# Fast way to write out the utility optimization problem for a Cobb-Douglas function?

In my last problem set, I had to solve both the Utility Maximization Problem (UMP) and Expenditure Minimization Problem (EMP) for a Cobb Douglas utility function. Recall, Cobb Douglas is defined as $$U(x,y) = x^\alpha y^\beta$$

I can compute the EMP or UMP. I can then find the other by "duality". But writing the problem was very tedious and messy. I ended up with an indirect utility function like this

$$v(p_x,p_y,m) = \left(\frac{\alpha m}{p_x}\right)^\alpha \left( \frac{\beta m }{p_y}\right)^\beta$$

and doing the Lagrangian was kind of clunky with all this multiplication going on.

Is there a simpler way to write the problem that can save me time / effort?

• This question could be improved by spelling out what a UMP and EMP are. While it is often the case that Cobb-Douglas production functions have exponents that sum to 1 it is not required. – BKay May 11 '15 at 23:37
• Take the natural logarithm. – Alecos Papadopoulos May 12 '15 at 0:30
• That's a good idea. That's a monotonic transform, right? But doesn't that change the utility function? – Stan Shunpike May 12 '15 at 0:33
• It changes the cardinality of the utility function but not the preferences over bundles of x and y (preserves ordinality). – BKay May 12 '15 at 12:43

Well all Cobb-Douglas have an structure, it is the utility that makes shares constants as budget shares independent of prices $p_{l}x^{CD}_{l}(p,w)/w=\alpha_{l}$ with $u^{CD}(x)=\sum_{l}\alpha_{l}log(x_{l})$ and $\sum_{l}\alpha_{l}=1$. then you can go back and compute $x_l(p,w)$, with this the indirect utility, then invert to obtain the expenditure and then derive to get the hicksian. Not sure if this is what you want but some utility functions have a defining property, the same with the leontief or CES.

This is a generic approach (find a transformation that make the utility problem easier to work with.

Assume a household has utility $$U(x,y) = x^\alpha y^\beta$$.

A utility function is a convenient way to represent preferences. However, we saw in the chapter that utility functions have many limitations. One limitation is that although utility functions tell us a lot about the ordinal rankings of goods, they reveal nothing about cardinal rankings. In other words, we know what a consumer prefers, but not the strength of that preference. Because of this, we can shift, stretch, or squeeze the utility function into any form as long as we don’t change the ordering of preferences over bundles, and we’ll still have the same representation of the relative utility of goods.

We'll take the log of this utility function because it is a positive monotonic transformation. It will give all the same preference relationships, so we can calculate the same demand functions.

$$u(x,y) = \log (U(x,y)) = \alpha \log (x) + \beta \log(y)$$.

Let's rewrite to optimization function $$\max_{s.t. x p_x + y p_y = m} \alpha \log (x) + \beta \log(y)$$

Which I find easier to understand as a Lagrangian.

$$\max \alpha \log (x) + \beta \log(y) - \lambda (x p_x + y p_y - m)$$

$$\frac{\partial u}{\partial x} = \frac{\alpha}{x} -\lambda p_x = 0$$

$$\frac{\partial u}{\partial y} = \frac{\beta}{y} -\lambda p_y = 0$$

$$\Rightarrow \frac{\alpha}{x p_x} = \lambda$$ and $$\frac{\beta}{y p_y} = \lambda$$

$$\Rightarrow \frac{\alpha}{x p_x} = \frac{\beta}{y p_y}$$ $$\Rightarrow y = \frac{\beta x p_x}{\alpha p_y}$$

Then use the budget equation:

$$m = x p_x + y p_y = x p_x + \frac{\beta x p_x}{\alpha p_y} p_y$$ $$\Rightarrow m = x p_x + \frac{\beta x p_x}{\alpha} = x p_x(1 + \frac{1}{\alpha})$$ $$\Rightarrow x^* = \frac{m}{p_x(1 + \frac{1}{\alpha})}$$ $$\Rightarrow y^* = \frac{\beta x^* p_x}{\alpha p_y} = \frac{\beta \frac{m}{p_x(1 + \frac{1}{\alpha})} p_x}{\alpha p_y} = \frac{\beta \frac{m}{(\alpha + 1)} }{ p_y}$$

Note that $$\frac{x p_x}{m}$$ is constant and function of the household's preferences but not the relative prices. This is the famous constant budget share result of Cobb Douglas preferences.

There is another approach which is Cobb-Douglas specific.

Note that if the utility function is $$U(x,y) = x^\alpha y^\beta$$ then the marginal utility functions are: $$U_x = \alpha x^{\alpha-1} y ^{\beta} = \alpha \frac{U}{x}$$ $$U_y = \alpha x^{\alpha} y ^{\beta - 1} = \beta\frac{U}{y}$$ Which means the first order condition with the Lagrangian is

$$\frac{\partial U}{\partial x} = \alpha \frac{U}{x} -\lambda p_x = 0$$ $$\frac{\partial U}{\partial y} = \beta\frac{U}{y} -\lambda p_y = 0$$ Which implies:

$$\beta\frac{U}{y p_y} = \alpha \frac{U}{x p_x} \Rightarrow \beta\frac{x p_x}{\alpha p_y} = y$$

Which is exactly what we got doing it the first way and will give the same $$x^*$$ and $$y^*$$.