A General Rule for the Consumption of Two Commodities

Solution
Q1a)
If we assume that Alice is consuming two commodities X and Y the utility function given is by;
u ( x , y )=x∝ Y 1−∝, x ≥ 0∧ y ≥ 0.
The general rule shows that we will have a constant budget share demand function of;
X¿= ( ∝
∝ +1−∝ )∗I
PX
, this can be represented as;
X¿=∝ ( I
px )
Similarly,
Y ¿= ( ∝
∝ +1−∝ )∗I
Py
, this can be represented as;
Y ¿=∝ ( I
py )
For this case, income (I) = 16 and px = py = 1. Where the initially the consumption bundle is
x¿=∝¿
(16
1 )=16 ∝ ¿ e initially the consumption bundleis 1. wheremand function of duction function
And y¿= ( 1−∝ )∗16
1 =16−16 ∝
Compensation variation is amount of money that will be given to consumer in order to keep original
bundle. There will be a price change of py from 1 to 4, the new bundle of y will be;
y∗¿ ( 1−∝ ) ∗16
4 =4−4 ∝
The original bundle was 16−16 ∝units of y
The new price of $4 per unit, the bundle will require $16 more than the current income; therefore the
compensating variation is $6.
The money that will be required to buy the old bundle at new price of 4*( 16−16 ∝¿, which is more
than the current income by $16 will be;

4*(16 - 16∝¿ +16 ∝−16=16
64 - 64 ∝+16 ∝−16=16
Collecting like terms together
64 - 64 ∝+16 ∝=16+16
64 - 64 ∝+16 ∝=32
48∝=32
∝= 2
3
b)
Equivalent variation is amount of money that is needed by the consumer in order to buy the same
bundle at the same old price old price that entails a new utility.
Equivalent variation = New income – old income
Demand function for x =
2
3∗I
Px
Demand function for y =
1
3∗I
Py
Utility function, U = X
2
3 ∗Y
1
3
U = [ 2
3∗I
Px ] 2
3
∗[ 1
3 ∗I
Py ] 1
3
= 2
9 ( I )∗( 1
Px ) 2
3∗( 1
Py ) 1
3
Uold income and new price = = 2
9∗16∗( 1
1 ) 2
3 ∗( 1
4 ) 1
3 = 2.2399
2.2399 = = 2
9 ( I (new) ) ∗( 1
1 ) 2
3∗( 1
1 ) 1
3

Inew = 10.07
≈ $ 10
Equivalent variation = $16 – $10 = $6
Q2)
Given
f(x1, x2) = min{x1, x2} + x2
Input 1 cost w1 > 0 per unit
Input 2 cost w2 > 0 per unit
a) The isoquant associated with an output of 4
f(x1, x2) = min{x1, x2} + x2
Case (i): x1 < x2
y = f(x1, x2)
= x1 + x2
X1 + x2 = 4
X2
4 x1 + x2 = 4
Slope = 1
4 X1

Case (ii)
x1 > x2
f(x1, x2) = x2 + x2 = 2x2
f(x1, x2) = 2x2
4 = 2x2
X2 = 2
x2
2 slope = ∞
X1
X2 kink (input x2 = input x1)
4 slope = 1
2 slope = ∞
4 X1
b) Long run cost function
c = w1x1 + w2x2
c = w1(y – x2) + w2(y – x1)
Case (i)
c = w1(y – x2) + w2(y – x1)