Answer :
To maximize the objective function [tex]\( z = 5x + 3y \)[/tex] given the constraints:
1. [tex]\( x + 4y \leq 28 \)[/tex]
2. [tex]\( 6x + y \leq 30 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
let’s follow the steps to solve this linear programming problem:
### Step 1: Identify the Constraints
The constraints define the feasible region. They are:
- [tex]\( x + 4y \leq 28 \)[/tex]
- [tex]\( 6x + y \leq 30 \)[/tex]
- [tex]\( x \geq 0 \)[/tex]
- [tex]\( y \geq 0 \)[/tex]
### Step 2: Graph the Constraints
To graph the constraints in a two-dimensional plane:
1. [tex]\( x + 4y = 28 \)[/tex]
- When [tex]\( x = 0 \)[/tex]: [tex]\( y = 7 \)[/tex]
- When [tex]\( y = 0 \)[/tex]: [tex]\( x = 28 \)[/tex]
2. [tex]\( 6x + y = 30 \)[/tex]
- When [tex]\( x = 0 \)[/tex]: [tex]\( y = 30 \)[/tex]
- When [tex]\( y = 0 \)[/tex]: [tex]\( x = 5 \)[/tex]
These lines, along with the non-negativity constraints [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex], define a polygonal feasible region on the xy-plane.
### Step 3: Determine the Feasible Region
The feasible region is bounded by the lines [tex]\( x + 4y = 28 \)[/tex], [tex]\( 6x + y = 30 \)[/tex], and the coordinate axes [tex]\( x = 0 \)[/tex] and [tex]\( y = 0 \)[/tex]. This region is a convex polygon.
### Step 4: Identify the Corner Points
The vertices (corner points) of the feasible region are found by solving the system of linear equations given by the intersections of the constraints:
1. Intersection of [tex]\( x + 4y = 28 \)[/tex] and [tex]\( 6x + y = 30 \)[/tex]
2. Intersection of [tex]\( x + 4y = 28 \)[/tex] and [tex]\( y = 0 \)[/tex]
3. Intersection of [tex]\( 6x + y = 30 \)[/tex] and [tex]\( x = 0 \)[/tex]
4. The origin (0,0)
By solving these intersections, the vertices (corner points) are determined. These are:
1. [tex]\( (4, 6) \)[/tex]
2. [tex]\( (0, 7) \)[/tex]
3. [tex]\( (5, 0) \)[/tex]
### Step 5: Evaluate the Objective Function at Each Corner Point
Evaluate [tex]\( z = 5x + 3y \)[/tex] at each corner point:
1. At [tex]\( (4, 6) \)[/tex]:
- [tex]\( z = 5(4) + 3(6) = 20 + 18 = 38 \)[/tex]
2. At [tex]\( (0, 7) \)[/tex]:
- [tex]\( z = 5(0) + 3(7) = 0 + 21 = 21 \)[/tex]
3. At [tex]\( (5, 0) \)[/tex]:
- [tex]\( z = 5(5) + 3(0) = 25 + 0 = 25 \)[/tex]
### Step 6: Identify the Maximum Value
The maximum value of [tex]\( z \)[/tex] is obtained at the point where [tex]\( z = 38 \)[/tex].
### Conclusion
The maximum value of [tex]\( z \)[/tex] is [tex]\( 38 \)[/tex] at the point:
[tex]\[ \begin{array}{l} x = 4 \\ y = 6 \end{array} \][/tex]
So the final answer is:
Maximum is [tex]\( 38 \)[/tex] at
[tex]\[ \begin{array}{l} x = 4 \\ y = 6 \end{array} \][/tex]
1. [tex]\( x + 4y \leq 28 \)[/tex]
2. [tex]\( 6x + y \leq 30 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
let’s follow the steps to solve this linear programming problem:
### Step 1: Identify the Constraints
The constraints define the feasible region. They are:
- [tex]\( x + 4y \leq 28 \)[/tex]
- [tex]\( 6x + y \leq 30 \)[/tex]
- [tex]\( x \geq 0 \)[/tex]
- [tex]\( y \geq 0 \)[/tex]
### Step 2: Graph the Constraints
To graph the constraints in a two-dimensional plane:
1. [tex]\( x + 4y = 28 \)[/tex]
- When [tex]\( x = 0 \)[/tex]: [tex]\( y = 7 \)[/tex]
- When [tex]\( y = 0 \)[/tex]: [tex]\( x = 28 \)[/tex]
2. [tex]\( 6x + y = 30 \)[/tex]
- When [tex]\( x = 0 \)[/tex]: [tex]\( y = 30 \)[/tex]
- When [tex]\( y = 0 \)[/tex]: [tex]\( x = 5 \)[/tex]
These lines, along with the non-negativity constraints [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex], define a polygonal feasible region on the xy-plane.
### Step 3: Determine the Feasible Region
The feasible region is bounded by the lines [tex]\( x + 4y = 28 \)[/tex], [tex]\( 6x + y = 30 \)[/tex], and the coordinate axes [tex]\( x = 0 \)[/tex] and [tex]\( y = 0 \)[/tex]. This region is a convex polygon.
### Step 4: Identify the Corner Points
The vertices (corner points) of the feasible region are found by solving the system of linear equations given by the intersections of the constraints:
1. Intersection of [tex]\( x + 4y = 28 \)[/tex] and [tex]\( 6x + y = 30 \)[/tex]
2. Intersection of [tex]\( x + 4y = 28 \)[/tex] and [tex]\( y = 0 \)[/tex]
3. Intersection of [tex]\( 6x + y = 30 \)[/tex] and [tex]\( x = 0 \)[/tex]
4. The origin (0,0)
By solving these intersections, the vertices (corner points) are determined. These are:
1. [tex]\( (4, 6) \)[/tex]
2. [tex]\( (0, 7) \)[/tex]
3. [tex]\( (5, 0) \)[/tex]
### Step 5: Evaluate the Objective Function at Each Corner Point
Evaluate [tex]\( z = 5x + 3y \)[/tex] at each corner point:
1. At [tex]\( (4, 6) \)[/tex]:
- [tex]\( z = 5(4) + 3(6) = 20 + 18 = 38 \)[/tex]
2. At [tex]\( (0, 7) \)[/tex]:
- [tex]\( z = 5(0) + 3(7) = 0 + 21 = 21 \)[/tex]
3. At [tex]\( (5, 0) \)[/tex]:
- [tex]\( z = 5(5) + 3(0) = 25 + 0 = 25 \)[/tex]
### Step 6: Identify the Maximum Value
The maximum value of [tex]\( z \)[/tex] is obtained at the point where [tex]\( z = 38 \)[/tex].
### Conclusion
The maximum value of [tex]\( z \)[/tex] is [tex]\( 38 \)[/tex] at the point:
[tex]\[ \begin{array}{l} x = 4 \\ y = 6 \end{array} \][/tex]
So the final answer is:
Maximum is [tex]\( 38 \)[/tex] at
[tex]\[ \begin{array}{l} x = 4 \\ y = 6 \end{array} \][/tex]
