Here, cases are considered which contain
$x=0$ or
$y=0$ as
optimum. For more complex problems with boundary solutions for more than two
variables, there is the so-called Kuhn-Tucker method, which, however, would go
beyond the scope of this text. Explanations can be found in corresponding
textbooks.

In the cases treated so far, we were always able to find points where the
marginal rate of substitution (MRS) equaled the price ratio. However, if
the MRS (i.e. the marginal utility ratio) is limited, i.e. there are values
$\alpha $ and/or
$\beta $ to
which applies,

$$\alpha <\mathit{MRS}<\beta $$ |

it means that for price ratios $\frac{{p}_{x}}{{p}_{y}}$
which are smaller than $\alpha $
or larger than $\beta $ there is no
solution for the equation $\mathit{MRS}=\frac{{p}_{x}}{{p}_{y}}$.
There are no quantities of goods for which the slope of the budget line equals the
slope of the isoquant.

If the MRS is greater than the price ratio
$\frac{{p}_{x}}{{p}_{y}}$,

$$\mathit{MRS}>\frac{{p}_{x}}{{p}_{y}}$$ |

it means that $x$
per additional unit brings more benefit than you lose by giving up as many units
$y$ as you need to save the
money for one unit $x$.
So by substituting $x$
for $y$
one could increase the benefit. This is possible until either the MRS has
adapted to the price level by changing the goods ratios or until all
$y$ have been
saved, i.e. until $y=0$
angelangt ist. Since $y$
cannot become negative, the solution in this case is to consume only
$x$.

Usually, the isoquants in our models are convex, i.e. the marginal
utility of a good decreases the more it is consumed. Thus, the MRS
always changes in the direction of the price ratio due to substitution.
However, if the price ratio is outside the value range of the MRS, i.e.
$\frac{{p}_{x}}{{p}_{y}}<\alpha $ or
$\beta <\frac{{p}_{x}}{{p}_{y}}$, then
only a partial adjustment can take place, which ends when one of the two goods
has been fully substituted.

If the MRS is smaller than the price ratio
$\frac{{p}_{x}}{{p}_{y}}$,
the analogous consideration leads to the boundary point
$x=0$.

In the graph below, the MRS as the slope of the red tangent is directly
comparable with the slope of the budget line. By increasing one of the
prices, the price ratio can be adjusted in such a way that a boundary
solution comes into play. Please note, a change of the budget, and thus a
change of the achievable isoquant, alters the limits of the MRS and can
thereby trigger a switch between boundary solution and normal solution.

The following example with $c=3$ was used in the diagram.

$$U=\sqrt{\left(x+c\right)\left(y+c\right)}$$ |

The MRS results in

$$\begin{array}{llll}\hfill \mathit{GRS}& =\frac{\frac{d}{\mathit{dx}}U\left(x,y\right)}{\frac{d}{\mathit{dy}}U\left(x,y\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{2\sqrt{\left(y+c\right)\left(y+c\right)}}{2\sqrt{\left(x+c\right)\left(x+c\right)}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{y+c}{x+c}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$When equating with the price ratio and using the constraint, it results in:

$$\begin{array}{llll}\hfill \frac{\frac{d}{\mathit{dx}}U\left(x,y\right)}{\frac{d}{\mathit{dy}}U\left(x,y\right)}& =\frac{y+c}{x+c}=\frac{{p}_{x}}{{p}_{y}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y{p}_{y}& ={p}_{x}\left(x+c\right)-c{p}_{y}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x{p}_{x}+{p}_{x}\left(x+c\right)-c{p}_{y}& =B\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =\frac{B+c\left({p}_{y}-{p}_{x}\right)}{2{p}_{x}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =\frac{B+c\left({p}_{x}-{p}_{y}\right)}{2{p}_{y}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$The functional equation of the isoquant is

$$\begin{array}{llll}\hfill U& =\sqrt{\left(x+c\right)\left(y+c\right)}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =\frac{{U}^{2}}{x+c}-c\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\left(\frac{B+c\left({p}_{y}-{p}_{x}\right)}{2{p}_{x}}+c\right)\left(\frac{B+c\left({p}_{x}-{p}_{y}\right)}{2{p}_{y}}+c\right)}{x+c}-c\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{{\left(B+c\left({p}_{x}+{p}_{y}\right)\right)}^{2}}{4{p}_{x}{p}_{y}\left(c+x\right)}-c\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Boundary solution $x=0$ and derivative at $x=0$

$$\begin{array}{llll}\hfill y{p}_{y}-B& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =\frac{B}{{p}_{y}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \mathit{GR}{S}_{x=0}& ={\left.\frac{y+c}{x+c}\right|}_{x=0}=\frac{\frac{B}{{p}_{y}}+c}{c}=\frac{B}{c{p}_{y}}+1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$So if $\frac{B}{c{p}_{y}}+1\le \frac{{p}_{x}}{{p}_{y}}$, then the boundary solution $x=0$ applies. Boundary solution $y=0$ and derivative at $y=0$

$$\begin{array}{llll}\hfill x{p}_{x}-B& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =\frac{B}{{p}_{x}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \mathit{GR}{S}_{y=0}& ={\left.\frac{y+c}{x+c}\right|}_{x=0}=\frac{c}{\frac{B}{{p}_{x}}+c}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$So if $\frac{c}{\frac{B}{{p}_{x}}+c}\ge \frac{{p}_{x}}{{p}_{y}}\iff \frac{{p}_{y}}{{p}_{x}}\ge \frac{B}{c{p}_{x}}+1$, then the
boundary solution $y=0$
applies.

(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