Answer
We prove:
$\frac{{{\rm{Marginal{\ }utility{\ }of{\ }Good{\ }1}}}}{{{\rm{Marginal{\ }utility{\ }of{\ }Good{\ }2}}}} = \frac{{{U_{{x_1}}}\left( {a,b} \right)}}{{{U_{{x_2}}}\left( {a,b} \right)}} = \frac{{{p_1}}}{{{p_2}}}$
Work Step by Step
Given a budget of $L$ dollars, we have the constraint:
${x_1}{p_1} + {x_2}{p_2} = L$,
where ${p_j}$ are dollar prices per unit of good $j$, and ${x_j}$ are number of good $j$ consumed.
Our task is to maximize the utility function $U\left( {{x_1},{x_2}} \right)$ subject to the constraint $g\left( {{x_1},{x_2}} \right) = {x_1}{p_1} + {x_2}{p_2} - L = 0$, where ${x_1},{x_2} \ge 0$.
Using Theorem 1, the Lagrange condition $\nabla U = \lambda \nabla g$ yields
$\left( {{U_{{x_1}}},{U_{{x_2}}}} \right) = \lambda \left( {{p_1},{p_2}} \right)$
So, the Lagrange equations are
${U_{{x_1}}} = \lambda {p_1}$, ${\ \ \ }$ ${U_{{x_2}}} = \lambda {p_2}$
$\lambda = \frac{{{U_{{x_1}}}}}{{{p_1}}} = \frac{{{U_{{x_2}}}}}{{{p_2}}}$
So, $\frac{{{U_{{x_1}}}}}{{{U_{{x_2}}}}} = \frac{{{p_1}}}{{{p_2}}}$.
By definition ${U_{{x_j}}}$ is the marginal utility of the $j$th good. So, at the consumption level $\left( {a,b} \right)$ we obtain:
$\frac{{{\rm{Marginal{\ }utility{\ }of{\ }Good{\ }1}}}}{{{\rm{Marginal{\ }utility{\ }of{\ }Good{\ }2}}}} = \frac{{{U_{{x_1}}}\left( {a,b} \right)}}{{{U_{{x_2}}}\left( {a,b} \right)}} = \frac{{{p_1}}}{{{p_2}}}$
