Maximize `Z = 5x + 3y` subject to `2x + 5y le 10, x , y ge 0`.

Answer

To solve the linear programming problem of maximizing \( Z = 5x + 3y \) subject to the constraints \( 2x + 5y \leq 10 \) and \( x, y \geq 0 \), we can follow these steps:

### Step 1: Identify the Constraints
We have two constraints:
1. \( 2x + 5y \leq 10 \)
2. \( x \geq 0 \)
3. \( y \geq 0 \)

### Step 2: Graph the Constraints
To graph the inequality \( 2x + 5y \leq 10 \), we first find the boundary line by converting the inequality into an equation:
\[ 2x + 5y = 10 \]

#### Finding Intercepts:
- **When \( x = 0 \)**:
\[ 2(0) + 5y = 10 \implies 5y = 10 \implies y = 2 \]
Thus, the y-intercept is \( (0, 2) \).

- **When \( y = 0 \)**:
\[ 2x + 5(0) = 10 \implies 2x = 10 \implies x = 5 \]
Thus, the x-intercept is \( (5, 0) \).

### Step 3: Draw the Line
Plot the points \( (0, 2) \) and \( (5, 0) \) on the graph and draw a line through these points. Since the inequality is \( \leq \), shade the area below this line.

### Step 4: Identify the Feasible Region
The feasible region is bounded by the axes (since \( x \geq 0 \) and \( y \geq 0 \)) and the line \( 2x + 5y = 10 \). The feasible region is a triangle formed by the points:
- \( (0, 0) \) (origin)
- \( (0, 2) \)
- \( (5, 0) \)

### Step 5: Evaluate the Objective Function at the Corner Points
The corner points of the feasible region are \( (0, 0) \), \( (0, 2) \), and \( (5, 0) \). We will evaluate \( Z = 5x + 3y \) at these points.

1. **At \( (0, 0) \)**:
\[ Z = 5(0) + 3(0) = 0 \]

2. **At \( (0, 2) \)**:
\[ Z = 5(0) + 3(2) = 6 \]

3. **At \( (5, 0) \)**:
\[ Z = 5(5) + 3(0) = 25 \]

### Step 6: Determine the Maximum Value
Now, we compare the values of \( Z \):
- At \( (0, 0) \), \( Z = 0 \)
- At \( (0, 2) \), \( Z = 6 \)
- At \( (5, 0) \), \( Z = 25 \)

The maximum value of \( Z \) is \( 25 \) at the point \( (5, 0) \).

### Conclusion
The maximum value of \( Z \) is \( 25 \) at the point \( (5, 0) \).

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