If A > 0, B > 0, and A + B = `pi/3` then the maximum value of tan A tan B is

Explanation

To find the maximum value of \( \tan A \tan B \) given that \( A + B = \frac{\pi}{3} \) and \( A > 0, B > 0 \), we can follow these steps:

### Step 1: Use the identity for \( \tan(A + B) \)
Given that \( A + B = \frac{\pi}{3} \), we can use the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Substituting \( A + B = \frac{\pi}{3} \) into the formula gives:
\[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]
Thus, we have:
\[ \sqrt{3} = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

### Step 2: Let \( x = \tan A \) and \( y = \tan B \)
We can rewrite the equation as:
\[ \sqrt{3}(1 - xy) = x + y \]
Rearranging this gives:
\[ \sqrt{3} - \sqrt{3}xy = x + y \]
or
\[ x + y + \sqrt{3}xy - \sqrt{3} = 0 \]

### Step 3: Express \( y \) in terms of \( x \)
From the equation above, we can express \( y \) in terms of \( x \):
\[ y = \frac{\sqrt{3} - x}{\sqrt{3}x + 1} \]

### Step 4: Find the product \( xy \)
Now, we want to maximize \( xy \):
\[ xy = x \cdot \frac{\sqrt{3} - x}{\sqrt{3}x + 1} \]
This can be simplified to:
\[ xy = \frac{x(\sqrt{3} - x)}{\sqrt{3}x + 1} \]

### Step 5: Differentiate to find the maximum
To find the maximum value of \( xy \), we can differentiate the expression with respect to \( x \) and set the derivative equal to zero. However, a simpler approach is to use the AM-GM inequality.

### Step 6: Apply AM-GM Inequality
By the AM-GM inequality:
\[ \frac{A + B}{2} \geq \sqrt{AB} \]
Thus:
\[ \frac{\frac{\pi}{3}}{2} \geq \sqrt{AB} \]
Squaring both sides, we get:
\[ \left(\frac{\pi}{6}\right)^2 \geq AB \]
This means:
\[ AB \leq \frac{\pi^2}{36} \]

### Step 7: Find the maximum value of \( \tan A \tan B \)
Since \( A + B = \frac{\pi}{3} \), we can find the maximum value of \( \tan A \tan B \) when \( A = B = \frac{\pi}{6} \):
\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]
Thus:
\[ \tan A \tan B = \tan\left(\frac{\pi}{6}\right) \tan\left(\frac{\pi}{6}\right) = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]

### Conclusion
The maximum value of \( \tan A \tan B \) when \( A + B = \frac{\pi}{3} \) is:
\[ \boxed{\frac{1}{3}} \]