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Suppose u(x, y, z, w) = xy + wz. Find the demand function for x ?
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Is that all the info we have?
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Yes, Ofcourse the solution has to be in general terms as a function of (M,px,py,pw,pz)
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This post was updated on Jun 14, 2017; 7:38am.
Lagrangian optimization gives me d(x)= I*p(y)/2(px*py plus pw*pz) where I is income.
I'm not able to prove whether this is a maxima or a minima succintly. I could argue for that utility is sum of 2 quasiconcave functions, hence is quasiconcave itself. So this makes the Lagrangian method necessary and sufficient for evaluating maxima. Also I have to make sure that the demand of x is less than maximum possible expenditure on x i.e. I/p(x) should be greater than d(x). Which is trivial once you work with the inequality. |
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In reply to this post by RichaS
the bundle xy and wz are perfect substitutes
let us assume the price of xy is a=Px+Py and price of wz is b= Pw+Pz if a>b then U=wz and hence it becomes x=0, y=0 w=I/2pw and z=I/2pz if a<b U=xy, x=I/2px, y= I/2py w=0 and z=0 If a=b x belongs to {0,I/2px} and likewise for others. I thought it like this :/ Not sure |
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Price of xy is dynamic. px plus py is price of 1 unit of x and 1 unit of y. if xy=4, than you can get it for 2px plus 2py, or 4py and 1px. So it varies.
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In reply to this post by Algaeconomics
I think we should compare the products instead:
If (M/2px)*(M/2py)>(M/2pw)*(M/2pz) then U=xy, x=I/2px, y= I/2py w=0 and z=0 i.e if px*py< pz*pw |
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Problem here is you're comparing two situations and ignoring all other possibilities. You're saying the consumer will only consume x,y or w,z. However that may not be true.
If your condition on optimal x,y giving more utility than optimal w,z holds, it does not imply consuming x,y is overall optimal. It just says that consuming x,y and not any of w,z is "better" than consuming only w,z and no x,y. Obviously it does not mean that its the best choice. |
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In reply to this post by Asd1995
I tried substituting different values for Px Py Pz and Pw and I and then maximizing as substitutes.
the relation I think is a= Px+Py and b = Pw+Pz |
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In reply to this post by Asd1995
This is a case of perfect substitutes.
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In reply to this post by RichaS
The solution isn't complete because again you've found one situation better than the other, not better than every other infinite alternatives. Having said that your solution is probably correct.
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Check this!!
let us assume u = a +b where a =xy, b = wz, pa = pxX+pyY, pb = pwW+pzZ if pa < pb, demand of a = M/pa, budget equation pxX+pyY = M, if you solve these two you will get xy =1 therefore y = 1/x, when pa<pb. substitute this in the above budget equation pxX+pyY = M and solve for x. which will give us the demand for x. |
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In reply to this post by Algaeconomics
I would agree with you if you can give me constants k1,k2 such that
k1*xy plus k2*wz=I |
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In reply to this post by shiva
your prices don't satisfy the budget constraint. according to your work,
(px*x plus py*y)(xy) plus (pz*z plus pw*w)(wz)=I which is clearly untrue. |
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In reply to this post by RichaS
Looking like a perfect substitute, but is actually a max. case where either X,Y or W,Z can be consumed, not all together..
2 cases: px+ py< Pw+pz, then x= M/px,Y=M/Py, and vice-versa for the second case That should be the ans, plz correct if anything seems wrong.. |
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In reply to this post by Asd1995
According to my explanation. we will get xy = 1 & wz = 0 in one case, or xy =0 & wz = 1, if you substitute them the budget equation is being satisfied.
(pxX+pyY)xy + (PwW+pzZ)wz =M in first case, this will become pxX +pyY =M in second case, the above equation turns to pwW + pzZ =M that is what I have used even while solving. so it satisfies. |
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I don't know man, you're equating price to revenue in your assumption.
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In reply to this post by RichaS
Solving even for U(x,y,z)= xy + z, for certain values of prices and income the result is either z is not consumed at all or only z is consumed. But I arrived at it by the long lagrangian critical points evaluation method.
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In reply to this post by RichaS
If U= (x+y) + (z+w), we can say since both terms in parenthesis are perfect subsitutes, consumer will consume either only x,y or only z,w.
However U= (x+z) + (y+w) we can say again both are perfect substitutes, consumer will consume only x,z or y,w. I hope someone can see my argument now. |
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i don't think parenthesis matter in such a case..(in case of U=(x+y)+(z+w))
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