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 $U\left(x,y\right)=\alpha \ln <!--\; FUNCTION\; APPLICATION\; -->x+\beta \sqrt{y}$ for a given budget $B$, when the prices of goods are $p_x$ and $p_y$, i.e.

$$\underset{x,y}{\max <!--\; FUNCTION\; APPLICATION\; -->}\alpha \ln <!--\; FUNCTION\; APPLICATION\; -->x+\beta \sqrt{y}\text{ under the condition that }x{p}_x+y{p}_y=B.$$

We form the Lagrange function

$$𝕃\left(x,y,\lambda \right)=\alpha \ln <!--\; FUNCTION\; APPLICATION\; -->x+\beta \sqrt{y}+\lambda \left(x{p}_x+y{p}_y-B\right)$$

and the first order conditions:

$$\begin{array}{lll}\hfill \frac{d}{\mathit{dx}}\left(\alpha \ln <!--\; FUNCTION\; APPLICATION\; -->x+\beta \sqrt{y}\right)+\lambda p_x& =\frac{\alpha }{x}+\lambda p_x=0& \hfill \text{(9.14)}\\ \hfill \frac{d}{\mathit{dy}}\left(\alpha \ln <!--\; FUNCTION\; APPLICATION\; -->x+\beta \sqrt{y}\right)+\lambda p_y& =\beta \frac{1}{2\sqrt{y}}+\lambda p_y=0& \hfill \text{(9.15)}\\ \hfill x{p}_x+y{p}_y-B& =0& \hfill \text{(9.16)}\end{array}$$

The FOC 3 represents the second order condition. We transform the other two by adding $-\lambda p_x$ and $-\lambda p_y$, respectively

$$\begin{array}{lll}\hfill \frac{\alpha }{x}& =-\lambda p_x& \hfill \text{(9.17)}\\ \hfill \frac{\beta }{2\sqrt{y}}& =-\lambda p_y& \hfill \text{(9.18)}\end{array}$$

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:

$$\begin{array}{llll}\hfill x{p}_x+y{p}_y& =B& \hfill & \\ \hfill \frac{2\alpha p_y}{\beta }\sqrt{y}& =x{p}_x& \hfill & \end{array}$$

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

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