Constrained Optimization
An example of applying the Lagrangian technique.
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
- Start with the thing you’re trying to optimize.
- 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:
\[\boxed{ \begin{align*} x &=10\cdot\alpha\\ y &=10\cdot\left(1-\alpha\right) \end{align*}}\]What about MRS=MRT?
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.