help starting problem

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bensmyname

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hey this is the problem:
Suppose you have the following utility function: U(x,y)=6x^0.5+y. The price of x is px and the price of y is 1.
Your income is Y=24. find the uncompensated demand for good x. That is, find the amount of x which maximises the consumers utility, subject to affordability. You can use any method you want. Do not worry about corner solutions.

My problem is finding px, I think i can do the problem from then on but i dont even have a clue how to get px?
Cheers for any help
Ben
 
Since y is constant, why isn't this an applicaation of a derivative in the remaining variable, x?
 
bensmyname said:
hey this is the problem:
Suppose you have the following utility function: U(x,y)=6x^0.5+y. The price of x is px and the price of y is 1.
Your income is Y=24. find the uncompensated demand for good x. That is, find the amount of x which maximises the consumers utility, subject to affordability. You can use any method you want. Do not worry about corner solutions.

My problem is finding px, I think i can do the problem from then on but i dont even have a clue how to get px?
Cheers for any help
Ben
Click to expand...

Do you mean that the price per unit of x is p? If the price is px for x units of x, then the price per unit of x is self-evidently 1.

Is there a distinction between Y and y?

If the problem should be stated that the utility function for x is U(x , y) subject to a budget constraint of y = 24, have you heard of Lagrangian multipliers?

As you have stated it, no one is going to be able to help you because the statement of the problem is ambiguous.
 
I apologise.
Yes Y is different to y, Y is the income, y is a good.
Do you mean that the price per unit of x is p? If the price is px for x units of x, then the price per unit of x is self-evidently 1.
Click to expand...
Yes i think so, P[sub:77w1orcf]x[/sub:77w1orcf] is the price of x, which is unknown. And this is where my problem lies, i think i can work out the rest if i have P[sub:77w1orcf]x[/sub:77w1orcf] (the price of x). Are you saying that the price is 1? I do not understand how, but if you believe it is I may go with it as I have no leads.
I understand Langrangian multipliers but I just could not work it out without having both prices.

Since y is constant, why isn't this an applicaation of a derivative in the remaining variable, x?
Click to expand...
Wouldnt the quantities X and Y and P[sub:77w1orcf]1[/sub:77w1orcf] all be variables relevant in the budget constraint sorta thing? I don't really know, quite confused. I just thought the first thing to working this out would be to find out P[sub:77w1orcf]1[/sub:77w1orcf] then do it through a langrangian method as normal.

Thank you for the help.
 
This what I done so far: (given lagrangian=L, lamda=# and partial differntiation=& and *=multiplication, not x) (sorry don't know how to do symbols)
L=6x[sup:1s1w2umb].5[/sup:1s1w2umb]+y+#(24-x*p[sub:1s1w2umb]x[/sub:1s1w2umb]-y)
&L/&x=3x[sup:1s1w2umb]-.5[/sup:1s1w2umb]-x*p[sub:1s1w2umb]x[/sub:1s1w2umb]*#
&L/&y=1-y*#
&L/&#=24-x*p[sub:1s1w2umb]x[/sub:1s1w2umb]-y

I semi get stuck here - I think i need to try and isolate p[sub:1s1w2umb]x[/sub:1s1w2umb]? So I try this:
By dividing &x and &y i get 3x[sup:1s1w2umb]-.5[/sup:1s1w2umb]=x*p[sub:1s1w2umb]x[/sub:1s1w2umb]/y - May have made errors here?

Rearranging to make y subject:
y=xp[sub:1s1w2umb]x[/sub:1s1w2umb]/3x[sub:1s1w2umb].5[/sub:1s1w2umb] - Didnt know what happened to the .5 whether it was negative or not, but figured outcome would make obvious. Maths errors completely likely.

Put this into constraint
24-x*p[sub:1s1w2umb]x[/sub:1s1w2umb]-(x*p[sub:1s1w2umb]x[/sub:1s1w2umb]/3x[sub:1s1w2umb].5[/sub:1s1w2umb])

Thought maybe i could remove the x*p[sub:1s1w2umb]x[/sub:1s1w2umb]?

Giving 24 - 3x[sub:1s1w2umb].5[/sub:1s1w2umb] = 0
So: Squareroot(24/3) = x

..Just realised ive removed p[sub:1s1w2umb]x[/sub:1s1w2umb]. Is this the right direction at all?
 
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