We are now solving the already known problem of the budget optimum: Maximize utility $U\left(x,y\right)$ with a given budget $B$, when the prices of goods are ${p}_{x}$ and ${p}_{y}$. For the utility function, non-saturation and concavity apply.

$$\underset{x,y}{\mathrm{max}}U\left(x,y\right)\text{undertheconditionthat}x{p}_{x}+y{p}_{y}=B.$$ |

We form the Lagrange function

$$\mathbb{L}\left(x,y,\lambda \right)=U\left(x,y\right)+\lambda \left(x{p}_{x}+y{p}_{y}-B\right)$$ |

and the first order conditions:

$$\begin{array}{lll}\hfill \frac{d}{\mathit{dx}}U\left(x,y\right)+\lambda {p}_{x}& =0\phantom{\rule{2em}{0ex}}& \hfill \text{(9.1)}\phantom{\rule{0.33em}{0ex}}\\ \hfill \frac{d}{\mathit{dy}}U\left(x,y\right)+\lambda {p}_{y}& =0\phantom{\rule{2em}{0ex}}& \hfill \text{(9.2)}\phantom{\rule{0.33em}{0ex}}\\ \hfill x{p}_{x}+y{p}_{y}-B=0& \phantom{\rule{2em}{0ex}}& \hfill \text{(9.3)}\phantom{\rule{0.33em}{0ex}}\end{array}$$The FOC 3 represents the second order constraint. The other two are transformed by adding $-\lambda {p}_{x}$ and $-\lambda {p}_{y}$, respectively

$$\begin{array}{lll}\hfill \frac{d}{\mathit{dx}}U\left(x,y\right)& =-\lambda {p}_{x}\phantom{\rule{2em}{0ex}}& \hfill \text{(9.4)}\phantom{\rule{0.33em}{0ex}}\\ \hfill \frac{d}{\mathit{dy}}U\left(x,y\right)& =-\lambda {p}_{y}\phantom{\rule{2em}{0ex}}& \hfill \text{(9.5)}\phantom{\rule{0.33em}{0ex}}\end{array}$$and then divide the equations with each other. This cancels $\lambda $.

$$\frac{\frac{d}{\mathit{dx}}U\left(x,y\right)}{\frac{d}{\mathit{dy}}U\left(x,y\right)}=\frac{{p}_{x}}{{p}_{y}}$$ | (9.6) |

The resulting equation represents the central point of the solution. It
represents a relationship between the marginal utility ratio and the price
ratio.

1) The price ratio $\frac{{p}_{x}}{{p}_{y}}$
is the slope of the budget line.

2) The marginal utility ratio $\frac{\frac{d}{\mathit{dx}}U\left(x,y\right)}{\frac{d}{\mathit{dy}}U\left(x,y\right)}$
is the marginal rate of substitution, i.e. the slope of the isoquant.

3) The maximum is reached where the slope of the budget line is equal to the
slope of the isoquant and lies on the budget line.

4) Now take equation (4) and the second order constraint and solve the system of two
equations for $x$
and $y$.

Equation (4) represents an intuitively very plausible heuristic. If good
$x$ has a
high marginal utility, then the willingness to pay (price) for this good is also high.
If the marginal utility is low, then the price paid for it is also low. Thus, the
higher the value that a consumer assigns to a good, the higher the price he is
willing to pay for it.

As a notation for the partial derivative, the indexing of the function
name with the variable in respect to which it is differentiated
$\frac{d}{\mathit{dx}}U\left(x,y\right)={U}_{x}\left(x,y\right)$, has
often been adopted. Then, equation (4) is written as

$$\frac{{U}_{x}\left(x,y\right)}{{U}_{y}\left(x,y\right)}=\frac{{p}_{x}}{{p}_{y}},$$ | (9.7) |

where the index of the function is to be read as a derivative and the index of the price variable as a simple classification.

The slope of the isoquant can be explained by the implicit function theorem.
This theorem states that the slope of an isoquant is the negative ratio of the
partial derivatives. To be more precise:

If $F\left(x,y\right)$ is a function
with $\frac{d}{\mathit{dy}}F\left({x}_{0},{y}_{0}\right)\ne 0$ and
$f\left(x\right)$ is a function that
keeps the level of $F$
constant, i.e. $F\left(x,f\left(x\right)\right)=c$,
then:

$${f}^{\prime}\left({x}_{0}\right)=-\frac{\frac{d}{\mathit{dx}}F\left({x}_{0},{y}_{0}\right)}{\frac{d}{\mathit{dy}}F\left({x}_{0},{y}_{0}\right)}$$ |

where ${y}_{0}=f\left({x}_{0}\right)$. The idea of the proof goes back to the multidimensional chain rule.

$$\begin{array}{llll}\hfill c& \equiv F\left(x,f\left(x\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 0& \equiv \frac{d}{\mathit{dx}}F\left(x,f\left(x\right)\right)+\frac{d}{\mathit{dy}}F\left(x,f\left(x\right)\right){f}^{\prime}\left(x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$(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