To determine the number of cars of each type [tex]\( c_1, c_2, \)[/tex] and [tex]\( c_3 \)[/tex] that can be produced using the given amounts of steel [tex]\( s_1, s_2, \)[/tex] and [tex]\( s_3 \)[/tex], we need to set up a system of linear equations based on the steel requirements and the available amounts of each type of steel.
Given:
- The steel requirements for each type of car:
- [tex]\( c_1 \)[/tex]: 2 tons of [tex]\( s_1 \)[/tex], 1 ton of [tex]\( s_2 \)[/tex], 3 tons of [tex]\( s_3 \)[/tex]
- [tex]\( c_2 \)[/tex]: 3 tons of [tex]\( s_1 \)[/tex], 1 ton of [tex]\( s_2 \)[/tex], 2 tons of [tex]\( s_3 \)[/tex]
- [tex]\( c_3 \)[/tex]: 4 tons of [tex]\( s_1 \)[/tex], 2 tons of [tex]\( s_2 \)[/tex], 1 ton of [tex]\( s_3 \)[/tex]
- The available amounts of each type of steel:
- [tex]\( s_1 \)[/tex]: 29 tons
- [tex]\( s_2 \)[/tex]: 13 tons
- [tex]\( s_3 \)[/tex]: 16 tons
We can translate these requirements into the following system of equations:
1. For [tex]\( s_1 \)[/tex] (total of 29 tons):
[tex]\[ 2c_1 + 3c_2 + 4c_3 = 29 \][/tex]
2. For [tex]\( s_2 \)[/tex] (total of 13 tons):
[tex]\[ c_1 + c_2 + 2c_3 = 13 \][/tex]
3. For [tex]\( s_3 \)[/tex] (total of 16 tons):
[tex]\[ 3c_1 + 2c_2 + c_3 = 16 \][/tex]
We now solve this system of equations:
1. Start with the given system:
[tex]\[
\begin{cases}
2c_1 + 3c_2 + 4c_3 = 29 \\
c_1 + c_2 + 2c_3 = 13 \\
3c_1 + 2c_2 + c_3 = 16
\end{cases}
\][/tex]
2. We solve this system using, for example, the method of substitution or elimination:
After solving, we find that:
[tex]\[
c_1 = 2
\][/tex]
[tex]\[
c_2 = 3
\][/tex]
[tex]\[
c_3 = 4
\][/tex]
Thus, the number of cars of each type that can be produced using 29 tons of [tex]\( s_1 \)[/tex], 13 tons of [tex]\( s_2 \)[/tex], and 16 tons of [tex]\( s_3 \)[/tex] are:
- [tex]\( c_1 = 2 \)[/tex] cars
- [tex]\( c_2 = 3 \)[/tex] cars
- [tex]\( c_3 = 4 \)[/tex] cars
