<Question>
If $-6\leq a \leq -2$ and $3 \leq b \leq 5$, what is the greatest possible value of $\displaystyle\left(a+\frac{1}{b}\right)\left(\frac{1}{b}-a\right) $?  Express your answer as a common fraction.
</Question>

<think>
<Plan_1>
Okay, let's see. I need to find the greatest possible value of the expression \(\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)\) given that \( -6 \leq a \leq -2 \) and \( 3 \leq b \leq 5 \). Hmm, alright. Let me start by understanding the problem step by step.

First, the expression is a product of two terms: \(\left(a + \frac{1}{b}\right)\) and \(\left(\frac{1}{b} - a\right)\). Maybe I can simplify this expression before worrying about the constraints on a and b. Let me expand the product to see if that helps.

<prompt_1.1> Expand the expression \(\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)\) to simplify it. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Expanding the terms: 

\[
\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right) = a \cdot \frac{1}{b} - a^2 + \frac{1}{b} \cdot \frac{1}{b} - \frac{a}{b}
\]

Simplifying that:

\[
= \frac{a}{b} - a^2 + \frac{1}{b^2} - \frac{a}{b}
\]

Oh, wait! The \(\frac{a}{b}\) and \(-\frac{a}{b}\) terms cancel each other out. So, the expression simplifies to:

\[
-a^2 + \frac{1}{b^2}
\]

So, the original expression simplifies to \(-a^2 + \frac{1}{b^2}\). That's much simpler!
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, the problem becomes finding the maximum value of \(-a^2 + \frac{1}{b^2}\) given the constraints on a and b. Let's analyze each term separately.

<prompt_2.1> Analyze the simplified expression \(-a^2 + \frac{1}{b^2}\) to find the conditions for maximizing it. </prompt_2.1>
-----

But let me think again: maybe there's a case where even though individually the terms are maximized at those points, the combination isn't? For example, sometimes when you have products or more complex expressions, the maximum isn't at the endpoints. But in this case, since the expression simplifies to \(-a^2 + \frac{1}{b^2}\), which is a straightforward function of a and b, and since both terms are independent—meaning, the expression is the sum of two terms each depending on a single variable—the maximum (or minimum) occurs when each term is optimized individually.

<prompt_2.2> Further verify that the expression \(-a^2 + \frac{1}{b^2}\) is maximized at the endpoints of the given intervals for \(a\) and \(b\). </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
First, consider \(-a^2\). Since \(a\) is between -6 and -2, squaring \(a\) will give positive values. The larger the absolute value of \(a\), the larger \(a^2\) becomes, so \(-a^2\) will be more negative. Conversely, if \(a\) is closer to zero, \(-a^2\) will be less negative (i.e., larger). Therefore, to maximize \(-a^2\), we need the smallest possible \(|a|\), which would occur when \(a\) is closest to zero. Given that \(a\) is in the interval \([-6, -2]\), the maximum value of \(-a^2\) occurs at the maximum value of \(a\), which is \(a = -2\). Because \(-(-2)^2 = -4\), whereas \(-(-6)^2 = -36\), which is much smaller (more negative). So, \(-a^2\) is maximized when \(a\) is at its maximum (least negative) value, which is -2. Then, \(-a^2 = -(-2)^2 = -4\).

Now, let's look at the \(\frac{1}{b^2}\) term. To maximize this term, we need to minimize \(b\), since \(b\) is in the denominator. The smaller \(b\) is, the larger \(\frac{1}{b^2}\) becomes. Given that \(b\) is between 3 and 5, the maximum of \(\frac{1}{b^2}\) occurs at the minimum \(b\), which is 3. So, \(\frac{1}{3^2} = \frac{1}{9}\).

Therefore, combining these two results, the expression \(-a^2 + \frac{1}{b^2}\) is maximized when \(a\) is as large as possible (i.e., -2) and \(b\) is as small as possible (i.e., 3). Plugging these values in, we get:

\[
-(-2)^2 + \frac{1}{3^2} = -4 + \frac{1}{9} = -\frac{36}{9} + \frac{1}{9} = -\frac{35}{9}
\]
</execution_2.1>

<execution_2.2>
Let me confirm that.

The term \(-a^2\) is a quadratic function in a. Since the coefficient is negative, it opens downward, so it attains its maximum at the vertex. However, the vertex of \(-a^2\) is at a = 0, which is not within the interval \(-6 \leq a \leq -2\). Therefore, on the interval \(-6 \leq a \leq -2\), the maximum of \(-a^2\) occurs at the point closest to 0, which is a = -2. Because as a becomes less negative (i.e., increases towards 0), \(-a^2\) increases. For example, at a = -6, \(-a^2 = -36\); at a = -2, \(-a^2 = -4\). So indeed, \(-a^2\) is maximized when a is at its maximum (closest to zero), which is a = -2.

Similarly, the term \(\frac{1}{b^2}\) is a decreasing function of b. As b increases, \(\frac{1}{b^2}\) decreases. Therefore, \(\frac{1}{b^2}\) is maximized when b is minimized. Since b is in the interval [3,5], the maximum occurs at b = 3, where \(\frac{1}{b^2} = \frac{1}{9}\).

