Maximize `z = 5x + 10y` subject to constraints `x + 2y le 10 , 3x + y le 12 , x ge 0 , y ge 0`

Answer

To maximize the function \( z = 5x + 10y \) subject to the constraints \( x + 2y \leq 10 \), \( 3x + y \leq 12 \), \( x \geq 0 \), and \( y \geq 0 \), we will follow these steps:

### Step 1: Identify the Constraints
We have the following constraints:
1. \( x + 2y \leq 10 \)
2. \( 3x + y \leq 12 \)
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. \( x + 2y = 10 \)
2. \( 3x + y = 12 \)

### Step 3: Find Intercepts of Each Line
- For \( x + 2y = 10 \):
- When \( x = 0 \): \( 2y = 10 \) → \( y = 5 \) (Intercept: \( (0, 5) \))
- When \( y = 0 \): \( x = 10 \) (Intercept: \( (10, 0) \))

- For \( 3x + y = 12 \):
- When \( x = 0 \): \( y = 12 \) (Intercept: \( (0, 12) \))
- When \( y = 0 \): \( 3x = 12 \) → \( x = 4 \) (Intercept: \( (4, 0) \))

### Step 4: Find the Intersection of the Lines
To find the intersection of the two lines, we solve the equations:
1. \( x + 2y = 10 \)
2. \( 3x + y = 12 \)

From the first equation, express \( x \):
\[ x = 10 - 2y \]

Substituting into the second equation:
\[ 3(10 - 2y) + y = 12 \]
\[ 30 - 6y + y = 12 \]
\[ 30 - 5y = 12 \]
\[ 5y = 18 \]
\[ y = \frac{18}{5} \]

Now substitute \( y \) back to find \( x \):
\[ x = 10 - 2\left(\frac{18}{5}\right) \]
\[ x = 10 - \frac{36}{5} \]
\[ x = \frac{50}{5} - \frac{36}{5} = \frac{14}{5} \]

So, the intersection point is \( \left(\frac{14}{5}, \frac{18}{5}\right) \).

### Step 5: Identify the Feasible Region
The feasible region is bounded by the lines and the axes. We will plot the points:
- \( (0, 0) \)
- \( (0, 5) \)
- \( (4, 0) \)
- \( \left(\frac{14}{5}, \frac{18}{5}\right) \)

### Step 6: Evaluate the Objective Function at Each Vertex
Now we evaluate \( z = 5x + 10y \) at each vertex:
1. At \( (0, 5) \):
\[ z = 5(0) + 10(5) = 50 \]
2. At \( (4, 0) \):
\[ z = 5(4) + 10(0) = 20 \]
3. At \( \left(\frac{14}{5}, \frac{18}{5}\right) \):
\[ z = 5\left(\frac{14}{5}\right) + 10\left(\frac{18}{5}\right) = 14 + 36 = 50 \]

### Step 7: Determine the Maximum Value
The maximum value of \( z \) occurs at both \( (0, 5) \) and \( \left(\frac{14}{5}, \frac{18}{5}\right) \) with \( z = 50 \).

### Conclusion
Thus, the maximum value of \( z \) is \( 50 \) at the points \( (0, 5) \) and along the line segment connecting \( (0, 5) \) to \( \left(\frac{14}{5}, \frac{18}{5}\right) \).

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