Answer :
Certainly! Let's solve the given linear programming problem step by step.
### Step 1: Problem Formulation
We are given the problem:
[tex]\[ \begin{array}{ll} \text{Maximize:} & z = 5x + 4y \\ \text{subject to:} & 2x + 4y \leq 8 \\ & 5x + y \leq 8 \\ & x \geq 0 \\ & y \geq 0 \end{array} \][/tex]
### Step 2: Convert Inequalities into Equations (Boundary Lines)
First, let's convert the inequalities into equations to find the boundary lines:
- [tex]\( 2x + 4y = 8 \)[/tex]
- [tex]\( 5x + y = 8 \)[/tex]
### Step 3: Find the Intersection Points of the Constraints
We'll find the intersection points by solving the equations:
1. Intersection of [tex]\( 2x + 4y = 8 \)[/tex] and [tex]\( 5x + y = 8 \)[/tex]:
- Multiply the second equation by 4 to align coefficients of [tex]\( y \)[/tex]:
[tex]\[ 4 \times (5x + y) = 4 \times 8 \implies 20x + 4y = 32 \][/tex]
- Subtract the first equation [tex]\( 2x + 4y = 8 \)[/tex] from the modified second equation:
[tex]\[ (20x + 4y) - (2x + 4y) = 32 - 8 \implies 18x = 24 \implies x = \frac{24}{18} = \frac{4}{3} \approx 1.333 \][/tex]
- Substitute [tex]\( x = \frac{4}{3} \)[/tex] back into [tex]\( 5x + y = 8 \)[/tex]:
[tex]\[ 5 \left(\frac{4}{3}\right) + y = 8 \implies \frac{20}{3} + y = 8 \implies y = 8 - \frac{20}{3} = \frac{24}{3} - \frac{20}{3} = \frac{4}{3} \approx 1.333 \][/tex]
So, the intersection point is [tex]\( \left(\frac{4}{3}, \frac{4}{3}\right) \)[/tex] or approximately [tex]\( (1.333, 1.333) \)[/tex].
### Step 4: Evaluate Objective Function at the Vertices of the Feasible Region
The feasible region is defined by the constraints [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex], and the boundary lines [tex]\( 2x + 4y \leq 8 \)[/tex] and [tex]\( 5x + y \leq 8 \)[/tex]. Let's evaluate the objective function [tex]\( z = 5x + 4y \)[/tex] at the vertices of the feasible region:
1. Vertex [tex]\( (0,0) \)[/tex]:
[tex]\[ z = 5(0) + 4(0) = 0 \][/tex]
2. Vertex [tex]\( (0,2) \)[/tex] (intersection of [tex]\( 2x + 4y = 8 \)[/tex] with [tex]\( x = 0 \)[/tex]):
[tex]\[ 2(0) + 4y = 8 \implies y = 2 \][/tex]
[tex]\[ z = 5(0) + 4(2) = 8 \][/tex]
3. Vertex [tex]\( (1.6, 0) \)[/tex] (intersection of [tex]\( 5x + y = 8 \)[/tex] with [tex]\( y = 0 \)[/tex]):
[tex]\[ 5x + 0 = 8 \implies x = 1.6 \][/tex]
[tex]\[ z = 5(1.6) + 4(0) = 8 \][/tex]
4. Vertex [tex]\( (1.333, 1.333) \)[/tex] (intersection found in Step 3):
[tex]\[ z = 5(1.333) + 4(1.333) = 6.665 + 5.332 \approx 12 \][/tex]
### Step 5: Determine the Maximum Value
Comparing these values:
- At [tex]\( (0,0) \)[/tex], [tex]\( z = 0 \)[/tex]
- At [tex]\( (0,2) \)[/tex], [tex]\( z = 8 \)[/tex]
- At [tex]\( (1.6, 0) \)[/tex], [tex]\( z = 8 \)[/tex]
- At [tex]\( (1.333, 1.333) \)[/tex], [tex]\( z = 12 \)[/tex]
The maximum value of [tex]\( z = 12 \)[/tex] occurs at [tex]\( (1.333, 1.333) \)[/tex].
Therefore, the maximum value is [tex]\(\boxed{12}\)[/tex].
