Question

A company manufactures two types of products, A and B.

Each unit of Product A requires 3 hours of labor and 2 units of raw material. Each unit of Product B requires 4 hours of labor and 3 units of raw material.

The company has a maximum of 120 hours of labor and 150 units of raw material available.

The profit from each unit of Product A is $30 and from each unit of Product B is $40. The production constraints are as follows:

Labor: 3x+4y ≤ 120

Raw material: 2x+3y ≤ 150

Non-negativity: x≥0, y≥0

Using P = 30x + 40y, find the maximum profit the company can achieve given the constraints.

(1 point)
Responses

$600
$600

$540
$540

$840
$840

$1200

Answer 1

To find the maximum profit, we need to solve the linear programming problem with the given constraints.

### Step 1: Define the constraints

1. Labor constraint: \(3x + 4y \leq 120\)
2. Raw material constraint: \(2x + 3y \leq 150\)
3. Non-negativity constraints: \(x \geq 0\), \(y \geq 0\)

### Step 2: Set up the objective function

The objective function to maximize is the profit:
\[ P = 30x + 40y \]

### Step 3: Graph the constraints

To find the feasible region, we can graph the constraints:

1. **Labor constraint**: \(3x + 4y = 120\)
- If \(x = 0\), \(4y = 120 \Rightarrow y = 30\)
- If \(y = 0\), \(3x = 120 \Rightarrow x = 40\)
- The line intersects the axes at points (40, 0) and (0, 30).

2. **Raw material constraint**: \(2x + 3y = 150\)
- If \(x = 0\), \(3y = 150 \Rightarrow y = 50\)
- If \(y = 0\), \(2x = 150 \Rightarrow x = 75\)
- The line intersects the axes at points (75, 0) and (0, 50).

### Step 4: Find the intersection of constraints

To find the intersection of the two lines \(3x + 4y = 120\) and \(2x + 3y = 150\), we can solve the equations simultaneously.

From \(3x + 4y = 120\):
\[ y = \frac{120 - 3x}{4} \]

Substituting into the second constraint:
\[ 2x + 3\left(\frac{120 - 3x}{4}\right) = 150 \]
\[ 2x + \frac{360 - 9x}{4} = 150 \]
\[ 8x + 360 - 9x = 600 \] (Multiplying through by 4)
\[ -x + 360 = 600 \]
\[ -x = 240 \]
\[ x = -240 \] (not feasible)

Instead, let's test the boundary points found:

### Step 5: Calculate profits at corner points

1. **Point (0, 30)**:
\[ P = 30(0) + 40(30) = 1200 \]

2. **Point (40, 0)**:
\[ P = 30(40) + 40(0) = 1200 \]

3. **Point (0, 50)** (not feasible due to raw materials).
4. Find intersection:
Substitute \(x = 20, y = 20\) into the constraints:
- \(3(20) + 4(20) = 60 + 80 = 140\) (not feasible).

Thus check corners:

### Final verdict

Among feasible points, max profit occurs when either corner:

$1200$ as maximum profit is at corners.

Hence, the correct answer is:

**$1200**