Answer :
To solve the given linear programming problem, we need to maximize the objective function subject to the constraints. Let’s go through this step by step.
### Step 1: Define the Objective Function
The objective function is to maximize [tex]\( z = 2x + 9y \)[/tex].
### Step 2: Define the Constraints
The constraints for the problem are:
1. [tex]\( x \geq 0 \)[/tex]
2. [tex]\( y \geq 0 \)[/tex]
3. [tex]\( x + y \leq 24 \)[/tex]
4. [tex]\( 4x + y \geq 24 \)[/tex]
5. [tex]\( x + 4y \geq 24 \)[/tex]
### Step 3: Convert Inequalities to Suitable Form for Graphical Solution
First, we would reorganize the inequalities if necessary to match typical [tex]\( \leq \)[/tex] or [tex]\( \geq \)[/tex] form for easier graphical representation or standard method solving.
### Step 4: Identify the Feasible Region
The feasible region is the set of all points [tex]\((x, y)\)[/tex] that satisfy all the constraints similar to below:
- The line [tex]\( x + y = 24 \)[/tex] forms an upper boundary.
- The line [tex]\( 4x + y = 24 \)[/tex] creates a lower boundary.
- The line [tex]\( x + 4y = 24 \)[/tex] also creates a lower boundary.
- [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex] limit the feasible region to the first quadrant.
### Step 5: Determine Intersection Points (Vertices)
We need to check either lines intersections points or manually solving as:
- Intersection of [tex]\( x + y = 24 \)[/tex] and [tex]\( 4x + y = 24 \)[/tex]:
[tex]\[ x + y = 24 \\ 4x + y = 24 \implies 3x = 0 \implies x = 0, y = 24 \][/tex]
- Intersection of [tex]\( x + y = 24 \)[/tex] and [tex]\( x + 4y = 24 \)[/tex]:
[tex]\[ x + y = 24 \\ x + 4y = 24 \implies 3y = 0 \implies y = 0, x = 24 \][/tex]
- Intersection of [tex]\( 4x + y = 24 \)[/tex] and [tex]\( x + 4y = 24 \)[/tex] needs less direct visibility, solved typically by substitution or elimination, provide solution but zero:
[tex]\[ y = \frac{24 - 4x}{4}; x=0; y = 6 and intersection not considered feasible. \][/tex]
### Step 6: Evaluate the Objective Function at Each Vertex
Evaluate [tex]\( z = 2x + 9y \)[/tex] for points
1. At [tex]\((0, 24)\)[/tex]
[tex]\[ z = 2(0) + 9(24) = 216 \][/tex]
2. At [tex]\((24, 0)\)[/tex]
[tex]\[ z = 2(24) + 9(0) = 48 \][/tex]
### Step 7: Choose the Maximum Value
Comparing the values, [tex]\( z = 216 \)[/tex] is the largest.
Thus, the values that maximize the objective function [tex]\( 2x + 9y \)[/tex] under the given constraints are:
- [tex]\( x = 0 \)[/tex]
- [tex]\( y = 24 \)[/tex]
- The maximum value [tex]\( z = 216 \)[/tex]
So, the solution to the given linear programming problem is [tex]\( x = 0 \)[/tex], [tex]\( y = 24 \)[/tex], and the maximum value of [tex]\( z = 216 \)[/tex].
### Step 1: Define the Objective Function
The objective function is to maximize [tex]\( z = 2x + 9y \)[/tex].
### Step 2: Define the Constraints
The constraints for the problem are:
1. [tex]\( x \geq 0 \)[/tex]
2. [tex]\( y \geq 0 \)[/tex]
3. [tex]\( x + y \leq 24 \)[/tex]
4. [tex]\( 4x + y \geq 24 \)[/tex]
5. [tex]\( x + 4y \geq 24 \)[/tex]
### Step 3: Convert Inequalities to Suitable Form for Graphical Solution
First, we would reorganize the inequalities if necessary to match typical [tex]\( \leq \)[/tex] or [tex]\( \geq \)[/tex] form for easier graphical representation or standard method solving.
### Step 4: Identify the Feasible Region
The feasible region is the set of all points [tex]\((x, y)\)[/tex] that satisfy all the constraints similar to below:
- The line [tex]\( x + y = 24 \)[/tex] forms an upper boundary.
- The line [tex]\( 4x + y = 24 \)[/tex] creates a lower boundary.
- The line [tex]\( x + 4y = 24 \)[/tex] also creates a lower boundary.
- [tex]\( x \geq 0 \)[/tex] and [tex]\( y \geq 0 \)[/tex] limit the feasible region to the first quadrant.
### Step 5: Determine Intersection Points (Vertices)
We need to check either lines intersections points or manually solving as:
- Intersection of [tex]\( x + y = 24 \)[/tex] and [tex]\( 4x + y = 24 \)[/tex]:
[tex]\[ x + y = 24 \\ 4x + y = 24 \implies 3x = 0 \implies x = 0, y = 24 \][/tex]
- Intersection of [tex]\( x + y = 24 \)[/tex] and [tex]\( x + 4y = 24 \)[/tex]:
[tex]\[ x + y = 24 \\ x + 4y = 24 \implies 3y = 0 \implies y = 0, x = 24 \][/tex]
- Intersection of [tex]\( 4x + y = 24 \)[/tex] and [tex]\( x + 4y = 24 \)[/tex] needs less direct visibility, solved typically by substitution or elimination, provide solution but zero:
[tex]\[ y = \frac{24 - 4x}{4}; x=0; y = 6 and intersection not considered feasible. \][/tex]
### Step 6: Evaluate the Objective Function at Each Vertex
Evaluate [tex]\( z = 2x + 9y \)[/tex] for points
1. At [tex]\((0, 24)\)[/tex]
[tex]\[ z = 2(0) + 9(24) = 216 \][/tex]
2. At [tex]\((24, 0)\)[/tex]
[tex]\[ z = 2(24) + 9(0) = 48 \][/tex]
### Step 7: Choose the Maximum Value
Comparing the values, [tex]\( z = 216 \)[/tex] is the largest.
Thus, the values that maximize the objective function [tex]\( 2x + 9y \)[/tex] under the given constraints are:
- [tex]\( x = 0 \)[/tex]
- [tex]\( y = 24 \)[/tex]
- The maximum value [tex]\( z = 216 \)[/tex]
So, the solution to the given linear programming problem is [tex]\( x = 0 \)[/tex], [tex]\( y = 24 \)[/tex], and the maximum value of [tex]\( z = 216 \)[/tex].
