The minimum value of the objective function `z = 2x + 10y` for linear constraints `x ge 0, y ge 0, x - y ge 0, x - 5y le -5` is

Options

A. 10
B. 15
C. 12
D. 8

Answer

To find the minimum value of the objective function \( z = 2x + 10y \) subject to the constraints:

1. \( x \geq 0 \)
2. \( y \geq 0 \)
3. \( x - y \geq 0 \) (or \( x \geq y \))
4. \( x - 5y \leq -5 \) (or \( x \leq 5y - 5 \))

we will follow these steps:

### Step 1: Graph the Constraints
First, we will graph the inequalities on the coordinate plane.

- **For \( x \geq 0 \)**: This represents the right half of the plane (including the y-axis).
- **For \( y \geq 0 \)**: This represents the upper half of the plane (including the x-axis).
- **For \( x - y \geq 0 \)**: This can be rewritten as \( y \leq x \). The line \( y = x \) divides the plane into two halves, and we will consider the area below this line.
- **For \( x - 5y \leq -5 \)**: This can be rewritten as \( x \leq 5y - 5 \). The line \( x = 5y - 5 \) can be rearranged to find its intercepts.

### Step 2: Find Intercepts for the Line \( x = 5y - 5 \)
To find the intercepts:
- Set \( y = 0 \): \( x = 5(0) - 5 = -5 \) (not in the feasible region)
- Set \( x = 0 \): \( 0 = 5y - 5 \) → \( 5y = 5 \) → \( y = 1 \)

Thus, the line intersects the y-axis at \( (0, 1) \).

### Step 3: Identify the Feasible Region
The feasible region is bounded by the lines:
- The x-axis and y-axis (first quadrant).
- The line \( y = x \) (above this line).
- The line \( x = 5y - 5 \) (below this line).

### Step 4: Find Intersection Points
Next, we need to find the intersection points of the lines:
1. \( y = x \)
2. \( x = 5y - 5 \)

Substituting \( y = x \) into \( x = 5y - 5 \):
\[ x = 5x - 5 \\ 4x = 5 \\ x = \frac{5}{4} \]
Thus, \( y = \frac{5}{4} \) as well. The intersection point is \( \left( \frac{5}{4}, \frac{5}{4} \right) \).

### Step 5: Evaluate the Objective Function at the Corner Points
Now we evaluate the objective function \( z = 2x + 10y \) at the corner point \( \left( \frac{5}{4}, \frac{5}{4} \right) \):
\[ z = 2\left(\frac{5}{4}\right) + 10\left(\frac{5}{4}\right) \\ = \frac{10}{4} + \frac{50}{4} \\ = \frac{60}{4} = 15 \]

### Conclusion
The minimum value of the objective function \( z = 2x + 10y \) subject to the given constraints is **15**.

---