 3x + 7

SOLUTION: Take IST derivative of X & equal it to zero

dy / dx = 10x – 3 = 0

dy / dx = 10x = 3

dy / dx = x = 3 / 10

dy / dx = X = 0. 3

Take 2nd derivative of x

d2y / d2x = 10 > 0

2nd derivative grater then zero so function is minimum at the

critical value of x = 0.3

QUESTION # 21: AC = 200 – 24Q + Q2

SOLUTION: Take IST derivative of Q & equal it to zero

dAC / dQ = - 24 + 2Q = 0

dAC / dQ = 2Q = 24

dAC / dQ = Q = 24 / 2

dAC / dQ = Q = 12

Taking 2nd derivative of Q

d2AC / d2Q = 2 > 0


critical value of Q = 12

QUESTION # 22: MC = 200 – 48Q + 3Q2

SOLUTION: Take IST derivative of Q & equal it to zero

dMC / dQ = - 48 + 6Q = 0

dMC / dQ = 6Q = 48

dMC / dQ = Q = 48 / 6

dMC / dQ = Q = 8

Taking 2nd derivative of Q

d2MC / d2Q = 6 > 0



QUESTION # 23: TR = 45Q – 0.5Q2

TC = Q3 – 8Q2 + 57Q + 2

SOLUTION: π = TR – TC

π = (45Q – 0.5Q2) – (Q3 – 8Q2 + 57Q + 2)

π = 45Q – 0.5Q2 – Q3 + 8Q2 - 57Q – 2

π = - Q3 + 7.5Q2 – 12Q – 2

Take IST derivative of Q

dπ / dQ = - 3Q + 15Q - 12

dπ / dQ = - 3 (Q2 + 5Q – 4) = 0

dπ / dQ = Q2 - 5Q + 4 = 0

dπ / dQ = Q2 –Q – 4Q + 4

dπ / dQ = Q(Q - 1) -4 (Q -1)

dπ / dQ = (Q – 1) (Q - 4)

Q – 1 = 0 , Q – 4 = 0

Q = 1 , Q = 4

Take 2nd derivative of Q

d2 π / d2Q = - 6Q + 15

If critical value is Q = 1 then putting the value in the 2nd

derivative equation

d2 π / d2Q = - 6Q + 15

d2 π / d2Q = - 6 (1) + 15

d2 π / d2Q = - 6 + 15

d2 π / d2Q = 9 > 0


critical value of Q = 1,


derivative equation

d2 π / d2Q = - 6Q + 15

d2 π / d2Q = - 6 (4) + 15

d2 π / d2Q = - 24 + 15

d2 π / d2Q = - 9 < 0

Then 2nd derivative smaller then zero so function is maximum

at the critical value of Q= 4

QUESTION # 24: TR = 220 – 0.5Q2

TC = 1/3Q3 – 8.5Q2 + 50Q + 90


π = (220 – 0.5Q2) – (1/3Q3 – 8.5Q2 + 50Q + 90)

π = 220 – 0.5Q2 – 1/3Q3 + 8.5Q2 - 50Q – 90

π = – 1/3Q3 + 8Q2 – 28Q – 90


dπ / dQ = - Q2 + 16Q – 28

Solve by factorization

dπ / dQ = - Q2 + 14Q + 2Q – 28

dπ / dQ = - Q (Q – 14) 2 (Q – 14)

dπ / dQ = (Q – 14) (2 – Q)

Q – 14 = 0 , 2 – Q = 0

Q = 14 , Q = 2

Take 2nd derivative of Q

d2 π / d2Q = - 2Q + 16


derivative equation

d2 π / d2Q = - 2Q + 16

d2 π / d2Q = - 2(14) + 16

d2 π / d2Q = - 28 + 16

d2 π / d2Q = - 12 < 0

2nd derivative smaller then zero so function is maximum at the

critical value of Q = 14,


derivative equation

d2 π / d2Q = - 2Q + 16

d2 π / d2Q = - 2 (2) + 16

d2 π / d2Q = - 4 + 16

d2 π / d2Q = 12 > 0



SOLUTION # 25: TR = 4Q

TC = 0.04Q3 – 0.9Q2 + 10Q + 5


π = (4Q) – (0.04Q3 – 0.9Q2 + 10Q + 5)