Therefore, combining these two, the expression \(-a^2 + \frac{1}{b^2}\) is maximized when \(-a^2\) is maximized (a = -2) and \(\frac{1}{b^2}\) is maximized (b = 3). Therefore, substituting these values gives:

\(-(-2)^2 + \frac{1}{3^2} = -4 + \frac{1}{9} = -\frac{36}{9} + \frac{1}{9} = -\frac{35}{9}\).
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, but let me check if that's correct. Let me verify by plugging in different values.

<prompt_3.1> Verify the maximum value of \(-a^2 + \frac{1}{b^2}\) by plugging in different values of \(a\) and \(b\). </prompt_3.1>
-----

Based on execution_2.2:
But let me check if there's a possibility that choosing a different combination of a and b could yield a higher value.

<prompt_3.2> Perform a final verification to ensure no other combinations of \(a\) and \(b\) could yield a higher value. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Let's take a = -2 and b = 3: then the original expression is \(\left(-2 + \frac{1}{3}\right)\left(\frac{1}{3} - (-2)\right)\). Let's compute that:

First term: \(-2 + 1/3 = -5/3\)

Second term: \(1/3 + 2 = 7/3\)

Multiplying them: \(-5/3 * 7/3 = -35/9 ≈ -3.888... \)

If I take a different a, say a = -6, then \(-a^2 = -36\), and if b is 5, \(\frac{1}{b^2} = 1/25\). So total expression would be -36 + 1/25 = -359/25 = -14.36, which is much smaller. If I take a = -2 and b = 5: \(-a^2 + 1/b^2 = -4 + 1/25 = -99/25 = -3.96\), which is actually slightly more than -35/9 (≈ -3.888... ), but wait, -99/25 is -3.96, which is less than -35/9 ≈ -3.888. Wait, no: -3.96 is more negative than -3.888, so -3.96 is smaller. So indeed, the maximum (least negative) value occurs at a = -2 and b = 3. But wait, but the problem is asking for the greatest possible value, which would be the least negative number here. But wait, is there a possibility that another combination gives a positive value? Let me check.

Wait, the expression is \(-a^2 + 1/b^2\). Since \(a\) is between -6 and -2, \(a^2\) is between 4 and 36, so \(-a^2\) is between -36 and -4. \(1/b^2\) is between 1/25 and 1/9, since b is between 3 and 5. So, the maximum value of the expression would be when \(-a^2\) is as large as possible (i.e., closest to zero) and \(1/b^2\) is as large as possible. So, combining those, as we did before, gives -4 + 1/9 = -35/9. But is there a way to get a positive value?

Let's see: suppose \(-a^2 + 1/b^2 > 0\). Then, \(1/b^2 > a^2\). But \(a^2\) is at least 4 (since a is at most -2 in absolute value, so a^2 is 4). But \(1/b^2\) is at most 1/9 (when b=3), which is approximately 0.111. Since 0.111 is less than 4, \(1/b^2\) can never be greater than \(a^2\). Therefore, \(-a^2 + 1/b^2\) is always negative, and the maximum value (i.e., the least negative) occurs when \(-a^2\) is maximized and \(1/b^2\) is maximized. Therefore, the maximum value is indeed -35/9.
</execution_3.1>

<execution_3.2>
For example, suppose we take a = -6 and b = 5. Then the expression would be \(-(-6)^2 + \frac{1}{5^2} = -36 + \frac{1}{25} = -36 + 0.04 = -35.96\), which is way worse (more negative). If we take a = -2 and b = 5: \(-(-2)^2 + \frac{1}{25} = -4 + 0.04 = -3.96\), which is still negative. If we take a = -6 and b = 3: \(-36 + 1/9 ≈ -36 + 0.111 ≈ -35.888\), which is still worse than -35/9 ≈ -3.888... 

Wait, but actually, -35/9 is approximately -3.888...., which is greater than -3.96 and -35.888. So, yes, the maximum occurs at a = -2 and b = 3. 

Another check: if we take a = -2 and b = 3, the original expression before simplification was \(\left(a + \frac{1}{b}\right)\left(\frac{1}{b} - a\right)\). Let's compute that to confirm:

\[
\left(-2 + \frac{1}{3}\right)\left(\frac{1}{3} - (-2)\right) = \left(-\frac{6}{3} + \frac{1}{3}\right)\left(\frac{1}{3} + \frac{6}{3}\right) = \left(-\frac{5}{3}\right)\left(\frac{7}{3}\right) = -\frac{35}{9}
\]

Which matches the simplified expression. So that's consistent. 

Therefore, after checking different points and verifying the calculation, I'm confident that the greatest possible value is \(-\frac{35}{9}\).
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
Wait, but the problem says "the greatest possible value", and since all other values give more negative results, this is indeed the maximum. So the answer should be \boxed{-\dfrac{35}{9}}.

Wait, but the problem says "greatest possible value", which in this case is the least negative number, which is correct. But let me just confirm once again if there's no other combination where the expression could be higher. For example, if a is at its minimum (-6), but b is at its minimum (3), but as we saw, that gives a much lower value. Similarly, if a is at its maximum (-2) and b at its maximum (5), we get a value of \(-4 + 1/25 = -99/25 ≈ -3.96\), which is still less than -35/9 ≈ -3.888. So indeed, the maximum occurs at a = -2 and b = 3.
</Plan_4>

<Execution_4>
No prompt to execute.
</Execution_4>

<Final_answer>
Therefore, the greatest possible value of the given expression, expressed as a common fraction, is $\boxed{-\frac{35}{9}}$.
</Final_answer>
</think>