# Profit Maximization Cobb Douglas Production function

• Feb 23rd 2011, 04:09 PM
reneebu
Profit Maximization Cobb Douglas Production function
I am having trouble with the algebra how do get from

y=(pay/w)^a * (pby/v)^b .... first I factor out y

y=(pa/w)^a * (pb/v)^b * y^(a+b) but how do I get to this result....

y=(pa/w)^a/(1-a-b) * (pb/v)^b/(1-a-b)

I know I could just assume the short cut...eliminate y and subtract each of its exponent variables from the denominator of the other variable exponents but I would really like to understand the long hand algebra... Thanks
• Feb 23rd 2011, 05:15 PM
Soroban
Hello, reneebu!

What an ugly problem!

Quote:

$\displaystyle y \:=\:\left(\dfrac{pay}{w}\right)^a\cdot\left(\dfra c{pby}{v}\right)^b$

$\displaystyle \text{First I factor out }y\!:\;\;y\:=\:\left(\dfrac{pa}{w}\right)^a\cdot\l eft(\dfrac{pb}{v}\right)^b\cdot y^{a+b}$ [1]

$\displaystyle \text{but how do I get to this result? }\;y\:=\:\left(\dfrac{pa}{w}\right)^{\frac{a}{1-a-b}} \cdot \left(\dfrac{pb}{v}\right)^{\frac{b}{1-a-b}}$ [2]

In [1] we have: .$\displaystyle y\:=\:\left(\dfrac{pa}{w}\right)^a\cdot\left(\dfra c{pb}{v}\right)^b\cdot y^{a+b}$

Divide by $\displaystyle y^{a+b}\!:$

. . $\displaystyle \displaystyle \frac{y}{y^{a+b}} \;=\;\left(\frac{pa}{w}\right)^a\cdot\left(\frac{p b}{v}\right)^b \quad\Rightarrow\quad y^{1-a-b} \;=\;\left(\frac{pa}{w}\right)^a\cdot\left(\frac{p b}{v}\right)^b$

Raise both sides to the power $\displaystyle \frac{1}{1-a-b}\!:$

. . $\displaystyle \displaystyle \left(y^{1-a-b}\right)^{\frac{1}{1-a-b}} \;=\;\bigg[\left(\frac{pa}{w}\right)^a\cdot\left(\frac{pb}{v} \right)^b\bigg]^{\frac{1}{1-a-b}}$

$\displaystyle \displaystyle\text{Simplify: }\qquad y \;=\;\left(\frac{pa}{w}\right)^{\frac{a}{1-a-b}}\cdot\left(\frac{pb}{v}\right)^{\frac{1}{1-a-b}}$