Minimum of `Z=5x+8y`
subject to `x+y ge 5, x le4, y le 2, x ge0, y ge0` is

Options

A. 28
B. 13
C. 0
D. 52

Answer

To solve the linear programming problem of minimizing \( Z = 5x + 8y \) subject to the constraints:

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

we will follow these steps:

### Step 1: Identify the constraints and plot them
We need to plot the inequalities on a graph to find the feasible region.

- For the constraint \( x + y \geq 5 \), we can rearrange it to \( y \geq 5 - x \). This line intersects the y-axis at (0, 5) and the x-axis at (5, 0).
- The line \( x = 4 \) is a vertical line that intersects the x-axis at (4, 0).
- The line \( y = 2 \) is a horizontal line that intersects the y-axis at (0, 2).
- The non-negativity constraints \( x \geq 0 \) and \( y \geq 0 \) restrict our feasible region to the first quadrant.

### Step 2: Determine the feasible region
By plotting these lines on a graph, we can identify the feasible region that satisfies all constraints. The feasible region will be bounded by the lines:

- \( x + y = 5 \)
- \( x = 4 \)
- \( y = 2 \)

### Step 3: Find the corner points of the feasible region
The corner points of the feasible region can be found by solving the equations of the lines where they intersect:

1. Intersection of \( x + y = 5 \) and \( x = 4 \):
\[ 4 + y = 5 \implies y = 1 \quad \text{(Point: (4, 1))} \]

2. Intersection of \( x + y = 5 \) and \( y = 2 \):
\[ x + 2 = 5 \implies x = 3 \quad \text{(Point: (3, 2))} \]

3. Intersection of \( x = 4 \) and \( y = 2 \):
\[ \text{Point: (4, 2)} \]

The corner points of the feasible region are (4, 1), (3, 2), and (4, 2).

### Step 4: Evaluate the objective function at each corner point
Now we will evaluate \( Z = 5x + 8y \) at each of the corner points:

1. At (4, 1):
\[ Z = 5(4) + 8(1) = 20 + 8 = 28 \]

2. At (3, 2):
\[ Z = 5(3) + 8(2) = 15 + 16 = 31 \]

3. At (4, 2):
\[ Z = 5(4) + 8(2) = 20 + 16 = 36 \]

### Step 5: Determine the minimum value
From the evaluations, we find:

- \( Z(4, 1) = 28 \)
- \( Z(3, 2) = 31 \)
- \( Z(4, 2) = 36 \)

The minimum value of \( Z \) occurs at the point (4, 1) with \( Z = 28 \).

### Final Answer
The minimum of \( Z = 5x + 8y \) subject to the given constraints is **28** at the point (4, 1).

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