<p>This is how you get from your first equation to your second.
your utility function is $u(x_1, x_2)=x_1^a x_2^b$
since $a+b=1$ I'll change it slightly to a and (1-a)
In order to optimise these two choices, you need to maximise utility, wrt your choice variables.</p>

<p>subject to $p_1x_1 + p_2x_2 = w$
using Walras Law. Basically, in order to optimise utility, all money will be spent.</p>

<p>Cobb-Douglas functions are typically difficult for optimisation problems. A monotonic transformation which preserves the ordinal properties of the function can be used.</p>

<p>$aln(x_1) + (1 − a)ln(x_2)$</p>

<p>This will be used instead. The same budget constraint will be applied.</p>

<p>The Lagrange and First Order Conditions are Below</p>

<p>$L = aln(x_1) + (1 − a)ln(x_2) − \lambda(w − p_1x_1 − p_2x_2)$</p>

<p>$\frac{δL} {δx_1}= \frac{a} {x_1} − \lambda p_1 = 0$</p>

<p>$\frac {δL} {δx_2}=\frac{1 − a} {x_2} − \lambda p_2 = 0$</p>

<p>manipulating the First order conditions result in</p>

<p>$\lambda = \frac{a} {x_1p_1}$</p>

<p>$\lambda =\frac{(1 − a)}{ x_2p_2}$</p>

<p>$\frac{a} {x_1p_1}=\frac{(1 − a)}{ x_2p_2}$</p>

<p>substituting in the budget constraint $p_2x_2 = w − p_1x_1$</p>

<p>$\frac{a}{ x_1p_1}=\frac{(1 − a)} {w − p_1x_1}$</p>

<p>$x_1 =\frac{wa}{ p_1}$</p>

<p>and</p>

<p>$p_1x_1 = w − p_2x_2$</p>

<p>$\frac{a} {w − p_2x_2}=\frac{(1 − a)}{ p_2x_2}$</p>

<p>$w =\frac{a}{(1 − α)}p_2x_2 + p_2x_2$</p>

<p>$w(1 − a) = p_2x_2$</p>

<p>$x_2=\frac{w(1 − a)}{p_2}$</p>

<p>Using these results, we can work out the optimal consumption bundles of $x_1$ and $x_2$ for a given price, wealth combination.</p>

<p>$x_1 =\frac{wa}{ p_1}$</p>

<p>$x_2=\frac{w(1 − a)}{p_2}$</p>

CC BY-SA 3.0
2015-04-09T01:47:43.623Z
2015-06-23T22:58:38.417Z
2015-06-23T22:58:38.417Z
Jamzy
<p>Since $a + b=1$ the equations are exactly the same. Substituting in for $a+b$ with $1$ in the third and fourth equations gives the first and second equations. </p>

CC BY-SA 3.0
2015-04-08T16:03:02.150Z
Can these formulas also be edited to work with a utility function like $u=5x_1^{0.5}*2x_2^{0.5}$? So with an additional number before $x_i$?
user1170330
CC BY-SA 3.0
2015-04-08T16:09:07.390Z
I suggest asking this as a new question.
BKay
CC BY-SA 3.0
2015-04-08T16:48:06.433Z
What if $a+b \neq 1$? Should I use formula 3 and 4 in this case?
user1170330
CC BY-SA 3.0
2015-04-08T22:37:42.567Z
@user1170330 if $a+b\neq1$ it still works
Jamzy
CC BY-SA 3.0
2015-04-08T15:44:52.250Z
2015-04-08T16:08:08.853Z
2015-04-08T16:08:08.853Z
BKay
<p>When trying maximize the utility having a cobb-douglas utility function $u=x_1^ax_2^b$, with $a+b = 1$, I found the following formulas (<a href="http://de.wikipedia.org/wiki/Marshallsche_Nachfragefunktion#Marshallsche_Nachfragefunktionen_f.C3.BCr_g.C3.A4ngige_Nutzenfunktionen">Wikipedia: Marshallian Demand</a>):</p>

<p>$x_1 = \frac{am}{p_1}\\
x_2 = \frac{bm}{p_2}$  </p>

<p>In one of my books I also find these formulas for the same purpose:</p>

<p>$x_1 = \frac{a}{a+b}\frac{m}{p_1} \\
x_2= \frac{b}{a+b}\frac{m}{p_2}$</p>

<p>With $p_i$: prices of the goods; $m$: budget</p>

<p>I tested all of them and they produced the same results.<br>
So are there any differences?</p>

