Let's define the following variables:

• $x$: the number of large tablets
• $y$: the number of small tablets

Step 1: Formulate the problem

We are given:

• Each large tablet weighs 40 milligrams.
• Each small tablet weighs 30 milligrams.
• The total amount of drug available is 600 milligrams.

Thus, the total weight constraint can be written as: $40x + 30y \leq 600$

Next, we are told that:

• At least twice the number of small tablets is needed compared to the large tablets. This gives the inequality: $y \geq 2x$

• At least 3 large tablets must be made, so: $x \geq 3$

Step 2: Define the profit function

• Each large tablet gives a profit of $2.
• Each small tablet gives a profit of $1.

The total profit, $P$, is: $P = 2x + y$

Step 3: Maximize the profit under constraints

We need to maximize $P = 2x + y$ subject to the following constraints:

1. $40x + 30y \leq 600$ (weight constraint)
2. $y \geq 2x$ (small tablet constraint)
3. $x \geq 3$ (minimum large tablets)

Step 4: Solve the system of inequalities

First, let's simplify the weight constraint:

$40x + 30y \leq 600$ Dividing the entire inequality by 10: $4x + 3y \leq 60$

Now, solve this system by testing the constraints step by step.

1. From the constraint $y \geq 2x$, substitute into the weight equation:

$4x + 3(2x) \leq 60 \implies 4x + 6x \leq 60 \implies 10x \leq 60 \implies x \leq 6$

2. So, $x$ can take values between 3 and 6. For each $x$, find the corresponding $y$ and the profit:

• For $x = 3$: $y \geq 2(3) = 6$ Plug into the weight constraint: $4(3) + 3(6) = 12 + 18 = 30 \leq 60$ The total weight is 30, which satisfies the constraint. The profit is: $P = 2(3) + 6 = 6 + 6 = 12$

• For $x = 4$: $y \geq 2(4) = 8$ Plug into the weight constraint: $4(4) + 3(8) = 16 + 24 = 40 \leq 60$ The total weight is 40, which satisfies the constraint. The profit is: $P = 2(4) + 8 = 8 + 8 = 16$

• For $x = 5$: $y \geq 2(5) = 10$ Plug into the weight constraint: $4(5) + 3(10) = 20 + 30 = 50 \leq 60$ The total weight is 50, which satisfies the constraint. The profit is: $P = 2(5) + 10 = 10 + 10 = 20$

• For $x = 6$: $y \geq 2(6) = 12$ Plug into the weight constraint: $4(6) + 3(12) = 24 + 36 = 60 \leq 60$ The total weight is exactly 60, which satisfies the constraint. The profit is: $P = 2(6) + 12 = 12 + 12 = 24$

Step 5: Conclusion

The maximum profit of 24 is obtained when 6 large tablets and 12 small tablets are produced.

Would you like more details on any part of this solution or further questions? Here are some related problems you might find interesting:

1. What if the weight limit was increased to 800 milligrams?
2. How would the problem change if the large tablet profit increased to $3?
3. What if no minimum number of large tablets were required?
4. How would the solution change if the small tablet profit was also $2?
5. What would the optimal solution be if only 500 milligrams of the drug were available?

Tip: Always check the feasibility of your solution by substituting values back into the constraints.