Consider the utility function U(x, y) = x 3 y 2 with the price of X equal to 3 a
ID: 1190674 • Letter: C
Question
Consider the utility function U(x, y) = x3y2 with the price of X equal to 3 and the price of Y equal to 4. Income devoted to the purchase of x and y is $36.
1) Find the equilibrium quantities of X and Y. (Use the LaGrange Method)
2) Using the budget constraint and utility function graph what this looks like.
3) What happens to equilibrium quantities of X and Y if income increases to $144. (Use the LaGrange Method)
4) What happens to equilibrium quantities of X and Y if the price of X increases to $12 and income remains at $144?
5) Does MRS = Price Ratio?
Explanation / Answer
1)
The utility function is
U(x, y) = x3y2.
The constrained maximization problem for each of the agents can be written as
maxX,Y U(x,y) subject to Pxx + Pyy = I
The Lagrangian for this problem is
L(x, y) = u(x, y) – (Pxx + Pyy – I)
i.e. L(x, y) = x3y2 – Pxx - Pyy + I
The three necessary conditions for the solution are:
(a) L(x, y)/x = 0
(b) L(x, y)/y = 0
(c) Pxx + Pyy = I
Solving (a), we have
L(x, y)/x = 0
3x2y2 – Px + 0 + 0 = 0
3x2y2 – Px = 0
3x2y2 = Px
= 3x2y2/Px …..(A)
Solving (b), we have
L(x, y)/y = 0
2x3y – Py + 0 + 0 = 0
2x3y – Py = 0
2x3y = Py
= 2x3y /Py …..(B)
From (A) and (B), we have
3x2y2/Px = 2x3y /PY
y = 2Pxx/3Py
Inserting the above result in the condition (c) solving for x, we have
Pxx + Py (2Pxx/3Py) = I
Pxx + 2Pxx/3 = I
5Pxx = 3I
x = 3I/5Px ....(i)
which is the demand function of good X.
Inserting (i) in the condition (c) and solving for y, we have
Pxy + Pyy = I
Px(3I/5Px) + Pyy = I
3I/5 + Pyy = I
Pyy = I - 3I/5
Pyy = 2I/5
y = 2I/5Py ....(ii)
which is the demand function of good Y.
When PX = 3, PY = 4, and I = 36, then the equilibrium quantity of good x is
x = 3(36)/5(3) = 7.20
and the equilibrium quantity of good y is
y = 2(36)/5(4) = 3.60
2)
y-intercept = I/PY = 36/4 = 9
x-intercept = I/PX = 36/3 = 12
3)
The utility function is
U(x, y) = x3y2.
The constrained maximization problem for each of the agents can be written as
maxX,Y U(x,y) subject to Pxx + Pyy = I
The Lagrangian for this problem is
L(x, y) = u(x, y) – (Pxx + Pyy – I)
i.e. L(x, y) = x3y2 – Pxx - Pyy + I
The three necessary conditions for the solution are:
(a) L(x, y)/x = 0
(b) L(x, y)/y = 0
(c) Pxx + Pyy = I
Solving (a), we have
L(x, y)/x = 0
3x2y2 – Px + 0 + 0 = 0
3x2y2 – Px = 0
3x2y2 = Px
= 3x2y2/Px …..(A)
Solving (b), we have
L(x, y)/y = 0
2x3y – Py + 0 + 0 = 0
2x3y – Py = 0
2x3y = Py
= 2x3y /Py …..(B)
From (A) and (B), we have
3x2y2/Px = 2x3y /PY
y = 2Pxx/3Py
Inserting the above result in the condition (c) solving for x, we have
Pxx + Py (2Pxx/3Py) = I
Pxx + 2Pxx/3 = I
5Pxx = 3I
x = 3I/5Px ....(i)
which is the demand function of good X.
Inserting (i) in the condition (c) and solving for y, we have
Pxy + Pyy = I
Px(3I/5Px) + Pyy = I
3I/5 + Pyy = I
Pyy = I - 3I/5
Pyy = 2I/5
y = 2I/5Py ....(ii)
which is the demand function of good Y.
When PX = 3, PY = 4, and I = 144, then the equilibrium quantity of good x is
x = 3(144)/5(3) = 28.80
and the equilibrium quantity of good y is
y = 2(144)/5(4) = 14.40
4)
When PX = 12, PY = 4, and I = 144, then the equilibrium quantity of good x is
x = 3(144)/5(12) = 7.20
and the equilibrium quantity of good y is
y = 2(144)/5(4) = 14.40
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