## Task 3 - 120613

In a room with a rectangular floor area of $11 \mathrm{~m}$ width and $36 \mathrm{~m}$ length, 6 machines with the following base areas are standing:

| Machine A: $15^{2}$ | Machine D: $60 \mathrm{~m}^{2}$ |
| :--- | :--- |
| Machine B: $5 \mathrm{~m}^{2}$ | Machine E: $18 \mathrm{~m}^{2}$ |
| Machine C: $18 \mathrm{~m}^{2}$ | Machine F: $50 \mathrm{~m}^{2}$ |

For the storage and preparation of the workpieces to be processed, additional areas with the following areas are required at the machines:
At Machine A: $14 \mathrm{~m}^{2} \quad$ At Machine D: $21 \mathrm{~m}^{2}$
At Machine B: $6 \mathrm{~m}^{2} \quad$ At Machine E: $13 \mathrm{~m}^{2}$
At Machine C: $15 \mathrm{~m}^{2} \quad$ At Machine F: $17 \mathrm{~m}^{2}$

The remaining floor area is to be used for transport routes.

a) Calculate (in $\mathrm{m}^{2}$) the area of the floor that is available for transport routes!

b) We assume that the arrangement of the machines and the storage areas allows the transport routes to be composed of rectangular areas of the same width. We will then refer to the sum of the lengths of these rectangular areas as the "total length of the transport routes."

How wide are these transport routes if they have a total length of $48 \mathrm{~m}$?
a) The area of the base for the machines is $166 \mathrm{~m}^{2}$, that for the storage of workpieces is $86 \mathrm{~m}^{2}$, and the total floor area is $11 \cdot 36=396$ in total $396 \mathrm{~m}^{2}$.

Thus, $396 - 166 - 86 = 144$ remains for the transport paths, which is $144 \mathrm{~m}^{2}$.

b) Let the width of the transport paths be $x \mathrm{~m}$. Then $48 \cdot x = 144$, from which we get $x = 144 : 48$, so $x = 3$. Therefore, the desired width is $3 \mathrm{~m}$.