It follows that \((\lambda p_{2},\lambda p_{1})\) is actually the unique global maximum of \(L\left(x_{1},x_{2}\right)\) so we have \(\left(x_{1}(\lambda),x_{2}(\lambda)\right)=(\lambda p_{2},\lambda p_{1})\).
Now let us find the value \(\lambda^{*}\) for \(\lambda\) for which the budget constraint holds with equality: \(p_{1}x_{1}\left(\lambda^{*}\right)+p_{2}x_{2}\left(\lambda^{*}\right)=y\) .
What is \(\lambda^{*}\)?
True: \(\lambda^{*}=\frac{y}{2\ p_{1}p_{2}}\)
False: \(\lambda^{*}=\frac{y}{p_{1}p_{2}}\)
Explanation:
Plugging \(\left(x_{1}(\lambda^{*}),x_{2}(\lambda^{*})\right)=(\lambda^{*}p_{2},\lambda^{*}p_{1})\) into \(p_{1}x_{1}\left(\lambda^{*}\right)+p_{2}x_{2}\left(\lambda^{*}\right)=y\) yields:
\(\begin{equation}p_{1}\lambda^{*}p_{2}+p_{2}\lambda^{*}\ p_{1}=y\nonumber \\ \end{equation}\) \(\begin{equation}\lambda^{*}=\frac{y}{2\ p_{1}p_{2}}\nonumber \\ \end{equation}\)