CC BY-SA 3.0
2015-04-08T05:15:14.343Z
does $a$ relate to $x_1$ exclusively? $b$ to $x_2$
Jamzy
CC BY-SA 3.0
2015-04-08T10:34:16.990Z
Can you straighten out some notation? In the second example, are a and b the exponents in the utility function x1 and x2? Do they sum to 1? Is y in the first problem the same as m in the second?
BKay
CC BY-SA 3.0
2015-04-08T15:29:56.450Z
@Jamzy: Yes, it does.
user1170330
CC BY-SA 3.0
2015-04-08T15:30:09.243Z
@BKay: Please see my updated notations.
user1170330
<microeconomics><consumer-theory><optimization><cobb-douglas>
Marshallian Demand for Cobb-Douglas
user1170330
# Marshallian Demand for Cobb-Douglas

(Asked by: username_0 on 2015-04-08)

When trying maximize the utility having a cobb-douglas utility function $u=x\_1^ax\_2^b$, with $a+b = 1$, I found the following formulas ([Wikipedia: Marshallian Demand](http://de.wikipedia.org/wiki/Marshallsche_Nachfragefunktion#Marshallsche_Nachfragefunktionen_f.C3.BCr_g.C3.A4ngige_Nutzenfunktionen)):

$x\_1 = \frac{am}{p\_1}\\
x\_2 = \frac{bm}{p\_2}$

In one of my books I also find these formulas for the same purpose:

$x\_1 = \frac{a}{a+b}\frac{m}{p\_1} \\
x\_2= \frac{b}{a+b}\frac{m}{p\_2}$

With $p\_i$: prices of the goods; $m$: budget

I tested all of them and they produced the same results.  
So are there any differences?

**Tags:** microeconomics, consumer-theory, optimization, cobb-douglas
> **Comments:**
> **username_1**: does $a$ relate to $x_1$ exclusively? $b$ to $x_2$  
> **username_2**: Can you straighten out some notation? In the second example, are a and b the exponents in the utility function x1 and x2? Do they sum to 1? Is y in the first problem the same as m in the second?  
> **username_0**: @Jamzy: Yes, it does.  
> **username_0**: @BKay: Please see my updated notations.  

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## 2 answers

### Answer by username_2 on 2015-04-08. Score: 13 | Accepted answer

Since $a + b=1$ the equations are exactly the same. Substituting in for $a+b$ with $1$ in the third and fourth equations gives the first and second equations.

> **Comments:**
> **username_0**: Can these formulas also be edited to work with a utility function like $u=5x_1^{0.5}*2x_2^{0.5}$? So with an additional number before $x_i$?  
> **username_2**: I suggest asking this as a new question.  
> **username_0**: What if $a+b \neq 1$? Should I use formula 3 and 4 in this case?  
> **username_1**: @user1170330 if $a+b\neq1$ it still works  

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### Answer by username_1 on 2015-04-09. Score: 7

This is how you get from your first equation to your second.
your utility function is $u(x\_1, x\_2)=x\_1^a x\_2^b$
since $a+b=1$ I'll change it slightly to a and (1-a)
In order to optimise these two choices, you need to maximise utility, wrt your choice variables.

subject to $p\_1x\_1 + p\_2x\_2 = w$
using Walras Law. Basically, in order to optimise utility, all money will be spent.

Cobb-Douglas functions are typically difficult for optimisation problems. A monotonic transformation which preserves the ordinal properties of the function can be used.

$aln(x\_1) + (1 − a)ln(x\_2)$

This will be used instead. The same budget constraint will be applied.

The Lagrange and First Order Conditions are Below

$L = aln(x\_1) + (1 − a)ln(x\_2) − \lambda(w − p\_1x\_1 − p\_2x\_2)$

$\frac{δL} {δx\_1}= \frac{a} {x\_1} − \lambda p\_1 = 0$

$\frac {δL} {δx\_2}=\frac{1 − a} {x\_2} − \lambda p\_2 = 0$

manipulating the First order conditions result in

$\lambda = \frac{a} {x\_1p\_1}$

$\lambda =\frac{(1 − a)}{ x\_2p\_2}$

$\frac{a} {x\_1p\_1}=\frac{(1 − a)}{ x\_2p\_2}$

substituting in the budget constraint $p\_2x\_2 = w − p\_1x\_1$

$\frac{a}{ x\_1p\_1}=\frac{(1 − a)} {w − p\_1x\_1}$

$x\_1 =\frac{wa}{ p\_1}$

and

$p\_1x\_1 = w − p\_2x\_2$

$\frac{a} {w − p\_2x\_2}=\frac{(1 − a)}{ p\_2x\_2}$

$w =\frac{a}{(1 − α)}p\_2x\_2 + p\_2x\_2$

$w(1 − a) = p\_2x\_2$

$x\_2=\frac{w(1 − a)}{p\_2}$

Using these results, we can work out the optimal consumption bundles of $x\_1$ and $x\_2$ for a given price, wealth combination.

$x\_1 =\frac{wa}{ p\_1}$

$x\_2=\frac{w(1 − a)}{p\_2}$


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