A manufacture produces two models of bike model X and model Y. Model X takes a 6 man hours to make per unit, while model Y takes 10 man hours per unit. There is a total of 450 man hour availbale per week. Handling and marketing costs are 2000 and 1000 per unit of model X and Y, respectively. The tota funds available for these purposes are 8000 per week. Profits per unit for models X and Y are 1000 and manufacturer produce, so, as to yeild a maximum profit? Find the maximum profit.

Explanation

To solve the problem of maximizing the profit from producing two models of bikes (Model X and Model Y), we will follow these steps:

### Step 1: Define the Variables
Let:
- \( x \) = number of units produced of Model X
- \( y \) = number of units produced of Model Y

### Step 2: Define the Objective Function
The profit per unit for Model X is 1000, and for Model Y, it is also 1000. Therefore, the total profit \( Z \) can be expressed as:
\[ Z = 1000x + 1000y \]

### Step 3: Define the Constraints
From the problem statement, we have the following constraints:

1. **Man-hours constraint**: Model X takes 6 man-hours per unit, and Model Y takes 10 man-hours per unit. The total available man-hours per week is 450. Thus, the constraint is:
\[ 6x + 10y \leq 450 \]

2. **Cost constraint**: The handling and marketing costs are 2000 per unit for Model X and 1000 per unit for Model Y, with a total budget of 8000. Thus, the constraint is:
\[ 2000x + 1000y \leq 8000 \]

3. **Non-negativity constraints**: Since we cannot produce negative units, we have:
\[ x \geq 0, \quad y \geq 0 \]

### Step 4: Simplify the Constraints
We can simplify the cost constraint by dividing the entire inequality by 1000:
\[ 2x + y \leq 8 \]

### Step 5: Graph the Constraints
To find the feasible region, we need to graph the constraints:

1. **For \( 6x + 10y = 450 \)**:
- When \( x = 0 \): \( 10y = 450 \) → \( y = 45 \) → Point (0, 45)
- When \( y = 0 \): \( 6x = 450 \) → \( x = 75 \) → Point (75, 0)

2. **For \( 2x + y = 8 \)**:
- When \( x = 0 \): \( y = 8 \) → Point (0, 8)
- When \( y = 0 \): \( 2x = 8 \) → \( x = 4 \) → Point (4, 0)

### Step 6: Identify the Feasible Region
The feasible region is determined by the area that satisfies all constraints. This will be the area below both lines and in the first quadrant (where \( x \) and \( y \) are non-negative).

### Step 7: Find the Corner Points
The corner points of the feasible region are where the constraints intersect:
1. (0, 0)
2. (0, 8)
3. (4, 0)
4. (75, 0) (not feasible as it does not satisfy the second constraint)

### Step 8: Evaluate the Objective Function at Each Corner Point
Now we will calculate the profit \( Z \) at each feasible corner point:

1. **At (0, 0)**:
\[ Z = 1000(0) + 1000(0) = 0 \]

2. **At (0, 8)**:
\[ Z = 1000(0) + 1000(8) = 8000 \]

3. **At (4, 0)**:
\[ Z = 1000(4) + 1000(0) = 4000 \]

### Step 9: Determine the Maximum Profit
The maximum profit occurs at the point (0, 8):
\[ \text{Maximum Profit} = 8000 \]

### Conclusion
The manufacturer should produce 0 units of Model X and 8 units of Model Y to yield a maximum profit of 8000.

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