Okay, let's tackle this problem step by step. So, we have two squares: one with corners at (0,0) and (2,2), and another with corners at (4,2) and (5,3). We need to find the maximum possible slope of a line connecting a point A in the first square and a point B in the second square. Hmm, interesting. Let me think about how to approach this.

First, I remember that the slope of a line between two points (x1, y1) and (x2, y2) is given by (y2 - y1)/(x2 - x1). So, to maximize the slope, we need to maximize the numerator (the difference in y-coordinates) and minimize the denominator (the difference in x-coordinates). That makes sense because a steeper line would have a larger rise over a smaller run.

Now, since points A and B can be anywhere within or on their respective squares, we need to consider the extremes. The maximum slope would likely come from the farthest apart points in terms of y and as close as possible in terms of x. Let me visualize the squares first.

The first square is from (0,0) to (2,2). So, it's a 2x2 square in the lower left corner. The second square has corners at (4,2) and (5,3). Wait, that's a 1x1 square shifted to the upper right. Let me confirm the coordinates. The second square's opposite corners are (4,2) and (5,3). So, the square extends from x=4 to x=5 and y=2 to y=3? Wait, actually, if the opposite corners are (4,2) and (5,3), then the square is bounded by x from 4 to 5 and y from 2 to 3. But depending on the orientation, maybe it's a rotated square? Wait, no, opposite corners are given, so the sides are aligned with the axes. Let me verify.

If the square has opposite corners at (4,2) and (5,3), then the sides are parallel to the axes. The length of each side would be the distance between (4,2) and (5,2), which is 1 unit, and between (4,2) and (4,3), which is 1 unit. So yes, the square is axis-aligned with corners at (4,2), (5,2), (5,3), and (4,3). Wait, but the problem says "opposite corners at points (4,2) and (5,3)". So, the square is indeed from (4,2) to (5,3), so the other corners would be (4,3) and (5,2). Got it.

So, the first square is from (0,0) to (2,2), and the second is from (4,2) to (5,3). Now, to maximize the slope, we need the line connecting A and B to have the maximum possible rise (difference in y) and minimum possible run (difference in x). So, the maximum rise would be from the lowest y in the first square to the highest y in the second square. The maximum run would be the smallest possible difference in x between the two squares.

Wait, let's think about that. The first square's y ranges from 0 to 2, and the second square's y ranges from 2 to 3. So, the maximum possible difference in y would be from the lowest point in the first square (which is (0,0)) to the highest point in the second square (which is (5,3)). That gives a rise of 3 - 0 = 3. The run would be from x=0 to x=5, which is 5 units. But that gives a slope of 3/5. But maybe we can get a steeper slope by choosing different points.

Alternatively, maybe choosing the highest y in the first square (which is 2) and the lowest y in the second square (which is 2). Wait, but then the rise would be 0, which is not helpful. So, that's not the way. Hmm.

Wait, maybe the maximum slope isn't just the maximum y in the second square minus the minimum y in the first square. Because if we take a point A with the highest possible y and a point B with the lowest possible y, the slope would be negative. But we want the maximum slope, which is the maximum possible positive slope. So, we need to maximize (y_B - y_A)/(x_B - x_A). So, to maximize the numerator and minimize the denominator.

So, the maximum y_B would be 3, and the minimum y_A would be 0. Then, the numerator is 3 - 0 = 3. The denominator is x_B - x_A. To minimize the denominator, we want x_B as small as possible and x_A as large as possible. The smallest x_B is 4 (from the second square) and the largest x_A is 2 (from the first square). So, x_B - x_A = 4 - 2 = 2. Therefore, the slope would be 3/2. But wait, is that possible? Let me check.

If point A is at (2,0) and point B is at (4,3), then the slope is (3 - 0)/(4 - 2) = 3/2. That seems higher than 3/5, which was the previous calculation. So, maybe 3/2 is the answer. But wait, let's confirm if those points are within the squares.

Point A is (2,0): the first square goes from (0,0) to (2,2), so (2,0) is on the square. Point B is (4,3): the second square goes from (4,2) to (5,3), so (4,3) is a corner point. So, yes, those points are within or on the squares. So, that gives a slope of 3/2. Is that the maximum?

Wait, but maybe there's a higher slope. Let's think. Suppose we take a point A higher up in the first square. For example, if A is at (2,2), which is the top-right corner of the first square, and B is at (4,3), then the slope is (3 - 2)/(4 - 2) = 1/2, which is less than 3/2. So, that's worse.

Alternatively, if A is at (0,0) and B is at (5,3), the slope is (3 - 0)/(5 - 0) = 3/5, which is 0.6, less than 3/2. So, that's worse. So, maybe the maximum occurs when A is at (2,0) and B is at (4,3).

But let's check other possibilities. Suppose we take A at (2,0), which is the rightmost point in the first square, and then take B as high as possible in the second square. The highest y in the second square is 3, achieved at (4,3) and (5,3). So, if we take B as (4,3), the slope is (3 - 0)/(4 - 2) = 3/2. If we take B as (5,3), the slope is (3 - 0)/(5 - 2) = 3/3 = 1, which is less. So, (4,3) gives a higher slope. So, that's better.

