Question

A company manufactures two types of leaf​ blowers: an electric Turbo model and a​ gas-powered Tornado model. The​ company's production plan calls for the production of at least 740 blowers per month. It costs ​$74 to produce each Turbo model and ​$111 to manufacture each Tornado​ model, and the company has at most ​$67 comma 340 per month to use for production. Find the number of units that should be produced to maximize profit for the​ company, and the maximum​ profit, if the profit on each Turbo model is ​$32 and the profit on each Tornado model is ​$40.
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Part 1
The maximum value is ​$
  
enter your response here by producing
  
enter your response here units of the Turbo model and
  
enter your response here units of the Tornado model. ​(Type integers or​ decimals.)

Answer 1

To solve the problem of maximizing profit for the company producing Turbo and Tornado leaf blowers, we need to set up the appropriate constraints and objective function.

### Decision Variables
Let:
- \( x \) = number of Turbo models produced
- \( y \) = number of Tornado models produced

### Objective Function
The profit for each model is as follows:
- Profit from Turbo = $32
- Profit from Tornado = $40

Thus, the total profit \( P \) can be expressed as:
\[
P = 32x + 40y
\]

### Constraints
We have the following constraints:

1. **Production Constraint**: The total production should be at least 740 units:
\[
x + y \geq 740
\]

2. **Cost Constraint**: The total production cost should not exceed $67,340:
\[
74x + 111y \leq 67,340
\]

3. **Non-negativity Constraints**:
\[
x \geq 0, \quad y \geq 0
\]

Now we use linear programming to solve this problem.

### Step 1: Set Up the System of Equations
The problem can be solved using the corner-point method. We first solve for intersections of the constraint lines:

1. From \( x + y = 740 \):
\[
y = 740 - x
\]

2. From \( 74x + 111y = 67,340 \):
\[
y = \frac{67,340 - 74x}{111}
\]

### Step 2: Find Intersection Points
Set the equations equal to each other to find \( x \):
\[
740 - x = \frac{67,340 - 74x}{111}
\]

Multiplying the whole equation by 111 to clear the denominator:
\[
111(740 - x) = 67,340 - 74x
\]
\[
82,740 - 111x = 67,340 - 74x
\]
Rearranging:
\[
82,740 - 67,340 = 111x - 74x
\]
\[
15,400 = 37x
\]
\[
x = \frac{15,400}{37} \approx 416.22
\]

Substituting \( x \) back into either equation to find \( y \). Using \( y = 740 - x \):
\[
y = 740 - 416.22 \approx 323.78
\]

### Step 3: Evaluating Other Corner Points
We also need to check feasible corner points which include:

- \( x = 0 \): Then from \( x + y = 740 \) gives \( y = 740 \)
- Costs: \( 74(0) + 111(740) = 82,140 \) (not feasible)

- \( y = 0 \): Then from \( x + y = 740 \) gives \( x = 740 \)
- Costs: \( 74(740) + 111(0) = 54,760 \) (feasible)

### Calculate Profits
Now let's evaluate profits at points:
1. At \( (740, 0) \):
\[
P = 32(740) + 40(0) = 23,680
\]

2. At \( (x, y) \approx (416.22, 323.78) \):
\[
P = 32(416.22) + 40(323.78) \approx 13,000 + 12,951 = 25,951
\]

3. Check other bound points based on cost: where that constraint binds:
Solve \( 74x + 111y = 67,340 \) for extremes \( y = 0 \):
\[
x = \frac{67,340}{74} \approx 909.19
\]

Now evaluate boundaries as well.

### Conclusion
Finally, compare scenario profits:
- \( (740, 0) = 23680 \)
- \( (0, 740) \)'s calculation pulls = infeasibility.

The maximum profit is thus at the variable-produced per feasibility constraint:
- Number of Turbo: Approximately \( x = 416 \)
- Number of Tornado: Approximately \( y = 324 \)

### Final Solution
Thus, the answers are as follows:
- Maximum Profit: \( \$25,951 \)
- Turbo models: \( 416 \)
- Tornado models: \( 324 \)

Reiterate realistically checking boundaries for potential optimization suggestions as this problem lives potentially at graphical output edges.