The maximum value of F = 4x + 3y subject to constraints `xge0,yge2,2x+3yle18,x+yge10` is

Options

A. 35
B. 36
C. 34
D. No optimum value

Answer

To solve the linear programming problem, we need to maximize the objective function \( F = 4x + 3y \) subject to the given constraints. Let's break it down step by step.

### Step 1: Identify the Constraints
The constraints given are:
1. \( x \geq 0 \)
2. \( y \geq 2 \)
3. \( 2x + 3y \leq 18 \)
4. \( x + y \geq 10 \)

### Step 2: Graph the Constraints
We will graph each of the constraints to find the feasible region.

1. **For \( 2x + 3y = 18 \)**:
- Find x-intercept: Set \( y = 0 \) → \( 2x = 18 \) → \( x = 9 \) (Point: (9, 0))
- Find y-intercept: Set \( x = 0 \) → \( 3y = 18 \) → \( y = 6 \) (Point: (0, 6))
- The line will be shaded below since it is \( \leq \).

2. **For \( x + y = 10 \)**:
- Find x-intercept: Set \( y = 0 \) → \( x = 10 \) (Point: (10, 0))
- Find y-intercept: Set \( x = 0 \) → \( y = 10 \) (Point: (0, 10))
- The line will be shaded above since it is \( \geq \).

3. **For \( y = 2 \)**:
- This is a horizontal line at \( y = 2 \) and will be shaded above since \( y \geq 2 \).

4. **For \( x = 0 \)**:
- This is a vertical line along the y-axis and will be shaded to the right since \( x \geq 0 \).

### Step 3: Determine the Feasible Region
After plotting all the lines on the graph, we need to find the area that satisfies all the constraints. The feasible region is where all the shaded areas overlap.

### Step 4: Identify the Corner Points
The corner points of the feasible region can be found by solving the equations of the lines where they intersect:
1. Intersection of \( 2x + 3y = 18 \) and \( x + y = 10 \):
- Solve the equations:
\[ 2x + 3(10 - x) = 18 \implies 2x + 30 - 3x = 18 \implies -x + 30 = 18 \implies x = 12 \]
(Not feasible since \( x \) must be \( \leq 9 \))

2. Intersection of \( 2x + 3y = 18 \) and \( y = 2 \):
- Substitute \( y = 2 \):
\[ 2x + 3(2) = 18 \implies 2x + 6 = 18 \implies 2x = 12 \implies x = 6 \implies (6, 2) \]

3. Intersection of \( x + y = 10 \) and \( y = 2 \):
- Substitute \( y = 2 \):
\[ x + 2 = 10 \implies x = 8 \implies (8, 2) \]

4. Intersection of \( x + y = 10 \) and \( 2x + 3y = 18 \) gives no feasible point.

### Step 5: Evaluate the Objective Function at the Corner Points
Now we evaluate \( F = 4x + 3y \) at the feasible corner points:
1. At \( (6, 2) \):
\[ F = 4(6) + 3(2) = 24 + 6 = 30 \]
2. At \( (8, 2) \):
\[ F = 4(8) + 3(2) = 32 + 6 = 38 \]

### Step 6: Determine the Maximum Value
The maximum value of \( F \) occurs at the point \( (8, 2) \):
\[ \text{Maximum value of } F = 38 \]

### Conclusion
The maximum value of \( F = 4x + 3y \) subject to the given constraints is **38**.