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### Engel's Law and the Consumption Function

A consumption function describes consumption expenditure C as a function of income Y.

The diagram here depicts two different graphs of rational functions that follow Engel's law.

Both of them have the form

$C(Y)=\large{\frac{{a \cdot Y+b}}{{2Y+5}}}$.

Since the numerator and denominator polynomials have the same degree, we know there must be a horizontal asymptote to start with.

a)

The horizontal asymptote is given by C=4, and the graph goes through the origin, i.e. through the point (0,0).
Determine the equation of this consumption function.

b)

The graph has a root at Y=5; at an income of 15, consumption is 1.
Determine the equation of this consumption function. Then, find the horizontal asymptote.

### The structure of the problem

The properties of the consumption function can be used to determine the parameters a and b.

The first thing we recognize is the type of function: it is a rational function, i.e. the equation is the quotient of two polynomials, in this case of two 1st degree polynomials. Don't get confused by the symbols: while in the script, variables were called x and y, here we have Y and C. This is so to make the variable names match their economic meaning (Y for income, like in economics textbooks, and C for consumption).

Thus we can make use of our theory of rational functions (see block II, learning sequence 2):

We now put this theory to work to determine the parameters a and b.

### The horizontal asymptote

Since the two polynomials are of the same degree, the graph muss have a horizontal asymptote. In both of the problems a) and b) the value of the horizontal asymptote (i.e. the saturation level) is an important aspect:
in a) it is one of the determining properties of the function, in b) we have to find its value.

The value of the horizontal asymptote depends only on the leading coefficients, highlighted in red:

$\frac{{\color {red} {a} \cdot Y+b}}{{\color {red} {2} \cdot Y+5}}\, \longrightarrow \, \frac{a}{2} \quad$ for ${Y \to \infty }$

So the horizontal asymptote reads: $\quad C=\frac{a}{2}$

(Note that the parameter b doesn't influence the saturation level.)

### Solution of a)

We start with the results above. Since for the function in a) the saturation level (horizontal asymptote) is C = 4, we conclude:

$\frac{a}{2}=4$, which means $a=8$

The second thing we know is that the graph goes through the origin, i.e. C has a root at Y = 0. This means nothing else but $C(0)=0$.

Our theory of rational functions (see here) tells us that the value of the numerator polynomial must be 0 for Y = 0. Actually, this is quite obvious, since 0 divided by any number must again be 0, provided the number you divide by is itself unequal 0. Thus, we have

$a \cdot 0+b=0 \quad \Rightarrow \quad b=0$

Thus, the functional equation reads: $C(Y)=\large{\frac{{8 Y}}{{2 Y+5}}}$

### Solution of b)

We know two pairs of values:

Root at Y=5: $\quad C(5)=\frac{{a \cdot 5+b}}{{2 \cdot 5+5}}=0$ $\quad \quad 5a+b=0$

For Y=15, C=1: $\quad C(15)=\frac{{a \cdot 15+b}}{{2 \cdot 15+5}}=1$ $\quad \quad 15a+b=35$

These two conditions for a and b must be simultaneously satisfied.
To put it differently, we have to solve the equation system

$\left\{ \; \begin{eqnarray}15a +b &=& 35 \\ 5a + b &=& 0 \end{eqnarray} \right.$.

One way of doing so is to solve both equations for b, and then to equate them. In our case, however, it seems easier to subtract one equation from the other (see also Systems of Equations in the Preparatory Course).

„First equation minus second equation“ yields

$10a=35 \quad$, i.e. $\quad a=3.5$

Inserting this value in the second equation allows us to find b:

$5 \cdot 3.5+b=0 \quad$$\quad b=-17.5$.

We now have the assignment: $\quad C(Y)=\large {\frac{{3.5Y-17.5}}{{2 Y+5}}} \;$ for $\;Y \ge 5$

The horizontal asymptote must be $\; C=1.75$

(for the same reason as in part a): here, a = 3.5, so we have ) $C=\frac{a}{2}=\frac{3.75}{2}=1.75 \;$).

The following simulation allows you to study the graphical effects of changing values of the parameters in

$C(Y)=\large {\frac{{a \cdot Y+b}}{{2 Y+c}}}$

In addition to a and b (as in the problem above) you can also change the value of c (which was fixed to 5 before).

### Solution of a)

For any a, we have:

$\frac{{a \cdot Y+b}}{{ 2 \cdot Y+5}}\, \longrightarrow \, \frac{a}{2} \quad$ für ${Y \to \infty }$

i.e.  $C= \frac{a}{2}$ is a horizontal asymptote.

Since the saturation level (horizontal asymptote) is C = 4, we conclude:

$\frac{a}{2}=4$, which means $a=8$

Root at Y=0:

$a \cdot 0+b=0 \quad \Rightarrow \quad b=0$

Thus, the functional equation reads: $C(Y)=\large{\frac{{8 Y}}{{2 Y+5}}}$

### Solution of b)

We know two pairs of values:

Root at Y=5: $\quad C(5)=\frac{{a \cdot 5+b}}{{2 \cdot 5+5}}=0$ $\quad \quad 5a+b=0$

For Y=15, C=1: $\quad C(15)=\frac{{a \cdot 15+b}}{{2 \cdot 15+5}}=1$ $\quad \quad 15a+b=35$

These two conditions for a and b must be simultaneously satisfied.
To put it differently, we have to solve the equation system

$\left\{ \; \begin{eqnarray}5a +b &=& 0 \\ 15a + b &=& 35 \end{eqnarray} \right.$.

It has the solutions $a=3.5$ and $b=-17.5$.

Thus, the functional equation reads: $\quad C(Y)=\large {\frac{{3.5Y-17.5}}{{2 Y+5}}}$

The horizontal asymptote must be $\; C=1.75$

(for the same reason as in part a): here, a = 3.5, so we have ) $C=\frac{a}{2}=\frac{3.75}{2}=1.75 \;$).