`Z = 7x + 4y`
`2x + y le 10, x + 2y le 12, x ge 0, y ge 0`

Answer

To solve the linear programming problem given by the objective function \( Z = 7x + 4y \) with constraints \( 2x + y \leq 10 \), \( x + 2y \leq 12 \), and \( x, y \geq 0 \), we will follow these steps:

### Step 1: Identify the Constraints
The constraints are:
1. \( 2x + y \leq 10 \)
2. \( x + 2y \leq 12 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert Inequalities to Equations
To find the boundary lines, we convert the inequalities into equations:
1. \( 2x + y = 10 \)
2. \( x + 2y = 12 \)

### Step 3: Find Intercepts
For the first equation \( 2x + y = 10 \):
- When \( x = 0 \): \( y = 10 \) (y-intercept)
- When \( y = 0 \): \( 2x = 10 \) → \( x = 5 \) (x-intercept)

For the second equation \( x + 2y = 12 \):
- When \( x = 0 \): \( 2y = 12 \) → \( y = 6 \) (y-intercept)
- When \( y = 0 \): \( x = 12 \) (x-intercept)

### Step 4: Plot the Constraints
Plot the lines on a graph:
- The line \( 2x + y = 10 \) intersects at points (5, 0) and (0, 10).
- The line \( x + 2y = 12 \) intersects at points (12, 0) and (0, 6).

### Step 5: Determine the Feasible Region
The feasible region is where the shaded areas of the inequalities overlap, which is bounded by the axes and the lines. Since both inequalities are less than or equal to, we shade towards the origin.

### Step 6: Find Corner Points
The corner points of the feasible region can be found by solving the equations:
1. Intersection of \( 2x + y = 10 \) and \( x + 2y = 12 \):
- From \( 2x + y = 10 \), we can express \( y = 10 - 2x \).
- Substitute into \( x + 2(10 - 2x) = 12 \):
\[ x + 20 - 4x = 12 \implies -3x = -8 \implies x = \frac{8}{3} \]
- Substitute \( x = \frac{8}{3} \) back into \( y = 10 - 2\left(\frac{8}{3}\right) \):
\[ y = 10 - \frac{16}{3} = \frac{30}{3} - \frac{16}{3} = \frac{14}{3} \]
- So one corner point is \( \left(\frac{8}{3}, \frac{14}{3}\right) \).

2. The other corner points from the intercepts are:
- (0, 0)
- (5, 0)
- (0, 6)

### Step 7: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( Z = 7x + 4y \) at each corner point:
1. At \( (0, 0) \):
\[ Z = 7(0) + 4(0) = 0 \]
2. At \( (5, 0) \):
\[ Z = 7(5) + 4(0) = 35 \]
3. At \( (0, 6) \):
\[ Z = 7(0) + 4(6) = 24 \]
4. At \( \left(\frac{8}{3}, \frac{14}{3}\right) \):
\[ Z = 7\left(\frac{8}{3}\right) + 4\left(\frac{14}{3}\right) = \frac{56}{3} + \frac{56}{3} = \frac{112}{3} \approx 37.33 \]

### Step 8: Determine Maximum and Minimum Values
- Maximum value of \( Z \) is \( 35 \) at \( (5, 0) \).
- Minimum value of \( Z \) is \( 0 \) at \( (0, 0) \).

### Final Solution
The maximum value of \( Z \) is \( 35 \) at the point \( (5, 0) \), and the minimum value of \( Z \) is \( 0 \) at the point \( (0, 0) \).