Problem Statement:

We need to find two positive real numbers whose product is maximized, given that their sum is 110. Guess and check isn't acceptable, so we must use calculus (optimization) or algebraic methods to find the maximum product.

Step-by-Step Solution:

Let the two numbers be $x$ and $y$.

1. Sum Constraint:
   We know that the sum of these two numbers is 110: x + y = 110 \tag{1} From equation (1), solve for $y$: y = 110 - x \tag{2}

2. Objective (Maximizing the Product):
   The product $P$ of the two numbers is: $P = x \cdot y$ Substitute $y = 110 - x$ into the product equation: $P(x) = x(110 - x)$ Simplify the expression: P(x) = 110x - x^2 \tag{3}

3. Maximization: To maximize the product, we take the derivative of $P(x)$ with respect to $x$ and set it equal to zero.

   First, find the derivative of $P(x)$: $P'(x) = 110 - 2x$ Set $P'(x) = 0$ to find the critical points: $110 - 2x = 0$ $2x = 110$ $x = 55$

4. Find the Corresponding $y$: Using equation (2), substitute $x = 55$ to find $y$: $y = 110 - 55 = 55$

5. Conclusion: The two numbers are both 55. The maximum product is: $P(55) = 55 \cdot 55 = 3025$ Therefore, the two positive real numbers whose product is maximized are 55 and 55, and their maximum product is 3025.

Justification:

We used calculus by finding the derivative of the product function and solving for the critical points. Since the second derivative $P''(x) = -2$ is negative, this confirms that $x = 55$ gives a maximum.

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Would you like more details on this solution or another approach?
Here are some related questions to explore further:

1. What if the sum of the two numbers was 200 instead of 110?
2. How would the solution change if the two numbers must be integers?
3. Can you generalize the formula for the two numbers whose product is maximized for any given sum $S$?
4. What happens if the sum constraint is changed to an inequality, like $x + y \leq 110$?
5. How can we solve this problem using only algebra without calculus?

Tip: In optimization problems with constraints, expressing the variables in terms of one another simplifies the problem and reduces the number of unknowns.