# Suggestion Lagrange Multiplier for Econometric Problem

For an essay required to complete my high school multivariable calc class I am wanting to write about an approach to an econometric problem (as I probably will study econometrics in uni). I really want to approach the problem using the Lagrange multipliers as I really like and understand well how the method works.

What I intend to do in my investigation is to create a formula in which you just input the constraint and all the constants after which it gives the optimum value for a variable. My best attempt in finding an econometric problem which could be approached using the Lagrange multipliers would be the Cobb-Douglas production function: Output = $A L^\alpha K^\beta$ with the constraint of $\alpha + \beta > 1$ (positive returns to scale). However as $\alpha$ and $\beta$ are constants and $L$ & $K$ are the variable this didn't work out because there was no possibility of $f(x,y)$ & $g(x,y)$.

Therefore I am wondering if any of you can help me with an economic/econometric problem that can be approached using the Lagrange multipliers to create a formula or something similar to maximise the variables!

What you need is to use the cost as a constrain, so for instance $C(L,K) = \omega L + \rho K$. The problem then becomes

\begin{eqnarray} \max&:&~ A L^\alpha K^\beta \\ {\rm s.t.} &:& C = \omega L + \rho K \end{eqnarray}

Using Lagrange multipliers you can then define the Lagrangian as

$$\mathcal{L}(L,K,\lambda) \stackrel{\rm def}{=} A L^\alpha K^\beta - \lambda(C - \omega L - \rho K)$$

And the problem is reduced to finding the solution to

\begin{eqnarray} \frac{\partial \mathcal{L}}{\partial L} &=& 0 \\ \frac{\partial \mathcal{L}}{\partial K} &=& 0 \\ \frac{\partial \mathcal{L}}{\partial \lambda} &=& 0 \end{eqnarray}

I will leave the rest to you

• Suggesting an edit, the $+\rho K$ should be $-\rho K$ right (in the Lagrangian)? – Václav Mordvinov Nov 13 '17 at 22:26
• @VáclavMordvinov Thanks! You're absolutely right, already fixed the typo – caverac Nov 13 '17 at 22:27
• Yes this worked thank you! However could I possibly use "alpha + beta > 1" as a second constraint? – Sven van Holten Nov 14 '17 at 12:07
• @SvenvanHolten You could, but then both $\alpha$ and $\beta$ would need to be included as dynamical variables and a new multiplier needs to be added $$\mathcal{L}(L, K, \color{red}{\alpha}, \color{red}{\beta}, \lambda, \color{red}{\mu})$$ – caverac Nov 14 '17 at 12:54
• @caverac pretty sure that that adding α and β as dynamic variables will complicate it by a lot. Isn't it better to find α and β in terms of K & L? Otherwise if it isn't possible, how exactly do you add a&b as dynamic variables – Sven van Holten Nov 14 '17 at 13:23