Question 6
So far we have shown that if we find a value \(\lambda^{*}\) such that the pair \(\left(x_{1}\left(\lambda^{*}\right),x_{2}\left(\lambda^{*}\right)\right)\) that maximizes \(L=x_{1}x_{2}-\lambda^{*}\left(p_{1}x_{1}+p_{2}x_{2}\right)\) satisfies the budget constraint with equality (i.e. such that \(p_{1}x_{1}\left(\lambda^{*}\right)+p_{2}x_{2}\left(\lambda^{*}\right)=y\) then \(\left(x_{1}\left(\lambda^{*}\right),x_{2}\left(\lambda^{*}\right)\right)\) is a solution to the original problem of maximizing \(u\left(x_{1},x_{2}\right)=x_{1}x_{2}\) under the budget constraint \(p_{1}x_{1}+p_{2}x_{2}\leq y\).
Now let us try to actually find this value \(\lambda^{*}\) for our particular problem with \(u\left(x_{1},x_{2}\right)=\log\left(x_{1}\right)+log(x_{2})\). First let us first find \(\left(x_{1}\left(\lambda\right),x_{2}\left(\lambda\right)\right)\) for any \(\lambda\). (As we have said above, \(\left(x_{1}\left(\lambda\right),x_{2}\left(\lambda\right)\right)\) denotes the solution to the problem of maximizing \(L=\log\left(x_{1}\right)+log(x_{2})-\lambda\left(p_{1}x_{1}+p_{2}x_{2}\right))\). Drag and drop the equations into the categories:
Category 1: First order condition for \(x_{1}\)
\(\frac{\partial L}{\partial x_{1}}=0\)
\(\frac{1}{x_{1}}-\lambda p_{1}=0\)
Category 1: First order condition for \(x_{1}\)
\(\frac{\partial L}{\partial x_{2}}=0\)
\(\frac{1}{x_{2}}-\lambda p_{2}=0\)