Description: In the coming capsules we will often be faced with
constraint optimization problems. In this capsule we will motivate,
define and practice the Lagrangian method for solving these problems.
Approximate duration: 25 minutes
Resource:
In the coming capsules we will often be faced with problems such as that
of finding which consumption bundles \(x\) maximise \(u(x)\) subject to \(p\ x\leq y\). We will follow the following procedure: We will write
down the Lagrangian \(L=u-\lambda\ p\ x\), differentiate with
respect to each component of \(x\) and then consider the set of
conditions
{\(\frac{\partial L}{\partial x_{1}}=0,\ \ldots,\ \frac{\partial L}{\partial x_{1}}=0,px=y\}\).
These are \(n+1\) conditions in the \(n+1\) unkwowns \(\{x_{1},\ldots,x_{n},\lambda\}\). We then find the solutions to this
system of equations. Typically (i.e. in the applications in economics),
the expressions \((x_{1},\ldots,x_{n})\) that we find in this way is a
global maximum of \(L=u-\lambda\ p\ x\). In this capsule we will see
why any such solution is also a solution to the original problem of
maximising \(u(x)\) subject to \(p\ x\leq y\).
For illustration we will at the end of this capsule study the special case
of the above problem where there are two goods and \(u\left(x_{1},x_{2}\right)=\log{(x_{1})}+log(x_{2})\).