Divide the commodity space into three regions.
1. x>2y
2. 2y>x>y/2
3. x<y/2
Region 1 - U= 2y+2x = 2(x+y) {since, x>2y => 2x>y => max{2x,y}=2x}
Region 2 - U=x+2x = 3x
Region 3 - U=x+y.
Eg - Plot U=4
Join in following sequence (2,0)--->(4/3,2/3)--->(4/3,8/3)--->(4,0)
Finding Demand. Take y to be numeraire. Let price of x=p.
Case 1 - p=1
Notice that you can maximise U if you are restricted in region 1. For same income level (m) utility in region 3 will be U/2.
Consumer problem - Max 2(x+y) s.t. x+y=m
x \in [2m/3,m] and y=m-x.
Case 2 - p<1
x=m/p; y=0
Case 3 - 1<p<=5/2
x=2m/(2p+1) and y=m/(2p+1)
Case 4 - p>=5/2
x=0; y=m. It'll lie in region 3.
Locked