π = 4Q – 0.04Q3 + 0.9Q2 - 10Q - 5

π = – 0.04Q3 + 0.9Q2 – 6Q – 5


dπ / dQ = - 0.12Q2 + 1.8Q – 6 = 0

dπ / dQ = - 0.12Q2 + 1.2Q + 0.6Q – 6 = 0

dπ / dQ = - 0.12Q (Q – 10) 0.6 (Q - 10)

dπ / dQ = (Q – 10) (0.6 – 0.12Q)

Q – 10 = 0 , 0.6 - 0.12Q = 0

Q = 10 , Q = 0.6/0.12 Q = 52ND derivative of Qd2 π / d2Q = - 0.24Q + 1.8


derivative equation

d2 π / d2Q = - 0.24Q + 1.8

d2 π / d2Q = - 0.24(5) + 1.8

d2 π / d2Q = - 1.2 + 1.8

d2 π / d2Q = 0.6 > 0

2nd derivative grater then zero so function is minimum at the critical value of Q = 5


derivative equation

d2 π / d2Q = - 0.24Q + 1.8

d2 π / d2Q = - 0.24(10) + 1.8

d2 π / d2Q = - 2.4 +1.8

d2 π / d2Q = - 0.6 < 0

2nd derivative smaller then zero so function is maximum at the


TWO VARIABLE

QUESTION # 26: π = 80x – 2x2 – xy – 3y2 + 100y

SOLUTION: Take partial derivative w.e.f x & y equate = 0

πx = 80 – 4x – y ------- (i)

πy = - x – 6y + 100 ------- (ii)

Solve equation by stimulatingly; multiply equation (ii) by 4

- 4x – y + 80

– 4x – 24y + 400

+ + -

+ 23y – 320 = 0

23y = 320

Y = 320/ 23

Y = 13.91

Putting the value of y in the equation of (i)

80 – 4x – y = 0

80 – 4x – 13.91 = 0

80 – 13.91 = 4x

66.09 = 4x

66.09 / 4 = x

16.52 = x

2nd order direct & cross partial derivative

π xx = -4 ,

π xy = - 1

π yx = -1 ,

π yy = - 6

RULE π xx . π yy > (π xy)2

(-4) . (-6) > (1)2

24 > 1

Here function is maximum if x = 16.52 & y = 13.91

QUESTION# 27: AC = x2 + 2y2 – 2xy – 2x – 6y + 20

SOLUTION: dACx = 2x – 2y – 2 --------- (i)

dAcy = 4y – 2x – 6 ------------- (ii)

2x – 2y – 2

2x + 4y – 6

2y – 8 = 0

2y = 8

Y = 8/2

Y = 4

Putting the value of y in the equation of (i)

2x – 2y – 2 = 0

2x – 2 (4) – 2 = 0

2x – 8 – 2 = 0

2x -10 = 0

2x = 10

X = 10 / 2

X = 5

2nd order direct & cross partial derivative

ACxx = 2 , ACxy = - 2

AC yx = -2 , AC yy = 4

RULE AC xx. AC yy > (AC xy) 2

(2). (-4) > (- 2)2

8 > 4

Here function is maximum if x =5 & y = 4

QUESTION # 28: π = 144x – 3x2 – xy – 2y2 + 120y – 35

SOLUTION: πx = - 6x – y + 144 --------- (i)πy = - 4y – x + 120 ------------- (ii)Solve by sim- 6x – y + 144- 6x – 24y + 720+ + -

23y – 576 = 0 Y = 576/23 Y = 25.04Putting the value of y in the equation (ii) - 4y – x + 120 = 0- 4(25.04) – x + 120 =0- 100.16 +120 - x = 0X = 19.842nd order direct & cross partial derivative

π xx = - 6 , π xy = - 1

π yx = -1 , π yy = - 4

RULE π xx . π yy > (π xy)2

(- 6). (- 4) > (1)2

24 > 1

Here function is maximum if x = 19.84 & y = 25.04

CONTRAINED VARIABLE

QYESTION # 29:π = 80x – 2x2 – xy - 3y2 + 100y

Subjective function x + y = 12

SOLUTION: Langrangian function = λ (x + y – 12)

π = 80x – 2x2 – xy - 3y2 + 100y + λ (x + y – 12)

Take derivative of optimize function . . .

