# Marshallian Demand for Cobb-Douglas

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):

$x_1 = \frac{am}{p_1}\\ x_2 = \frac{bm}{p_2}$

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?

• does $a$ relate to $x_1$ exclusively? $b$ to $x_2$ – Jamzy Apr 8 '15 at 5:15
• 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 Apr 8 '15 at 10:34
• @Jamzy: Yes, it does. – user1170330 Apr 8 '15 at 15:29
• @BKay: Please see my updated notations. – user1170330 Apr 8 '15 at 15:30

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.

• 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 Apr 8 '15 at 16:03
• I suggest asking this as a new question. – BKay Apr 8 '15 at 16:09
• What if $a+b \neq 1$? Should I use formula 3 and 4 in this case? – user1170330 Apr 8 '15 at 16:48
• @user1170330 if $a+b\neq1$ it still works – Jamzy Apr 8 '15 at 22:37

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.

$x_1 =\frac{wa}{ p_1}$

$x_2=\frac{w(1 − a)}{p_2}$