### Step 1: Problem Formulation
We are given the problem:
[tex]\[ \begin{array}{ll} \text{Maximize:} & z = 5x + 4y \\ \text{subject to:} & 2x + 4y \leq 8 \\ & 5x + y \leq 8 \\ & x \geq 0 \\ & y \geq 0 \end{array} \][/tex]
### Step 2: Convert Inequalities into Equations (Boundary Lines)
First, let's convert the inequalities into equations to find the boundary lines:
- [tex]\( 2x + 4y = 8 \)[/tex]
- [tex]\( 5x + y = 8 \)[/tex]
### Step 3: Find the Intersection Points of the Constraints
We'll find the intersection points by solving the equations:
1. Intersection of [tex]\( 2x + 4y = 8 \)[/tex] and [tex]\( 5x + y = 8 \)[/tex]:
- Multiply the second equation by 4 to align coefficients of [tex]\( y \)[/tex]:
[tex]\[ 4 \times (5x + y) = 4 \times 8 \implies 20x + 4y = 32 \][/tex]
- Subtract the first equation [tex]\( 2x + 4y = 8 \)[/tex] from the modified second equation:
[tex]\[ (20x + 4y) - (2x + 4y) = 32 - 8 \implies 18x = 24 \implies x = \frac{24}{18} = \frac{4}{3} \approx 1.333 \][/tex]
- Substitute [tex]\( x = \frac{4}{3} \)[/tex] back into [tex]\( 5x + y = 8 \)[/tex]:
[tex]\[ 5 \left(\frac{4}{3}\right) + y = 8 \implies \frac{20}{3} + y = 8 \implies y = 8 - \frac{20}{3} = \frac{24}{3} - \frac{20}{3} = \frac{4}{3} \approx 1.333 \][/tex]
So, the intersection point is [tex]\( \left(\frac{4}{3}, \frac{4}{3}\right) \)[/tex] or approximately [tex]\( (1.333, 1.333) \)[/tex].
### Step 4: Evaluate Objective Function at the Vertices of the Feasible Region
The feasible region is defined by the constraints [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex], and the boundary lines [tex]\( 2x + 4y \leq 8 \)[/tex] and [tex]\( 5x + y \leq 8 \)[/tex]. Let's evaluate the objective function [tex]\( z = 5x + 4y \)[/tex] at the vertices of the feasible region:
1. Vertex [tex]\( (0,0) \)[/tex]:
[tex]\[ z = 5(0) + 4(0) = 0 \][/tex]
2. Vertex [tex]\( (0,2) \)[/tex] (intersection of [tex]\( 2x + 4y = 8 \)[/tex] with [tex]\( x = 0 \)[/tex]):
[tex]\[ 2(0) + 4y = 8 \implies y = 2 \][/tex]
[tex]\[ z = 5(0) + 4(2) = 8 \][/tex]
3. Vertex [tex]\( (1.6, 0) \)[/tex] (intersection of [tex]\( 5x + y = 8 \)[/tex] with [tex]\( y = 0 \)[/tex]):
[tex]\[ 5x + 0 = 8 \implies x = 1.6 \][/tex]
[tex]\[ z = 5(1.6) + 4(0) = 8 \][/tex]
4. Vertex [tex]\( (1.333, 1.333) \)[/tex] (intersection found in Step 3):
[tex]\[ z = 5(1.333) + 4(1.333) = 6.665 + 5.332 \approx 12 \][/tex]
### Step 5: Determine the Maximum Value
Comparing these values:
- At [tex]\( (0,0) \)[/tex], [tex]\( z = 0 \)[/tex]
- At [tex]\( (0,2) \)[/tex], [tex]\( z = 8 \)[/tex]
- At [tex]\( (1.6, 0) \)[/tex], [tex]\( z = 8 \)[/tex]
- At [tex]\( (1.333, 1.333) \)[/tex], [tex]\( z = 12 \)[/tex]
The maximum value of [tex]\( z = 12 \)[/tex] occurs at [tex]\( (1.333, 1.333) \)[/tex].
Therefore, the maximum value is [tex]\(\boxed{12}\)[/tex].
