Let's solve the given linear programming problem step-by-step.
### Problem:
Maximize
[tex]\[
z = x + 5y
\][/tex]
subject to
[tex]\[
\begin{array}{l}
x + 4y \leq 24 \\
5x + 6y \leq 50 \\
x \geq 0 \\
y \geq 0 \\
\end{array}
\][/tex]
### Solution:
1. Set up the inequalities:
[tex]\[
\begin{array}{l}
x + 4y \leq 24 \\
5x + 6y \leq 50 \\
x \geq 0 \\
y \geq 0 \\
\end{array}
\][/tex]
2. Graph the feasible region:
We need to plot the inequalities on a graph to find the feasible region.
- For [tex]\( x + 4y \leq 24 \)[/tex]:
- When [tex]\( x = 0 \)[/tex], [tex]\( 4y = 24 \)[/tex] ⇒ [tex]\( y = 6 \)[/tex]
- When [tex]\( y = 0 \)[/tex], [tex]\( x = 24 \)[/tex]
So, the line passes through [tex]\((0, 6)\)[/tex] and [tex]\((24, 0)\)[/tex].
- For [tex]\( 5x + 6y \leq 50 \)[/tex]:
- When [tex]\( x = 0 \)[/tex], [tex]\( 6y = 50 \)[/tex] ⇒ [tex]\( y \approx 8.33 \)[/tex]
- When [tex]\( y = 0 \)[/tex], [tex]\( 5x = 50 \)[/tex] ⇒ [tex]\( x = 10 \)[/tex]
So, the line passes through [tex]\((0, 8.33)\)[/tex] and [tex]\((10, 0)\)[/tex].
3. Find the vertices of the feasible region:
The vertices of the feasible region are found at the intersections of the lines and the coordinate axes.
- The intersection of [tex]\( x + 4y = 24 \)[/tex] and [tex]\( 5x + 6y = 50 \)[/tex]:
Solve the system of equations:
[tex]\[
\begin{cases}
x + 4y = 24 \\
5x + 6y = 50
\end{cases}
\][/tex]
By solving, we get:
[tex]\[
x = 0 \quad \text{and} \quad y = 6
\][/tex]
- Other points of intersection:
- [tex]\( (0, 0) \)[/tex]
- [tex]\( (0, 8.33) \)[/tex]
- [tex]\( (10, 0) \)[/tex]
- [tex]\( (24, 0) \)[/tex]
4. Evaluate the objective function at each vertex:
The objective function is [tex]\( z = x + 5y \)[/tex].
- At [tex]\( (0, 6) \)[/tex]:
[tex]\[
z = 0 + 5(6) = 0 + 30 = 30
\][/tex]
- At [tex]\( (10, 0) \)[/tex]:
[tex]\[
z = 10 + 5(0) = 10 + 0 = 10
\][/tex]
5. Determine the maximum value:
The maximum value of [tex]\( z \)[/tex] from evaluating the vertices is [tex]\( 30 \)[/tex] at [tex]\( (0, 6) \)[/tex].
Therefore, the solution to the linear programming problem is:
[tex]\[
\text{Maximum is } 30 \text{ at }
\begin{array}{l}
x = 0 \\
y = 6
\end{array}
\][/tex]
