This problem involves optimizing the production of two products under constraints, making it a linear programming problem. Let's define the variables, constraints, and the objective function:

Variables:

• Let $x$ represent the number of Count Dracula Salamis produced.
• Let $y$ represent the number of Frankenstein Sausages produced.

Constraints:

1. Pork constraint: $x + 2y \leq 1000$ (Since each Dracula Salami uses 1 pound of pork, and each Frankenstein Sausage uses 2 pounds of pork)

2. Beef constraint: $3x + 2y \leq 2400$ (Since each Dracula Salami uses 3 pounds of beef, and each Frankenstein Sausage uses 2 pounds of beef)

3. Production ratio constraint: $x \geq \frac{1}{3}(x + y)$ (Since at least one-third of the total production should be Dracula Salamis)

   Simplifying, this becomes: $2x - y \geq 0 \quad \text{or} \quad y \leq 2x$

4. Non-negativity constraint: $x \geq 0, \quad y \geq 0$

Objective Function:

Maximize the profit function: $P = 1x + 7y$ where $P$ is the total profit, $1 profit per Dracula Salami and $7 profit per Frankenstein Sausage.

Steps to Solve:

1. Convert constraints to inequalities:

   • Pork: $x + 2y \leq 1000$
   • Beef: $3x + 2y \leq 2400$
   • Ratio: $y \leq 2x$

2. Find intersection points by solving the inequalities simultaneously to determine the feasible region.

3. Evaluate the objective function at the vertices of the feasible region to find the maximum profit.

Let's compute these steps.

1. Intersection Points

• Solve $x + 2y = 1000$ and $3x + 2y = 2400$:

  Subtract the first equation from the second: $2x = 1400 \quad \Rightarrow \quad x = 700$ Substitute $x = 700$ into $x + 2y = 1000$: $700 + 2y = 1000 \quad \Rightarrow \quad y = 150$ So, one intersection point is $(700, 150)$.

• Solve $y = 2x$ and $x + 2y = 1000$:

  Substitute $y = 2x$ into $x + 2y = 1000$: $x + 4x = 1000 \quad \Rightarrow \quad 5x = 1000 \quad \Rightarrow \quad x = 200$ Substitute $x = 200$ into $y = 2x$: $y = 400$ So, another intersection point is $(200, 400)$.

• Solve $y = 2x$ and $3x + 2y = 2400$:

  Substitute $y = 2x$ into $3x + 2y = 2400$: $3x + 4x = 2400 \quad \Rightarrow \quad 7x = 2400 \quad \Rightarrow \quad x = 343$ Substitute $x = 343$ into $y = 2x$: $y = 686$ So, the last intersection point is $(343, 686)$.

2. Evaluate the Objective Function

• At $(700, 150)$: $P = 1(700) + 7(150) = 700 + 1050 = 1750$
• At $(200, 400)$: $P = 1(200) + 7(400) = 200 + 2800 = 3000$
• At $(343, 686)$: $P = 1(343) + 7(686) = 343 + 4802 = 5145$

Conclusion

The maximum profit of $5145 occurs when you produce 343 Count Dracula Salamis and 686 Frankenstein Sausages.

Would you like more details on this solution or have any questions?

Here are some related questions:

1. How does changing the profit for Count Dracula Salami affect the optimal production?
2. What happens if the pork or beef availability increases or decreases?
3. How would the solution change if more than one-third of the production had to be Count Dracula Salami?
4. Can you describe the feasible region and its significance in this context?
5. What if the production costs are introduced? How would this affect the profit maximization?

Tip: In linear programming, the optimal solution often lies at one of the vertices of the feasible region.