NUMERICALS - easy project
MANEGERIAL
ECONOMICS
GROUP
ASSIGNMENT
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 = 27
QD + 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
Putting the value of P in the equation of QD = QS
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 – p
QS = 6p – 10
SOLUTION:
For p QD = QS
5 – P = 6p – 10
5 + 10 = 6p + p
15 = 7p
15 / 7 = p
2.14 = p
Putting the value of P in the equation of QD = QS
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 – 5p
QS = 4 + 3p
SOLUTION:
For p QD = QS
20 – 5p = 4 + 3p
20 – 4 = 3p + 5p
16 = 8p
16 / 8 = p
2 = p
Putting the value of P in the equation of QD = QS
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
SOLUTION:
For p QD = QS
15 – 0.2p = -1 + 0.6p
15 + 1 = 0.6p + 0.2p
16 = o.8p
16 / 0.8 = p
20 = p
Putting the value of P in the equation of QD = QS
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
SOLUTION:
For p QD = QS
8 - 2p = 2 + 2p
8 – 2 = 2p + 2p
6 = 4p
6 / 4 = p
1.5 = p
Putting the value of P in the equation of QD = QS
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
SOLUTION:
Arrange equation
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
Putting the value of P in the equation of QD = QS
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
SOLUTION:
For p QD = QS
20 + 3p = 160 – 2p
3p + 2p = 160 – 20
5p = 140
P = 140 / 5
P = 28
Putting the value of P in the equation of QD = QS
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
SOLUTION:
For p QD = QS
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
Putting the value of P in the equation of QD = QS
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
SOLUTION:
Arrange equation
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
As we know that equilibrium price is settled where QD = QS
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
SOLUTION:
For p QD = QS
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
Putting the value of P in the equation of QD = QS
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
Putting the value of P in the equation of QD = QS
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
As we know that equilibrium price is settled where QD = QS
- 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 / 4
Q = √ 36 – p
SOLUTION:
P = 6 + Q2 / 4
P = (24 + Q2) / 4
4p = 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
As we know that equilibrium price is settled where QD = QS
- P/2 = P – 4
- P = 2P – 8
-P – 2P = - 8
- 3P = - 8
P = - 8/ - 3
P = 2.66
Putting the value of P in the equation of QD = QS
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 – 6
P = 7
Putting the value of p in the equation of q
Q – 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
2nd derivative grater then zero so function is minimum at the 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
2nd derivative grater then zero so function is minimum at the critical value of Q = 8
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
2nd derivative grater then zero so function is minimum at the critical value of Q = 1,
If critical value is Q = 4 then putting the value in the 2nd 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
SOLUTION:
π = TR – TC
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
Take IST derivative of Q
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
If critical value is Q = 14 then putting the value in the 2nd 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,
If critical value is Q = 2 then putting the value in the 2nd derivative equation
d2 π / d2Q = - 2Q + 16
d2 π / d2Q = - 2 (2) + 16
d2 π / d2Q = - 4 + 16
d2 π / d2Q = 12 > 0
2nd derivative grater then zero so function is minimum at the critical value of Q = 2
SOLUTION # 25:
TR = 4Q
TC = 0.04Q3 – 0.9Q2 + 10Q + 5
SOLUTION:
π = TR – TC
π = (4Q) – (0.04Q3 – 0.9Q2 + 10Q + 5)
π = 4Q – 0.04Q3 + 0.9Q2 - 10Q - 5
π = – 0.04Q3 + 0.9Q2 – 6Q – 5
Take IST derivative of Q
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 = 5
2ND derivative of Q
d2 π / d2Q = - 0.24Q + 1.8
If critical value is Q = 5 then putting the value in the 2nd 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
If critical value is Q = 10 then putting the value in the 2nd 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 critical value of Q = 10
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 = 25.04
Putting the value of y in the equation (ii)
- 4y – x + 120 = 0
- 4(25.04) – x + 120 =0
- 100.16 +120 - x = 0
X = 19.84
2nd 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 equation
80 – 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 = 12
π λ = 5/3y + y = 12 + 20/3
π λ = (5y + 3y)/3 = (36 + 20)/3
π λ = 8y = 56
π λ = y = 56/8
π λ = y = 7
Putting the value of y = 7 in equation of (iii)
π λ = x + y – 12 = 0
π λ = x + 7 – 12 = 0
π λ = x – 5 = 0
π λ = x = 5
Putting the value of x = 5 , y = 7 in equation of λ for interpret
80 – 4x – y = - λ
80 – 4(5) – 7 = - λ
80 – 20 – 7 = - λ
53 = - λ
- 53 = λ
Prove of profit:
X + Y = 12
5 + 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)
Take derivative of optimize function . . .
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 = 7
Putting the value of y = 7 in equation of (iii)
Uλ = 5x + 10y – 105 = 0
Uλ = 5x + 10(7) = 105
Uλ = 5x + 70 = 105
Uλ = 5x = 105 – 70
Uλ = 5x = 35
X = 35/5
X = 7
Putting the value of x = 7, y = 7 in equation of λ for interpret
2xy +10λ = 0
2(7) (7) = - 10λ
98 = - 10 λ
- 98/10 = λ
- 9.8 = λ
Prove of profit:
5X + 10Y = 105
5(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)
Take derivative of optimize function . . .
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 = 0
Uλ = 10x + 20(10) = 400
Uλ = 10x + 200 = 400
Uλ = 10x = 400 – 200
Uλ = 10x = 200
X = 200/10
X = 20
Prove of profit:
10X + 20Y = 400
10(20) + 20(10) = 400
200 + 200 = 400
-----------------------
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 – A 1
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