Answer :
To solve this problem, we need to find the values of [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(a\)[/tex] such that the objective function [tex]\(F = 4x + 3y + a\)[/tex] reaches its maximum value of 26 under the given constraints. Let's go through this step-by-step.
### Step 1: Set Up the Objective Function and Constraints
The objective function to maximize is:
[tex]\[ F = 4x + 3y + a \][/tex]
The given constraints are:
1. [tex]\( x + y \leq 6 \)[/tex]
2. [tex]\( 2x - y \leq 3 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
### Step 2: Solve for Optimal Values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]
We first need to find the optimal values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] without considering [tex]\(a\)[/tex] (since [tex]\(a\)[/tex] is a constant that will be adjusted later). We will maximize the function:
[tex]\[ F' = 4x + 3y \][/tex]
Subject to the same constraints:
1. [tex]\( x + y \leq 6 \)[/tex]
2. [tex]\( 2x - y \leq 3 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
### Step 3: Determine Feasible Region and Vertices
The feasible region is determined by the intersection of the constraints. We need to find the points of intersection (vertices) that typically include:
- Intersection of [tex]\(x + y = 6\)[/tex] and [tex]\(2x - y = 3\)[/tex]
- Intersection of [tex]\(x + y = 6\)[/tex] and [tex]\(x = 0\)[/tex]
- Intersection of [tex]\(2x - y = 3\)[/tex] and [tex]\(y = 0\)[/tex]
Solving these intersections:
1. Intersection of [tex]\(x + y = 6\)[/tex] and [tex]\(2x - y = 3\)[/tex]:
[tex]\[ \begin{cases} x + y = 6 \\ 2x - y = 3 \end{cases} \][/tex]
Adding these two equations, we get:
[tex]\[ 3x = 9 \implies x = 3 \][/tex]
Substituting [tex]\(x = 3\)[/tex] into [tex]\(x + y = 6\)[/tex], we get:
[tex]\[ 3 + y = 6 \implies y = 3 \][/tex]
So, [tex]\((x, y) = (3, 3)\)[/tex].
2. Intersection of [tex]\(x + y = 6\)[/tex] and [tex]\(x = 0\)[/tex]:
[tex]\[ 0 + y = 6 \implies y = 6 \][/tex]
So, [tex]\((x, y) = (0, 6)\)[/tex].
3. Intersection of [tex]\(2x - y = 3\)[/tex] and [tex]\(y = 0\)[/tex]:
[tex]\[ 2x - 0 = 3 \implies x = 1.5 \][/tex]
So, [tex]\((x, y) = (1.5, 0)\)[/tex].
### Step 4: Evaluate [tex]\(F' = 4x + 3y\)[/tex] at the Vertices
1. At [tex]\((x, y) = (3, 3)\)[/tex]:
[tex]\[ F' = 4(3) + 3(3) = 12 + 9 = 21 \][/tex]
2. At [tex]\((x, y) = (0, 6)\)[/tex]:
[tex]\[ F' = 4(0) + 3(6) = 0 + 18 = 18 \][/tex]
3. At [tex]\((x, y) = (1.5, 0)\)[/tex]:
[tex]\[ F' = 4(1.5) + 3(0) = 6 + 0 = 6 \][/tex]
The optimal value of [tex]\(F'\)[/tex] is 21 at the vertex [tex]\((3, 3)\)[/tex].
### Step 5: Determine the Value of [tex]\(a\)[/tex]
Given that the maximum value of [tex]\(F = 4x + 3y + a\)[/tex] is 26, and we found that [tex]\(4x + 3y\)[/tex] equals 21 at the optimal point:
[tex]\[ 4(3) + 3(3) + a = 26 \][/tex]
[tex]\[ 21 + a = 26 \][/tex]
Solving for [tex]\(a\)[/tex]:
[tex]\[ a = 26 - 21 \][/tex]
[tex]\[ a = 5 \][/tex]
### Conclusion
The value of [tex]\(a\)[/tex] that makes the maximum value of the objective function [tex]\(F = 4x + 3y + a\)[/tex] equal to 26 is [tex]\( \boxed{5} \)[/tex].