π x = 80 – 4x – y + λ -------- (i)

πy = - x – 6y + 100 + λ --------(ii)

π λ = x + y – 12 -------------------- (iii)

Equate equation (i), (ii) & (iii) = 0

Equation ----- (i) 80 – 4x – y = - λ

Equation ------ (ii) - x – 6y + 100 = - λ

- λ = - λ

Solve λ algebraic equation80 – 4x – y = - x – 6y + 100 - 4x + x = y – 6y + 100 – 80 - 3x = - 5y + 20 X = 5/3y – 20/3 -------- (iv)Putting the value of x eq (iv) in equation (iii)π λ = x + y – 12 = 0 π λ = (5/3y – 20/3) + y = 12π λ = 5/3y + y = 12 + 20/3π λ = (5y + 3y)/3 = (36 + 20)/3π λ = 8y = 56π λ = y = 56/8π λ = y = 7

Putting the value of y = 7 in equation of (iii)π λ = x + y – 12 = 0π λ = x + 7 – 12 = 0π λ = x – 5 = 0

π λ = x = 5Putting the value of x = 5 , y = 7 in equation of λ for interpret80 – 4x – y = - λ80 – 4(5) – 7 = - λ80 – 20 – 7 = - λ 53 = - λ - 53 = λProve of profit:X + Y = 125 + 7 = 12 12 = 12

QUESTION # 30:

U = xy2

Subjective function 5x + 10y = 105

SOLUTION: Langrangian function = λ (5x + 10y - 105)

U = xy2 + λ (5x + 10y - 105)

U = xy2 + (5xλ + 10yλ - 105λ)


Ux = y2 + 5λ --------(i)

Uy = 2xy + 10λ --------(ii)

Uλ = 5x + 10y - 105 -------------------- (iii)

Equate equation (i), (ii) = 0

equation ----- (i) , Equation ------ (ii)

y2 + 5λ = 0 , 2xy + 10λ = 0

y2 = - 5λ , 2xy = - 10λ

y2 / 5 = - λ , 2xy/10 = - λ

- λ = - λ

y2 / 5 = 2xy/10

y = x

Putting the value of x eq (iii)Uλ = 5x + 10y = 105 5(y) + 10y = 105 5y + 10y = 105 15y = 105 Y = 105/15 Y = 7Putting the value of y = 7 in equation of (iii)

Uλ = 5x + 10y – 105 = 0Uλ = 5x + 10(7) = 105Uλ = 5x + 70 = 105Uλ = 5x = 105 – 70Uλ = 5x = 35 X = 35/5 X = 7Putting the value of x = 7, y = 7 in equation of λ for interpret2xy +10λ = 02(7) (7) = - 10λ 98 = - 10 λ - 98/10 = λ - 9.8 = λProve of profit:5X + 10Y = 1055(7) + 10(7) = 105 35 + 70 = 105

QUESTION # 31:

U = 24X1/2 Y1/2

Se = 10x + 20y = 400

SOLUTION: Langrangian function = λ (10x + 20y - 400)

U = 24x1/2 y1/2 + λ (10x + 20y - 400)

U = 24x1/2 y1/2 + (10xλ + 20yλ - 400λ)


Ux = 12x -1/2 y1/2 + 10λ --------(i)

Uy= 12x 1/2 y -1/2 + 20λ --------(ii)

Uλ = 10x + 20y - 400 -------------------- (iii)

Equate equation (i), (ii) = 0

Equation ----- (i) , Equation ------ (ii)

12x -1/2 y1/2 + 10λ = 0 , 12x 1/2 y -1/2 + 20λ = 0

12x -1/2 y1/2 = - 10λ , 12x -1/2 y -1/2 = - 20λ

12x -1/2 y1/2 /10 = - λ , 12x -1/2 y -1/2 /20 = - λ

- λ = - λ

12x -1/2 y1/2 /10 = 12x -1/2 y -1/2 /20

y/1 = x/2

2y = x

Putting the value of x eq (iii)Uλ = 10x + 20y = 400 10(2y) + 20y = 400 20y + 40y = 400 40y = 400 Y = 400/40 Y = 10

Putting the value of y = 10 in equation of (iii)Uλ = 10x + 20y – 400 = 0Uλ = 10x + 20(10) = 400Uλ = 10x + 200 = 400Uλ = 10x = 400 – 200Uλ = 10x = 200X = 200/10X = 20Prove of profit:10X + 20Y = 40010(20) + 20(10) = 400 200 + 200 = 400

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NUMERICALS - easy project | It is a place where people can …… · Web view · 2012-07-11NUMERICALS. MARKET MODEL. QUESTION # 1: QD = 105 - 2p. QS = 35 + 8p. SOLUTION: For p