We need to find two positive numbers, say $x$ and $y$, such that:

1. The sum of the first number and twice the second number is 200: $x + 2y = 200$
2. The product $P = x \cdot y$ is maximized.

Step 1: Express one variable in terms of the other

From the first equation, solve for $x$ in terms of $y$: $x = 200 - 2y$

Step 2: Write the product equation

The product $P$ is given by: $P = x \cdot y = (200 - 2y) \cdot y$ This simplifies to: $P = 200y - 2y^2$

Step 3: Maximize the product

To find the value of $y$ that maximizes $P$, take the derivative of $P$ with respect to $y$ and set it equal to zero: $\frac{dP}{dy} = 200 - 4y$ Set the derivative equal to zero: $200 - 4y = 0$ Solving for $y$: $4y = 200 \quad \Rightarrow \quad y = 50$

Step 4: Find the corresponding value of $x$

Substitute $y = 50$ into the equation $x = 200 - 2y$: $x = 200 - 2(50) = 100$

Conclusion

The two numbers are $x = 100$ and $y = 50$.

Thus, the two numbers are: $\boxed{100, 50}$

Would you like more details or have any questions?

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Relative Questions:

1. What if the condition was that the sum of the first and three times the second is 300?
2. How would you solve it if the goal was to minimize the product instead of maximizing it?
3. What happens if we constrain both numbers to be integers?
4. How do you graph the relationship between $x$, $y$, and the product $P$?
5. Can this problem be extended to more than two numbers with a similar condition?

Tip: To find maximum or minimum values in optimization problems, using derivatives is a key method known as the first derivative test.