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 \(u\left(x_{1},x_{2}\right)=\log{(x_{1})}+log(x_{2})\).