### Step 1: Set Up the Objective Function and Constraints
The objective function to maximize is:
[tex]\[ F = 4x + 3y + a \][/tex]
The given constraints are:
1. [tex]\( x + y \leq 6 \)[/tex]
2. [tex]\( 2x - y \leq 3 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
### Step 2: Solve for Optimal Values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]
We first need to find the optimal values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] without considering [tex]\(a\)[/tex] (since [tex]\(a\)[/tex] is a constant that will be adjusted later). We will maximize the function:
[tex]\[ F' = 4x + 3y \][/tex]
Subject to the same constraints:
1. [tex]\( x + y \leq 6 \)[/tex]
2. [tex]\( 2x - y \leq 3 \)[/tex]
3. [tex]\( x \geq 0 \)[/tex]
4. [tex]\( y \geq 0 \)[/tex]
### Step 3: Determine Feasible Region and Vertices
The feasible region is determined by the intersection of the constraints. We need to find the points of intersection (vertices) that typically include:
- Intersection of [tex]\(x + y = 6\)[/tex] and [tex]\(2x - y = 3\)[/tex]
- Intersection of [tex]\(x + y = 6\)[/tex] and [tex]\(x = 0\)[/tex]
- Intersection of [tex]\(2x - y = 3\)[/tex] and [tex]\(y = 0\)[/tex]
Solving these intersections:
1. Intersection of [tex]\(x + y = 6\)[/tex] and [tex]\(2x - y = 3\)[/tex]:
[tex]\[ \begin{cases} x + y = 6 \\ 2x - y = 3 \end{cases} \][/tex]
Adding these two equations, we get:
[tex]\[ 3x = 9 \implies x = 3 \][/tex]
Substituting [tex]\(x = 3\)[/tex] into [tex]\(x + y = 6\)[/tex], we get:
[tex]\[ 3 + y = 6 \implies y = 3 \][/tex]
So, [tex]\((x, y) = (3, 3)\)[/tex].
2. Intersection of [tex]\(x + y = 6\)[/tex] and [tex]\(x = 0\)[/tex]:
[tex]\[ 0 + y = 6 \implies y = 6 \][/tex]
So, [tex]\((x, y) = (0, 6)\)[/tex].
3. Intersection of [tex]\(2x - y = 3\)[/tex] and [tex]\(y = 0\)[/tex]:
[tex]\[ 2x - 0 = 3 \implies x = 1.5 \][/tex]
So, [tex]\((x, y) = (1.5, 0)\)[/tex].
### Step 4: Evaluate [tex]\(F' = 4x + 3y\)[/tex] at the Vertices
1. At [tex]\((x, y) = (3, 3)\)[/tex]:
[tex]\[ F' = 4(3) + 3(3) = 12 + 9 = 21 \][/tex]
2. At [tex]\((x, y) = (0, 6)\)[/tex]:
[tex]\[ F' = 4(0) + 3(6) = 0 + 18 = 18 \][/tex]
3. At [tex]\((x, y) = (1.5, 0)\)[/tex]:
[tex]\[ F' = 4(1.5) + 3(0) = 6 + 0 = 6 \][/tex]
The optimal value of [tex]\(F'\)[/tex] is 21 at the vertex [tex]\((3, 3)\)[/tex].
### Step 5: Determine the Value of [tex]\(a\)[/tex]
Given that the maximum value of [tex]\(F = 4x + 3y + a\)[/tex] is 26, and we found that [tex]\(4x + 3y\)[/tex] equals 21 at the optimal point:
[tex]\[ 4(3) + 3(3) + a = 26 \][/tex]
[tex]\[ 21 + a = 26 \][/tex]
Solving for [tex]\(a\)[/tex]:
[tex]\[ a = 26 - 21 \][/tex]
[tex]\[ a = 5 \][/tex]
### Conclusion
The value of [tex]\(a\)[/tex] that makes the maximum value of the objective function [tex]\(F = 4x + 3y + a\)[/tex] equal to 26 is [tex]\( \boxed{5} \)[/tex].
