The objective function `z=15x+30y` subject to `3x+yle12,x+2yle10,xge0,yge0,` can be maximized

Options

A. at infinite number of points
B. at two points only
C. at one points only
D. at three points only

Answer

To solve the linear programming problem of maximizing the objective function \( z = 15x + 30y \) subject to the constraints:

1. \( 3x + y \leq 12 \)
2. \( x + 2y \leq 10 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

we will follow these steps:

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

### Step 2: Find the Intercepts of the Constraints
For the first constraint \( 3x + y = 12 \):
- When \( x = 0 \), \( y = 12 \) (y-intercept).
- When \( y = 0 \), \( x = 4 \) (x-intercept).

For the second constraint \( x + 2y = 10 \):
- When \( x = 0 \), \( y = 5 \) (y-intercept).
- When \( y = 0 \), \( x = 10 \) (x-intercept).

### Step 3: Graph the Constraints
Plot the lines on a graph:
- The line \( 3x + y = 12 \) will pass through the points \( (4, 0) \) and \( (0, 12) \).
- The line \( x + 2y = 10 \) will pass through the points \( (10, 0) \) and \( (0, 5) \).

### Step 4: Determine the Feasible Region
The feasible region is where the shaded areas of the constraints overlap, considering the non-negativity constraints \( x \geq 0 \) and \( y \geq 0 \). This region will be in the first quadrant.

### Step 5: Identify the Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:
1. Intersection of \( 3x + y = 12 \) and \( x + 2y = 10 \):
- Multiply the second equation by 3: \( 3x + 6y = 30 \)
- Subtract the first equation from this: \( 5y = 18 \) → \( y = \frac{18}{5} \)
- Substitute \( y \) back into \( 3x + \frac{18}{5} = 12 \):
\[ 3x = 12 - \frac{18}{5} = \frac{60 - 18}{5} = \frac{42}{5} \implies x = \frac{14}{5} \]
- So one corner point is \( \left(\frac{14}{5}, \frac{18}{5}\right) \).

2. The other corner points are:
- \( (0, 0) \)
- \( (0, 5) \)
- \( (4, 0) \)

### Step 6: Evaluate the Objective Function at Each Corner Point
Now we evaluate \( z = 15x + 30y \) at each corner point:
1. At \( (0, 0) \):
\[ z = 15(0) + 30(0) = 0 \]
2. At \( (0, 5) \):
\[ z = 15(0) + 30(5) = 150 \]
3. At \( (4, 0) \):
\[ z = 15(4) + 30(0) = 60 \]
4. At \( \left(\frac{14}{5}, \frac{18}{5}\right) \):
\[ z = 15\left(\frac{14}{5}\right) + 30\left(\frac{18}{5}\right) = \frac{210}{5} + \frac{540}{5} = \frac{750}{5} = 150 \]

### Step 7: Determine the Maximum Value
The maximum value of \( z \) occurs at the points \( (0, 5) \) and \( \left(\frac{14}{5}, \frac{18}{5}\right) \), both yielding \( z = 150 \).

### Conclusion
The maximum value of the objective function \( z = 15x + 30y \) is \( 150 \) at the points \( (0, 5) \) and \( \left(\frac{14}{5}, \frac{18}{5}\right) \).

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