The point at which the maximum value of `3x+2y` subject to the constraints `x+y le 2, x le 2, y ge 0`, obtained is:

Options

A. (0,0)
B. (1.5,1.5)
C. (2,0)
D. (0,2)

Answer

To solve the problem of finding the point at which the maximum value of \(3x + 2y\) occurs given the constraints \(x + y \leq 2\), \(x \leq 2\), and \(y \geq 0\), we will follow these steps:

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

### Step 2: Graph the constraints
We will graph the inequalities on a coordinate plane:
- The line \(x + y = 2\) intersects the axes at points (2, 0) and (0, 2).
- The line \(x = 2\) is a vertical line that intersects the x-axis at (2, 0).
- The line \(y = 0\) is the x-axis itself.

### Step 3: Determine the feasible region
The feasible region is the area that satisfies all the constraints.
- For \(x + y \leq 2\), we shade below the line segment connecting (2, 0) and (0, 2).
- For \(x \leq 2\), we shade to the left of the line \(x = 2\).
- For \(y \geq 0\), we shade above the x-axis.

The feasible region will be a triangle formed by the points (0, 0), (2, 0), and (0, 2).

### Step 4: Identify the corner points
The corner points of the feasible region are:
1. (0, 0)
2. (2, 0)
3. (0, 2)

### Step 5: Evaluate the objective function at each corner point
Now we will evaluate the objective function \(z = 3x + 2y\) at each corner point:
- At (0, 0):
\[ z = 3(0) + 2(0) = 0 \]
- At (2, 0):
\[ z = 3(2) + 2(0) = 6 \]
- At (0, 2):
\[ z = 3(0) + 2(2) = 4 \]

### Step 6: Determine the maximum value
Comparing the values of \(z\):
- At (0, 0): \(z = 0\)
- At (2, 0): \(z = 6\)
- At (0, 2): \(z = 4\)

The maximum value of \(z\) is 6, which occurs at the point (2, 0).

### Conclusion
The point at which the maximum value of \(3x + 2y\) occurs is \((2, 0)\).

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