Question 10
Now let us carry out the following procedure:
- Write down \(L=u-\lambda\left(p_{1}x_{1}+\ldots+p_{n}x_{n}\right)\).
(\(L\) is often called ‘the Lagrangian’ and \(\lambda\) is called ‘a
Lagrange multiplier’)
- For each \(i\in\{1,\ldots,n\}\) compute \(\frac{\partial L}{\partial x_{\text{i\ }}}=0\), i.e. \(\frac{\partial u}{\partial x_{i}}-\lambda p_{i}=0\) (These
equations are called the first order conditions for \(x_{1},\ldots x_{n}\))
- Solve the system of \(n+1\) equations \(\{\frac{\partial u}{\partial x_{1}}-\lambda p_{1}=0,\ldots,\frac{\partial u}{\partial x_{n}}-\lambda p_{n}=0,p_{1}x_{1}+\ldots+p_{n}x_{n}=y\}\) in the \(n+1\) unknowns.
To solve this system, use the following procedure:
- Solve \(\frac{\partial u}{\partial x_{1}}-\lambda p_{1}=0\) for \(x_{1}\), solve \(\frac{\partial u}{\partial x_{2}}-\lambda p_{2}=0\) for \(x_{2}\ldots,\ \text{solve\ }\frac{\partial u}{\partial x_{n}}-\lambda p_{n}=0\text{\ for\ }x_{n}\)
- substitute all the expression obtained in step 3.1 for \(x_{1},\ x_{2},\ldots,x_{n}\) into \(p_{1}x_{1}+\ldots+p_{n}x_{n}=y\).
- Then solve for \(\lambda\) the equation obtained in step 3.2
- Substitute the expression obtained for \(\lambda\) in step 3.3 into
the expression obtained for \(x_{1}\) in step 3.1. This gives an
expression for \(x_{1}\) not involving λ. Similarly,
proceed for \(x_{2},\ldots x_{n}\).
Applying this procedure to our problem with \(u\left(x_{1},x_{2}\right)=\log{\left(x_{1}\right)+log(x_{2})}\),
what do we obtain?
True \(x_{1}=\frac{y}{2p_{1}}\)
False \(x_{1}=\frac{y}{p_{1}}\)
False \(x_{1}=\frac{p_{2}}{p_{1}+p_{2}}\frac{y}{p_{1}}\)
Explanation
We follow the steps of the procedure:
- \(L=\log\left(x_{1}\right)+\log\left(x_{2}\right)-\lambda\left(p_{1}x_{1}+p_{2}x_{2}\right)\)
- \(\frac{\partial L}{\partial x_{1}}=0,\frac{\partial L}{\partial x_{2}}=0\) becomes \(\frac{1}{x_{1}}-\lambda p_{1}=0,\ \frac{1}{x_{2}}-\lambda p_{2}=0\)
- We have the following system of \(3\) equations in the 3 unkwowns \((x_{1},\ x_{2},\lambda)\):
\(\begin{equation}\frac{1}{x_{1}}-\lambda p_{1}=0,\ \frac{1}{x_{2}}-\lambda p_{2}=0,p_{1}x_{1}+p_{2}x_{2}=y\nonumber \\ \end{equation}\)
Then one can repeat this procedure for these remaining \(n-1\) equations. In the case of consumer choice problems one usually use the
following faster procedure::
- Solving \(\frac{1}{x_{1}}-\lambda p_{1}\) for \(x_{1}\) gives \(x_{1}=\frac{1}{\lambda p_{1}}\). Similarly, we obtain \(x_{2}=\frac{1}{\lambda p_{2}}\).
- Now substituting this into \(p_{1}x_{1}+p_{2}x_{2}=y\) gives: \(p_{1}\frac{1}{\lambda p_{1}}+p_{2}\frac{1}{\lambda p_{2}}=y\)
- Solving \(p_{1}\frac{1}{\lambda p_{1}}+p_{2}\frac{1}{\lambda p_{2}}=y\) for \(\lambda\) gives: \(\lambda=\frac{2}{y}\)
- Substituting \(\lambda=\frac{2}{y}\) into \(x_{1}=\frac{1}{\lambda p_{1}}\) gives \(x_{1}=\frac{y}{2p_{1}}\)