Demand Funtions

[Devika]

 (a) u(x,y) = max{min{3x,3y}, x + y}

 (b) u(x,y) = [x+y], where [] denotes the greatest integer function.

 (c) u(x,y) = max{x,2y}

p=(1,2) and m=100. Derive the demand functions for x and y. All three parts. These questions intimidate me! Any sort of help/hint will be hugely appreciated.

Thank you

[Amit Goyal]

Given the price vector (1, 2) and income 100, we will solve for utility maximization problem (UMP) of the consumer in the following three cases:
(a) u(x,y) = max{min{3x,3y}, x + y}
There are two solutions to the UMP:
(100/3, 100/3)
and (100, 0)

(b) u(x,y) = [x+y], where [] denotes the greatest integer function.
There is a unique solution to the UMP:
(100, 0)

(c) u(x,y) = max{x,2y}
There are two solutions to the UMP:
(0, 50)
and (100, 0)

[Devika]

Thank you so much Amit Sir, but could you please explain part (a) a little more elaborately?

Thank you so much :)

[maahi]

hey can u xplain d c part when we hv to maximise utility why r we choosing max = 50 when we can get 100

[Sumit]

Hey Maahi,
(0,50) means X=0 & Y=50....N Utility in this case too is maximizing that's why this questions have two solutions...

[Devika]

Sumit, can you explain part (a)?

[Sumit]

Hey Devika,
look at fn
u(x,y) = max{min{3x,3y}, x + y}

In this type of fn we are concern only to Max one of the two i.e either  min{3x,3y} or x + y....n then look which case will give Max utility.

Now, First Consider case one where we try to max [min{3x,3y}]...this could be done if we take x=100/3 n Y=100/3 by taking these values of x n y notice we also satisfying M=100..n also only these values provide us max [min{3x,3y}]=100.
Now consider utility fn u(x,y) = max{min{3x,3y}, x + y} n put  x=100/3 n y=100/3....you will see Max{100,66.66}=100, which is equal to max utility we can get by max [min{3x,3y}].

Now, consider the case in which we try to Max[x + y], this could be done if we take x=100 (notice we take x=100 n y=0 bcoz if we only take y the max qty we can purchase will be equal to y=50 coz price of y=2 but in this case we can't get Max[x + y])

Now, in this case look at u(x,y) = max{min{3x,3y}, x + y} n put x=100 n y=0 the utility u get u(x,y) = max{0, 100}=100 which is same as case one above..
thats y we have two solutions to the UMP..
1)(100/3, 100/3) .
2)(100, 0) ...

[Devika]

Thanks a ton sumit!!

[AnonymousNow]

Why 100/3 ??