Chapter 15 - Differentiation in Several Variables - 15.8 Lagrange Multipliers: Optimizing with a Constraint - Exercises - Page 833: 51

(a) we show that the maximum of $U\left( {{x_1},{x_2}} \right)$ subject to the budget constraint is equal to $\frac{{{c^2}}}{{4{p_1}{p_2}}}$. (b) $\lambda = \frac{c}{{2{p_1}{p_2}}}$ (c) we prove the interpretation: $\lambda$ is the rate of increase in utility per unit increase in total budget $c$.

(a) We are given the utility function $U\left( {{x_1},{x_2}} \right) = {x_1}{x_2}$. Our task is to find the maximum of $U\left( {{x_1},{x_2}} \right)$ subject to the budget constraint $g\left( {{x_1},{x_2}} \right) = {p_1}{x_1} + {p_2}{x_2} - c = 0$. Using Theorem 1, the Lagrange condition $\nabla U = \lambda \nabla g$ yields $\left( {{x_2},{x_1}} \right) = \lambda \left( {{p_1},{p_2}} \right)$ So, the Lagrange equations are ${x_2} = \lambda {p_1}$, ${\ \ \ \ }$ ${x_1} = \lambda {p_2}$ We may assume ${x_1},{x_2} \ne 0$. So, $\lambda = \frac{{{x_2}}}{{{p_1}}} = \frac{{{x_1}}}{{{p_2}}}$ So, ${x_2} = {x_1}\frac{{{p_1}}}{{{p_2}}}$. Substituting ${x_2} = {x_1}\frac{{{p_1}}}{{{p_2}}}$ in the constraint $g$ gives ${p_1}{x_1} + {p_2}\left( {{x_1}\frac{{{p_1}}}{{{p_2}}}} \right) - c = 0$ $2{p_1}{x_1} = c$ So, ${x_1} = \frac{c}{{2{p_1}}}$. Using ${x_2} = {x_1}\frac{{{p_1}}}{{{p_2}}}$, we obtain the critical point $\left( {{x_1},{x_2}} \right) = \left( {\frac{c}{{2{p_1}}},\frac{c}{{2{p_2}}}} \right)$. The extreme value corresponding to $\left( {\frac{c}{{2{p_1}}},\frac{c}{{2{p_2}}}} \right)$ is $U\left( {\frac{c}{{2{p_1}}},\frac{c}{{2{p_2}}}} \right) = \frac{{{c^2}}}{{4{p_1}{p_2}}}$ Since ${x_1},{x_2} > 0$, the utility function $U\left( {{x_1},{x_2}} \right) = {x_1}{x_2}$ is increasing. So, we conclude that the maximum of $U\left( {{x_1},{x_2}} \right)$ subject to the budget constraint is equal to $\frac{{{c^2}}}{{4{p_1}{p_2}}}$. (b) From part (a) we obtain $\lambda = \frac{{{x_2}}}{{{p_1}}} = \frac{{{x_1}}}{{{p_2}}}$. At the critical point $\left( {{x_1},{x_2}} \right) = \left( {\frac{c}{{2{p_1}}},\frac{c}{{2{p_2}}}} \right)$, we have $\lambda = \frac{c}{{2{p_1}{p_2}}}$. (c) From part (a) we obtain the maximum of $U\left( {{x_1},{x_2}} \right)$ subject to the budget constraint is equal to $\frac{{{c^2}}}{{4{p_1}{p_2}}}$. If we consider the maximum of $U\left( {{x_1},{x_2}} \right)$ as a function of total budget $c$ and write: $W\left( c \right) = \frac{{{c^2}}}{{4{p_1}{p_2}}}$, then the rate of increase in utility per unit increase in total budget $c$ is just the derivative of $W$ with respect to $c$. That is, $W'\left( c \right) = \frac{c}{{2{p_1}{p_2}}}$ Comparing $W'\left( c \right)$ with $\lambda = \frac{c}{{2{p_1}{p_2}}}$ in part (b), we may interpret that $\lambda$ is the rate of increase in utility per unit increase in total budget $c$.