To solve the problem of maximizing the objective function [tex]\( z = 5x + y \)[/tex] subject to the given constraints:
1. [tex]\( 2x + 7y \leq 49 \)[/tex]
2. [tex]\( 5x + 2y \leq 45 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
we take the following steps:
### Step 1: Identify the constraints and plot them
We need to identify the feasible region defined by these constraints.
1. Constraint 1: [tex]\(2x + 7y \leq 49 \)[/tex]
- Rearrange to [tex]\( y \leq \frac{49 - 2x}{7} \)[/tex]
- This is a line with intercepts [tex]\( x = \frac{49}{2} = 24.5 \)[/tex] on the x-axis and [tex]\( y = 7 \)[/tex] on the y-axis.
2. Constraint 2: [tex]\(5x + 2y \leq 45 \)[/tex]
- Rearrange to [tex]\( y \leq \frac{45 - 5x}{2} \)[/tex]
- This is a line with intercepts [tex]\( x = 9 \)[/tex] on the x-axis and [tex]\( y = 22.5 \)[/tex] on the y-axis.
3. Non-negativity constraints [tex]\(x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex]
- These constraints simply mean that our feasible region is in the first quadrant.
### Step 2: Determine the feasible region
The feasible region is where the lines intersect within the first quadrant. Let's find these points of intersection by solving the equations simultaneously:
- Intersection of [tex]\( 2x + 7y = 49 \)[/tex] and [tex]\( 5x + 2y = 45 \)[/tex]
Using elimination or substitution, we solve these linear equations. We get the possible vertices and potential points within the feasible region:
### Step 3: Evaluate the objective function at each vertex
The feasible region is a polygon bounded by intersections of lines and points along the axes:
Vertices:
1. Intersection of [tex]\( x = 0 \)[/tex] and [tex]\( y \leq \frac{49}{7} \)[/tex] : [tex]\( (0, 7) \)[/tex]
2. Intersection of [tex]\( y = 0 \)[/tex] and [tex]\( x \leq \frac{45}{5} \)[/tex]: [tex]\( (9, 0) \)[/tex]
3. Intersection of the two constraints as solved:
### Step 4: Calculating the value of [tex]\( z \)[/tex] at each vertex
- [tex]\( z(0, 7) = 5(0) + 7 = 7 \)[/tex]
- [tex]\( z(9, 0) = 5(9) + 0 = 45 \)[/tex]
Among these evaluations, we see that [tex]\( z \)[/tex] is maximized at [tex]\( x = 9 \)[/tex] and [tex]\( y = 0 \)[/tex] with the objective value of [tex]\( z = 45 \)[/tex].
### Conclusion
The maximum value of [tex]\( z \)[/tex], subject to the given constraints, is [tex]\( \boxed{45} \)[/tex]. The optimal solution occurs at [tex]\( (x, y) = (9, 0) \)[/tex].
