# Answer to Question #51129 in Microeconomics for Paul Muchira

Question #51129

Question 1

Mr. Hassanâ€™s demand function for rice is given by

X = 15 + M (10P) -1

Where X = amount of rice demanded, M = income of the consumer, P = price of rice.

Originally, the income of Mr. Hassan is $ 4,800 per month and the price of rice is $120/kg. If the price falls to $ 100/kg, calculate to total effect (TE), substitution effect (SE) and Income effect (IE) emanating from this change in price.

Question 2

(a) Given the following monotonically transformed utility function faced by the consumer

U(X1X2) = X_1^0.5 X_2^0.5

The price of good X1 is P1 and the price of good X2 is P2. Derive the optimal demand (Marshallian demand) function for X1 and for X2.

Question 3

Under a perfect competition the price as sh. 6 per unit has been determined. An individual firm has a total cost function given by C=10+15Q - 5Q^2+Q^3/3. Find:

i) Revenue function

ii)The quantity produced at which profit will be maximum profit

iii)Maximum profit

Mr. Hassanâ€™s demand function for rice is given by

X = 15 + M (10P) -1

Where X = amount of rice demanded, M = income of the consumer, P = price of rice.

Originally, the income of Mr. Hassan is $ 4,800 per month and the price of rice is $120/kg. If the price falls to $ 100/kg, calculate to total effect (TE), substitution effect (SE) and Income effect (IE) emanating from this change in price.

Question 2

(a) Given the following monotonically transformed utility function faced by the consumer

U(X1X2) = X_1^0.5 X_2^0.5

The price of good X1 is P1 and the price of good X2 is P2. Derive the optimal demand (Marshallian demand) function for X1 and for X2.

Question 3

Under a perfect competition the price as sh. 6 per unit has been determined. An individual firm has a total cost function given by C=10+15Q - 5Q^2+Q^3/3. Find:

i) Revenue function

ii)The quantity produced at which profit will be maximum profit

iii)Maximum profit

Expert's answer

Question 1

Mr. Hassan’s demand function for rice is given by

X = 15 + M (10P)^-1

M = $ 4,800 per month, P = $120/kg.

If the price falls to $ 100/kg:

1) the total effect TE = X2 - X1 = 15 + 4800/(10*120) - 15 - 4800/(10*150) = 0.8 kg

2) substitution effect SE = Xh - X0 = 15 + (4800*150/120)/(10*150) - 15 - 4800/(10*120) = 0 kg

3) income effect IE = X1 - Xh = 15 + 4800/(10*150) - 15 - (4800*150/120)/(10*120) = 0.8 kg

Question 2

(a) U(X1X2) = X1^0.5 X2^0.5

The price of good X1 is P1 and the price of good X2 is P2.

Optimal demand (Marshallian demand) function for X1 and for X2 will be:

X = (0.5I/P1, 0.5I/P2)

Question 3

P = 6 per unit

C=10+15Q - 5Q^2+Q^3/3.

i) Revenue function is:

TR = P*Q = 6Q

ii) The quantity produced at which profit will be maximum profit is in the point, where marginal revenue equals marginal cost: MR = MC

MR = TR' = 6

MC = C' = 15 - 10Q + Q^2

15 - 10Q + Q^2 = 6

Q^2 - 10Q + 9 = 0

Q1 = 9 units, Q2 = 1 unit (may not be profit maximizing).

iii) Maximum profit is:

TP1 = TR - TC = 6*1 - (10+15-5+1/3) = -$14.33

TP2 = TR - TC = 6*9 - (10 + 15*9 - 5*81 + 729/3) = 54 - 17 = $37

Mr. Hassan’s demand function for rice is given by

X = 15 + M (10P)^-1

M = $ 4,800 per month, P = $120/kg.

If the price falls to $ 100/kg:

1) the total effect TE = X2 - X1 = 15 + 4800/(10*120) - 15 - 4800/(10*150) = 0.8 kg

2) substitution effect SE = Xh - X0 = 15 + (4800*150/120)/(10*150) - 15 - 4800/(10*120) = 0 kg

3) income effect IE = X1 - Xh = 15 + 4800/(10*150) - 15 - (4800*150/120)/(10*120) = 0.8 kg

Question 2

(a) U(X1X2) = X1^0.5 X2^0.5

The price of good X1 is P1 and the price of good X2 is P2.

Optimal demand (Marshallian demand) function for X1 and for X2 will be:

X = (0.5I/P1, 0.5I/P2)

Question 3

P = 6 per unit

C=10+15Q - 5Q^2+Q^3/3.

i) Revenue function is:

TR = P*Q = 6Q

ii) The quantity produced at which profit will be maximum profit is in the point, where marginal revenue equals marginal cost: MR = MC

MR = TR' = 6

MC = C' = 15 - 10Q + Q^2

15 - 10Q + Q^2 = 6

Q^2 - 10Q + 9 = 0

Q1 = 9 units, Q2 = 1 unit (may not be profit maximizing).

iii) Maximum profit is:

TP1 = TR - TC = 6*1 - (10+15-5+1/3) = -$14.33

TP2 = TR - TC = 6*9 - (10 + 15*9 - 5*81 + 729/3) = 54 - 17 = $37

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