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.

Question

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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.