Isoquant and minimum cost combination

A company wants to make 400 sun umbrellas. For this it needs labour ($x$ labour hours) and machines ($y$ machine hours). One labour hour costs 60 €, one machine hour 20 €. There are many recipes $(x|y)$ for the 400 umbrellas. Which one is the cheapest?

The cheapest recipe is called the minimum cost combination (MCC for short). Here it is $(3|7)$: 3 labour hours and 7 machine hours for 320 €. With less money you cannot make the 400 umbrellas; more money would be wasted.

The isoquant

First the curve of all recipes. The company knows its recipes for 400 umbrellas as a function:

You put in the labour hours $x$. $I(x)$ itself stands for $y$: it tells you how many machine hours $y$ you then need. In the table: on the left $x$, in the middle the denominator $x-2$, on the right $y=I(x)$.

This curve is called an isoquant ("equal quantity"): every point gives the same 400 umbrellas. You also write $I_{400}(x)$.

Exercise 1: Determine the domain and the range of the isoquant function.

The function is a rational function. It has two branches:

Mathematically, only $x=2$ must not be put in. Economically, only the right branch counts. This is the economic domain:

In short: $D_{\ddot{o}k}=(2;\infty )$, $W_{\ddot{o}k}=(4;\infty )$. The round bracket means: the boundary does not belong to it.

The isocost line

Now the lines. A recipe $(x|y)$ costs:

Here: $K=60x+20y$. This cost function should go into the same picture as the isoquant. For this, $y$ must stand alone on the left. Each time you do the opposite of the operation:

Rearranged:

With 60 and 20:

An isocost line ("equal cost") contains all recipes that cost exactly $K$ euros.

With its previous recipe the company pays 450 €. An example for 450 € is $(5|7.5)$: $60\cdot 5+20\cdot 7.5=450€$. It does not lie on the isoquant: it needs 7.5 machine hours; for 400 umbrellas 5 are enough, because $I(5)=5$.

Exercise 2: So far, costs of 450 € have arisen. Give the isocost line for the costs so far.

This is the 450 € line from the family: start at 22.5, then 1 to the right and 3 down.

The slope $-\frac{p_x}{p_y}=-3$ depends only on the prices. The intercept $\frac{K}{p_y}$ depends on the budget: 450 € gives 22.5, 350 € gives 17.5, 320 € gives 16.

Cutting or touching

Why is the point where the line touches the curve the cheapest recipe? Above, the budget grew from 280 € to 320 €. What happens if it grows even further?

Check:

- $(2.5|10)$: $150+200=350€$ - $(4|5.5)$: $240+110=350€$ - $(3|7)$: $180+140=320€$

How do you recognise the point of contact when you calculate?

Calculating the minimum cost combination

Exercise 3: Determine the minimum cost combination and the minimum production costs for it. Give the isocost line for the minimum costs.

For $I^{\prime }(x)$ you write the fraction as a power:

Now set them equally steep: $I^{\prime }(x)=-3$.

Result:

- Minimum cost combination: MCC $(3|7)$, so 3 labour hours and 7 machine hours. - Minimum costs: $K_{min}=60\cdot 3+20\cdot 7=320€$. - Isocost line for the minimum costs: $y=-3x+16$. Check: $-3\cdot 3+16=7$. ✓

The method

60 € and 20 € become $p_x$ and $p_y$, and $\frac{3}{x-2}+4$ becomes any isoquant $I(x)$.

The minimum cost combination lies where the isocost line touches the isoquant. There:

The isocost line for it:

Other prices

Same isoquant, but labour is cheap: $p_x=20$ €, $p_y=60$ €. We write prices as a pair labour : machine, here 20 : 60.

The calculation:

$x-2=3$ gives $x=5$; $x-2=-3$ gives $x=-1$, not in $D_{\ddot{o}k}$.

Two more price pairs for the same isoquant:

- $p_x=90$ €, $p_y=30$ €: the slope is $-3$ again. So the same point of contact $(3|7)$, only more expensive: $270+210=480$ €. - $p_x=80$ €, $p_y=60$ €: the slope is $-\frac{4}{3}$, so $(x-2{)}^2=\frac{9}{4}$ and $x=3.5$, $y=6$. Costs: $280+360=640$ €.

Same price ratio, same MCC. Different ratio, different MCC.

Another isoquant

$I(x)=\frac{12}{x-1}+2$ with $p_x=30$ €, $p_y=10$ €. Here $D_{\ddot{o}k}=(1;\infty )$, so $x>1$.

$x-1=2$ gives $x=3$; $x-1=-2$ gives $x=-1$, not in $D_{\ddot{o}k}$.

For isoquants of this form the equation gives two solutions. One lies outside $D_{\ddot{o}k}$ (on the left branch), the other is the cheapest recipe: the point where the isocost line touches the isoquant.