The maximum value of `z=15x+30y` subject to `3x+yle12,x+2yle10,xge0,yge0` is

Options

A. 60
B. 150
C. 160
D. 100

Answer

To solve the problem of maximizing \( z = 15x + 30y \) subject to the constraints \( 3x + y \leq 12 \), \( x + 2y \leq 10 \), \( x \geq 0 \), and \( y \geq 0 \), we will follow these steps:

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

### Step 2: Convert the inequalities to equations
To graph the constraints, we first convert the inequalities into equations:
1. \( 3x + y = 12 \)
2. \( x + 2y = 10 \)

### Step 3: Find the intercepts for each line
For \( 3x + y = 12 \):
- When \( x = 0 \), \( y = 12 \) (y-intercept)
- When \( y = 0 \), \( 3x = 12 \) → \( x = 4 \) (x-intercept)

So the intercepts are (0, 12) and (4, 0).

For \( x + 2y = 10 \):
- When \( x = 0 \), \( 2y = 10 \) → \( y = 5 \) (y-intercept)
- When \( y = 0 \), \( x = 10 \) (x-intercept)

So the intercepts are (0, 5) and (10, 0).

### Step 4: Graph the lines and identify the feasible region
- Draw the lines \( 3x + y = 12 \) and \( x + 2y = 10 \) on a graph.
- Shade the areas that satisfy the inequalities \( 3x + y \leq 12 \) and \( x + 2y \leq 10 \).
- The feasible region is where the shaded areas overlap, bounded by the axes.

### Step 5: Find the vertices of the feasible region
The vertices of the feasible region can be found at the intersections of the lines and the axes:
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: \( (3x + 6y) - (3x + y) = 30 - 12 \)
- This simplifies to \( 5y = 18 \) → \( y = \frac{18}{5} \)
- Substitute \( y \) back into \( x + 2y = 10 \):
- \( x + 2(\frac{18}{5}) = 10 \)
- \( x + \frac{36}{5} = 10 \)
- \( x = 10 - \frac{36}{5} = \frac{14}{5} \)
- So the intersection point is \( C\left(\frac{14}{5}, \frac{18}{5}\right) \).

2. The other vertices are:
- A(0, 0)
- B(0, 5)
- D(4, 0)

### Step 6: Evaluate the objective function at each vertex
Now we will evaluate \( z = 15x + 30y \) at each vertex:
1. At A(0, 0):
\[ z_A = 15(0) + 30(0) = 0 \]
2. At B(0, 5):
\[ z_B = 15(0) + 30(5) = 150 \]
3. At C\(\left(\frac{14}{5}, \frac{18}{5}\right)\):
\[ z_C = 15\left(\frac{14}{5}\right) + 30\left(\frac{18}{5}\right) = \frac{210}{5} + \frac{540}{5} = \frac{750}{5} = 150 \]
4. At D(4, 0):
\[ z_D = 15(4) + 30(0) = 60 \]

### Step 7: Determine the maximum value
From the evaluations:
- \( z_A = 0 \)
- \( z_B = 150 \)
- \( z_C = 150 \)
- \( z_D = 60 \)

The maximum value of \( z \) is \( 150 \) at points B(0, 5) and C\(\left(\frac{14}{5}, \frac{18}{5}\right)\).

### Conclusion
The maximum value of \( z = 15x + 30y \) subject to the given constraints is **150**.

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