Utility functions

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April 8th 2011, 10:38 AM
GordonComstock

Utility functions

Hi!

I've got the following utility function:

$U = (y_1 - 2) (y_2 - 4) = y_1y_2 - 4y_1-2y_2-8$

prices are

$p_1 = 2 , p_2=4$

$p_1$ is the price of good 1
$p_2$ is the price of good 2

$y_1$ is the quantity of good 1
$y_2$ is the quantity of good 2

The budget costraint is:

$100 = 2y_1 + 4y_2$

Now, we know that we have equilibrium when we have

$\frac { \delta U / \delta y_1} {p_1} = \frac { \delta U / \delta y_2} {p_2}$

because weighted marginal utilities are equal.

Ok, now this is where i get stuck! The book "says" that

Quote:

with our data, we have:

$\frac { \delta U / \delta y_1} {p_1} = \frac {y_2 - 4} {2}$ and $\frac { \delta U / \delta y_2} {p_2} = \frac {y_1 - 2} {4}$

My question is: why are $\delta U / \delta y_1$ equal to $y_2 - 4$ and $\delta U / \delta y_2$ equal to equal to $y_1 - 2$ ?

Thank you in advance!
April 8th 2011, 11:15 AM
SpringFan25

$\displaystyle \frac{\delta U}{\delta y_1}$ is the partial differential of U with respect to y1. This means you differentiate the function with respect to y1, treating all other variables as constants

$\displaystyle \frac{\delta U}{\delta y_1}=\frac{\delta (y_1 y_2)}{\delta y_1} + \frac{\delta (-4 y_1) }{\delta y_1} + \frac{\delta ( -2 y_2) }{\delta y_1} + \frac{\delta (-8)}{\delta y_1}$

Taking the first term, remembering to treat y2 as a constant when differentiating, this is
$\displaystyle \frac{\delta (y_1 y_2)}{\delta y_1} = y_2$

Now all the other terms
$\displaystyle \frac{\delta U}{\delta y_1}=\frac{\delta (y_1 y_2)}{\delta y_1} + \frac{\delta (-4 y_1) }{\delta y_1} + \frac{\delta ( -2 y_2) }{\delta y_1} + \frac{\delta (-8)}{\delta y_1}$

$\displaystyle \frac{\delta U}{\delta y_1}= y_2 - 4 + 0 + 0$

$\displaystyle \frac{\delta U}{\delta y_1}= y_2 - 4$

You try the other one.
April 8th 2011, 11:48 AM
GordonComstock

$\displaystyle \frac{\delta U}{\delta y_2}=\frac{\delta (y_1 y_2)}{\delta y_2} + \frac{\delta (-4 y_1) }{\delta y_2} + \frac{\delta ( -2 y_2) }{\delta y_2} + \frac{\delta (-8)}{\delta y_2}$

$\displaystyle \frac{\delta (y_1 y_2)}{\delta y_2} = y_1$

$\displaystyle \frac{\delta (-4 y_1) }{\delta y_2} = 0$ (because $y_1$ is a constant and its derivative is 0, multiplied by -4 gives 0, right?)

$\displaystyle \frac{\delta ( -2 y_2) }{\delta y_2} = -2$ (because $y_2$ 's derivative is 1, multiplied by -2 gives -2, right?)

$\displaystyle \frac{\delta (-8)}{\delta y_2} = 0$ (because 8 is a constant and its derivative is 0?)

$\displaystyle \frac{\delta U}{\delta y_2} = y_1 - 2$

Ok, thank you. Please, tell me if I did something wrong!

Thank you again!
April 8th 2011, 11:55 AM
SpringFan25

perfect!
April 8th 2011, 12:44 PM
GordonComstock

Quote:

Originally Posted by SpringFan25

perfect!

Ok, thanks again ;)
April 9th 2011, 04:40 AM
GordonComstock

Hi, again. I've got another problem here. I have to maximize the following Utility function: $max U = U(y_1, y_2)$

subject to the following budget costraint: $R = y_1 p_1 + y_2 p_2$

Of course $\displaystyle y_1=\frac {R - y_2 p_2} {p_1}$

Then $max U = U(\frac {R - y_2 p_2} {p_1}, y_2)$

Clear till here... now he says:

Quote:

By differentiating this expression and making it equal to zero, we get

$\displaystyle \frac {\delta U}{\delta y_1} \frac {\delta y_1}{\delta y_2} +\frac {\delta U}{\delta y_2} = 0$

Of course I don't want to take advantage of your helpfulness again, but could you please just tell me if I have to do the same things we've seen in my first post? Thanks again!