### 9.5 The Lagrange formalism for the example of another utility function

We now solve the well-known problem of the household optimum for the utility function $U\left(x,y\right)=\alpha \mathrm{ln}x+\beta \sqrt{y}$ for a given budget $B$, when the prices of goods are ${p}_{x}$ and ${p}_{y}$, i.e.

We form the Lagrange function

 $𝕃\left(x,y,\lambda \right)=\alpha \mathrm{ln}x+\beta \sqrt{y}+\lambda \left(x{p}_{x}+y{p}_{y}-B\right)$

and the first order conditions:

$\begin{array}{lll}\hfill \frac{d}{\mathit{dx}}\left(\alpha \mathrm{ln}x+\beta \sqrt{y}\right)+\lambda {p}_{x}& =\frac{\alpha }{x}+\lambda {p}_{x}=0\phantom{\rule{2em}{0ex}}& \hfill \text{(9.14)}\phantom{\rule{0.33em}{0ex}}\\ \hfill \frac{d}{\mathit{dy}}\left(\alpha \mathrm{ln}x+\beta \sqrt{y}\right)+\lambda {p}_{y}& =\beta \frac{1}{2\sqrt{y}}+\lambda {p}_{y}=0\phantom{\rule{2em}{0ex}}& \hfill \text{(9.15)}\phantom{\rule{0.33em}{0ex}}\\ \hfill x{p}_{x}+y{p}_{y}-B& =0\phantom{\rule{2em}{0ex}}& \hfill \text{(9.16)}\phantom{\rule{0.33em}{0ex}}\end{array}$

The FOC 3 represents the second order condition. We transform the other two by adding $-\lambda {p}_{x}$ and $-\lambda {p}_{y}$, respectively

$\begin{array}{lll}\hfill \frac{\alpha }{x}& =-\lambda {p}_{x}\phantom{\rule{2em}{0ex}}& \hfill \text{(9.17)}\phantom{\rule{0.33em}{0ex}}\\ \hfill \frac{\beta }{2\sqrt{y}}& =-\lambda {p}_{y}\phantom{\rule{2em}{0ex}}& \hfill \text{(9.18)}\phantom{\rule{0.33em}{0ex}}\end{array}$

and then dividing the equations with each other. This cancels $\lambda$.

 $\frac{\frac{\alpha }{x}}{\frac{\beta }{2\sqrt{y}}}=\frac{{p}_{x}}{{p}_{y}}$ (9.19)

The resulting equation represents the central point of the solution. It represents a relationship between the marginal utility ratio and the price ratio. We now summarize this equation with the constraint, where we have slightly transformed both:

$\begin{array}{llll}\hfill x{p}_{x}+y{p}_{y}& =B\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{2\alpha {p}_{y}}{\beta }\sqrt{y}& =x{p}_{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

We insert and obtain

 $B=\frac{2\alpha {p}_{y}}{\beta }\sqrt{y}+y{p}_{y}$

and therefore the only positive solution (Note: Consider this as a quadratic equation in $\sqrt{y}$!)

 $y={\left(-\frac{\alpha }{\beta }+\sqrt{{\left(\frac{\alpha }{\beta }\right)}^{2}+\frac{B}{{p}_{y}}}\right)}^{2}$

$x$ then results as

 $x=\frac{1}{{p}_{x}}\frac{2\alpha {p}_{y}}{\beta }\sqrt{y}=\frac{{p}_{y}}{{p}_{x}}\frac{2\alpha }{\beta }\left(-\frac{\alpha }{\beta }+\sqrt{{\left(\frac{\alpha }{\beta }\right)}^{2}+\frac{B}{{p}_{y}}}\right).$

(c) by Christian Bauer
Prof. Dr. Christian Bauer
Chair of monetary economics
Trier University
D-54296 Trier
Tel.: +49 (0)651/201-2743
E-mail: Bauer@uni-trier.de
URL: https://www.cbauer.de