The maximum value of z = 4x +3y subject to the constraints `3x+2yge160,5x+2yge200,x+2yge80,x,yge0` is

Options

A. 320
B. 300
C. 230
D. None of these

Answer

To solve the problem of maximizing \( z = 4x + 3y \) subject to the given constraints, we will follow these steps:

### Step 1: Identify the Constraints
The constraints given are:
1. \( 3x + 2y \geq 160 \)
2. \( 5x + 2y \geq 200 \)
3. \( x + 2y \geq 80 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To graph the constraints, we convert the inequalities into equations:
1. \( 3x + 2y = 160 \)
2. \( 5x + 2y = 200 \)
3. \( x + 2y = 80 \)

### Step 3: Find Intercepts
For each equation, we find the x-intercept and y-intercept.

1. For \( 3x + 2y = 160 \):
- x-intercept: Set \( y = 0 \) → \( 3x = 160 \) → \( x = \frac{160}{3} \approx 53.33 \)
- y-intercept: Set \( x = 0 \) → \( 2y = 160 \) → \( y = 80 \)

2. For \( 5x + 2y = 200 \):
- x-intercept: Set \( y = 0 \) → \( 5x = 200 \) → \( x = 40 \)
- y-intercept: Set \( x = 0 \) → \( 2y = 200 \) → \( y = 100 \)

3. For \( x + 2y = 80 \):
- x-intercept: Set \( y = 0 \) → \( x = 80 \)
- y-intercept: Set \( x = 0 \) → \( 2y = 80 \) → \( y = 40 \)

### Step 4: Graph the Constraints
Plot the lines on a graph using the intercepts found above. The feasible region will be the area that satisfies all the inequalities.

### Step 5: Determine the Feasible Region
Since all inequalities are of the form \( \geq \), the feasible region will be above the lines. The feasible region is bounded by the lines and the axes.

### Step 6: Find 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 \( 3x + 2y = 160 \) and \( 5x + 2y = 200 \):
- Subtract the first from the second:
\[ (5x + 2y) - (3x + 2y) = 200 - 160 \implies 2x = 40 \implies x = 20 \]
Substitute \( x = 20 \) into \( 3x + 2y = 160 \):
\[ 3(20) + 2y = 160 \implies 60 + 2y = 160 \implies 2y = 100 \implies y = 50 \]
So, one corner point is \( (20, 50) \).

2. Intersection of \( 3x + 2y = 160 \) and \( x + 2y = 80 \):
- Subtract the second from the first:
\[ (3x + 2y) - (x + 2y) = 160 - 80 \implies 2x = 80 \implies x = 40 \]
Substitute \( x = 40 \) into \( x + 2y = 80 \):
\[ 40 + 2y = 80 \implies 2y = 40 \implies y = 20 \]
So, another corner point is \( (40, 20) \).

3. Intersection of \( 5x + 2y = 200 \) and \( x + 2y = 80 \):
- Subtract the second from the first:
\[ (5x + 2y) - (x + 2y) = 200 - 80 \implies 4x = 120 \implies x = 30 \]
Substitute \( x = 30 \) into \( x + 2y = 80 \):
\[ 30 + 2y = 80 \implies 2y = 50 \implies y = 25 \]
So, the third corner point is \( (30, 25) \).

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( z = 4x + 3y \) at each corner point:
1. At \( (20, 50) \):
\[ z = 4(20) + 3(50) = 80 + 150 = 230 \]
2. At \( (40, 20) \):
\[ z = 4(40) + 3(20) = 160 + 60 = 220 \]
3. At \( (30, 25) \):
\[ z = 4(30) + 3(25) = 120 + 75 = 195 \]

### Step 8: Determine the Maximum Value
The maximum value of \( z \) occurs at the point \( (20, 50) \) with \( z = 230 \).

### Conclusion
The maximum value of \( z = 4x + 3y \) subject to the given constraints is **230**.

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