# Constrained Optimization

An example of applying the Lagrangian technique.

Posted: Sep 11th, 2023 - Modified: Sep 11th, 2023

## ¶ The Problem:

\begin{align*} \max_{x,y}\,\,\,&x^{\alpha}y^{1-\alpha}\\ \text{s.t.\,\,\,}&x\geq0\\ &y\geq0\\ &x+y\leq10 \end{align*}

## ¶ The Solution

### ¶ Setting up the Lagrangian:

To set up the lagrangian, we add in a term for each binding constraint. (“Binding” here means, that at the optimum, the constraint will hold with equality.)

For this problem, the budget constraint ($x+y\leq10$) will be binding, and the non-negativity constraints won’t be. (Why? That argument is left as an exercise for the reader.)

$\mathcal{L}=x^{\alpha}y^{1-\alpha}-\lambda\cdot\left[x+y-10\right]$

The basic recipe for the Lagrangian is

2. For each binding constraint
• Take the constraint and move everything to one side.
• Multiply the constraint by a new variable, which is called the “Lagrange Multiplier” for that constraint.

If you have multiple binding constraints, you’ll need a lagrange multiplier for each one. $\lambda$ is the typical symbol used for these variables, and that’s what I’ve used above.

Note also that we could write the above equation like so:

$\mathcal{L}=x^{\alpha}y^{1-\alpha}+\lambda\cdot\left[10-x-y\right]$

This is equivalent.

### ¶ First Order Conditions:

Now we take the partial derivatives of the Lagrangian, and set each derivative equal to zero:

$\frac{\partial\mathcal{L}}{\partial x}=\alpha x^{\alpha-1}y^{1-\alpha}-\lambda=0\tag{x}$ $\frac{\partial\mathcal{L}}{\partial y}=\left(1-\alpha\right)x^{\alpha}y^{-\alpha}-\lambda=0\tag{y}$ $\frac{\partial\mathcal{L}}{\partial\lambda}=10-x-y=0\tag{λ}$

Next, we can rearrange these equations to get:

$\alpha x^{\alpha-1}y^{1-\alpha}=\lambda\tag{x}$ $\left(1-\alpha\right)x^{\alpha}y^{-\alpha}=\lambda\tag{y}$ $x+y=10\tag{λ}$

The solution to this system of equations gives us the critical points for $\mathcal{L}$, and thus potential solutions to the original problem.

Note that the last equation is just our budget constraint! This is the magic behind the Lagrangian: it transforms constraints into FOCs.

### ¶ Solving the System of Equations:

If we want to solve for x and y, we need to start by getting rid of that extra variable \lambda we added. In this case, that’s easy. We just combine the first two FOCs like so:

$\alpha x^{\alpha-1}y^{1-\alpha}=\lambda=\left(1-\alpha\right)x^{\alpha}y^{-\alpha}$ $\alpha x^{\alpha-1}y^{1-\alpha}=\left(1-\alpha\right)x^{\alpha}y^{-\alpha}$

The next step is to solve for one of the two remaining variables in terms of the other. We can start by cancelling x from one side and y from the other.

\begin{align*} \alpha x^{\alpha-1}y^{1-\alpha}&=\left(1-\alpha\right)x^{\alpha}y^{-\alpha}\\ \frac{\alpha\cancel{x^{\alpha-1}}y^{1-\alpha}}{\cancel{x^{\alpha-1}}y^{-\alpha}}&=\frac{\left(1-\alpha\right)x^{\alpha}\cancel{y^{-\alpha}}}{x^{\alpha-1}\cancel{y^{-\alpha}}}\\ \frac{\alpha y^{1-\alpha}}{y^{-\alpha}}&=\frac{\left(1-\alpha\right)x^{\alpha}}{x^{\alpha-1}} \end{align*}

Next we can simplify and rearrange a bit more to get:

\begin{align*} \frac{\alpha y^{1-\alpha}}{y^{-\alpha}}&=\frac{\left(1-\alpha\right)x^{\alpha}}{x^{\alpha-1}}\\ \alpha y^{1-\alpha-(-\alpha)}&=\left(1-\alpha\right)x^{\alpha-(\alpha-1)}\\ \alpha y^{1}&=\left(1-\alpha\right)x^{1}\\ y&=\frac{1-\alpha}{\alpha}\cdot x \end{align*}

Substitute this expression of y into the budget and solve for x:

\begin{align*} 10 &=x+y\\ 10 &=x+\left(\frac{1-\alpha}{\alpha}\cdot x\right)\\ 10 &=\left(1+\frac{1-\alpha}{\alpha}\right)\cdot x\\ 10 &=\left(\frac{\alpha}{\alpha}+\frac{1-\alpha}{\alpha}\right)\cdot x\\ 10 &=\left(\frac{1}{\alpha}\right)\cdot x\\ \alpha\cdot10 &=x \end{align*}

Fantastic! We’ve solved for x. It’s fine that $\alpha$ is still in there because $\alpha$ is a constant, not a variable. The last step is to plug our solution for x back into the expression for y:

\begin{align*} y &=\frac{1-\alpha}{\alpha}\cdot x\\ y &=\frac{1-\alpha}{\cancel{\alpha}}\cdot\left(\cancel{\alpha}10\right)\\ y &=\left(1-\alpha\right)10\\ \end{align*}

### ¶ And We’ve Found the Solution:

\boxed{ \begin{align*} x &=10\cdot\alpha\\ y &=10\cdot\left(1-\alpha\right) \end{align*}}

You may remember from micro that finding the solution to a constrained optimization problem involves the condition that at the optimum, the marginal rate of substitution (MRS) is equal to the marginal rate of transformation (MRT).

In this problem, where $u(x,y)=x^{\alpha}y^{1-\alpha}$, we can find the MRS by dividing one marginal utility by the other:

$MRS\equiv\frac{(\partial u/\partial y)}{(\partial u/\partial x)}=\frac{\left(1-\alpha\right)x^{\alpha}y^{-\alpha}}{\alpha x^{\alpha-1}y^{1-\alpha}}=\frac{1-\alpha}{\alpha}\cdot\frac{x}{y}$

And in the budget constraint, $x$ and $y$ each “cost” one unit, so the MRT is $\frac{1}{1}=1$.

Setting MRT=MRS, we get:

$1 =\frac{1-\alpha}{\alpha}\cdot\frac{x}{y}$

which we can rearrange to get

$y=\frac{1-\alpha}{\alpha}\cdot x$

Now wait a second. Is that the same equation we go by combining our FOCs?

It is!

This is, in fact, where the MRS=MRT rule comes from. And the same process can be used to derive similar “characterizing equations” for other constrained optimization problems.