To solve this problem, let's break down the information given:

We have two positive numbers $x$ and $y$ such that:

1. The product of the numbers is 9, so $x \cdot y = 9$.
2. We want to find the maximum possible value of $x + y$.

To find this maximum, we can express $y$ in terms of $x$ using the constraint: $y = \frac{9}{x}$

Now, we want to maximize $x + y = x + \frac{9}{x}$.

To find the maximum, let's differentiate $f(x) = x + \frac{9}{x}$ with respect to $x$ and set the derivative to zero.

1. Differentiate: $f'(x) = 1 - \frac{9}{x^2}$

2. Set $f'(x) = 0$: $1 - \frac{9}{x^2} = 0$ $\frac{9}{x^2} = 1$ $x^2 = 9$ $x = 3 \quad (\text{since } x \text{ is positive})$

3. Substitute $x = 3$ back to find $y$: $y = \frac{9}{3} = 3$

So, $x = 3$ and $y = 3$ maximize $x + y$, and their sum is: $x + y = 3 + 3 = 6$

Answer

The largest possible value of $x + y$ is $\boxed{6}$.

Would you like further explanation, or have any additional questions?

Here are some related questions:

1. How can we find the minimum possible value of $x + y$ if $x \cdot y = 9$?
2. How do we determine whether a critical point gives a maximum or minimum?
3. Can the AM-GM inequality be used to solve this type of problem?
4. What if the product constraint was a different number, like 16?
5. How does changing one of the numbers while keeping the product constant affect their sum?

Tip: Problems involving maximizing or minimizing sums under a product constraint can often be tackled by using calculus or the AM-GM inequality.