Alternatively, if we take A as (2,0), and B as (4,3), that's 3/2. What if we take A as (2,0) and B as (5,3)? Then, the slope is (3 - 0)/(5 - 2) = 1, which is lower. So, no gain there.

Alternatively, take A as (0,0) and B as (4,3). The slope is (3 - 0)/(4 - 0) = 3/4, which is 0.75, still lower than 3/2.

Alternatively, take A as (0,0) and B as (5,3). The slope is 3/5, as before. So, still lower.

Alternatively, take A as (2,0) and B as (4,3). That's 3/2. What if we take A as (1,0) and B as (4,3). Then, the slope is (3 - 0)/(4 - 1) = 3/3 = 1, which is lower. So, moving A to the left from (2,0) decreases the slope.

Alternatively, take A as (2,0) and B as (4,3). That's 3/2. Is there a way to get a higher slope?

Wait, perhaps if we take A not at the corner but somewhere else. For example, if A is somewhere else in the first square, can we get a higher slope? Let's see.

Suppose A is at (2,0), the rightmost point, and B is at (4,3). Slope is 3/2. If we take A slightly to the left, say (2 - ε, 0), and B slightly to the right, say (4 + ε, 3). Then, the slope is (3 - 0)/(4 + ε - (2 - ε)) = 3/(2 + 2ε). As ε approaches 0, the slope approaches 3/2. So, moving A and B closer to (2,0) and (4,3) respectively gives a slope approaching 3/2. So, 3/2 is the maximum in that direction.

Alternatively, could we take A in the first square with a higher y-coordinate and B in the second square with a lower x-coordinate? Wait, but in the second square, the lowest x is 4, so to get a lower x than 4, there's nothing. The first square's x goes up to 2, so the maximum difference in x is 5 - 0 = 5. But if we take A at (0,0) and B at (5,3), the slope is 3/5, which is lower. So, that's worse.

Alternatively, if A is somewhere in the first square with a higher y and B is at (4,3). Let's take A at (0,2), the top-left corner of the first square. Then, the slope is (3 - 2)/(4 - 0) = 1/4, which is 0.25. That's worse. If A is at (2,2), the top-right corner, the slope is (3 - 2)/(4 - 2) = 1/2. Still worse than 3/2.

Alternatively, if A is somewhere else. Let's say A is at (x1, y1) in the first square, and B is at (x2, y2) in the second square. To maximize (y2 - y1)/(x2 - x1), we need to maximize y2 and minimize y1, and also minimize x2 - x1. So, the maximum y2 is 3, the minimum y1 is 0. The minimum x2 is 4, the maximum x1 is 2. So, the difference x2 - x1 is 4 - 2 = 2. So, the maximum slope would be (3 - 0)/(4 - 2) = 3/2. So, that seems to be the maximum.

But wait, let's confirm. Suppose we take A as (2,0) and B as (4,3). Then, the slope is 3/2. Is there any point in the first square where y1 is 0 and x1 is 2, and in the second square x2 is 4 and y2 is 3. So, those points are on the boundary of their respective squares.

But let's check if there's any other combination that could give a higher slope. For example, if A is at (2,0) and B is at (4,3). Suppose we take a point A slightly above (2,0), say (2, ε), and a point B slightly to the right of (4,3), say (4 + ε, 3 + ε). Then, the slope would be (3 + ε - ε)/(4 + ε - 2) = 3/(2 + ε). As ε approaches 0, the slope approaches 3/2. So, that's still 3/2. So, even if we take points near the corners, the maximum slope is 3/2.

Alternatively, if we take A at (2,0) and B at (4,3), which are both on the boundary, that gives the maximum slope of 3/2. If we take any other points, either the slope decreases or we can't get a higher slope.

Wait a second, but maybe there's another way. Suppose we take a point A in the first square with a higher y-coordinate and a point B in the second square with a lower x-coordinate. But in the first square, the maximum y is 2, and in the second square, the minimum x is 4. So, the difference in x is 4 - x1, where x1 is in [0,2]. To minimize the denominator, x2 - x1 should be as small as possible. The smallest possible x2 is 4, and the largest possible x1 is 2, so 4 - 2 = 2. If we take x1 smaller, x2 - x1 would be larger, which would make the slope smaller. So, to minimize the denominator, we need x2 as small as possible and x1 as large as possible. Therefore, 4 - 2 = 2 is the smallest possible denominator. So, the maximum slope is 3/2.

Therefore, the greatest possible slope is 3/2. But let me check if that's indeed the case.

Wait, but let's visualize the two squares. The first square is from (0,0) to (2,2), and the second is from (4,2) to (5,3). The line connecting (2,0) to (4,3) passes through (2,0) and (4,3), which are both corners. The slope is 3/2. If we take any other points, say A somewhere inside the first square and B somewhere inside the second square, the slope would be less than or equal to 3/2. Because the rise would be less than 3 (if B is not at (5,3)) and the run would be more than 2 (if A is not at (2,0) or B is not at (4,3)). Wait, but if A is somewhere else, maybe the rise is larger but the run is not. Let me check.

Suppose A is at (0,2), which is the top-left corner of the first square, and B is at (5,3), the top-right corner of the second square. Then, the slope is (3 - 2)/(5 - 0) = 1/5, which is 0.2. That's worse. If A is at (1,1), which is the center of the first square, and B is at (4,3), the slope is (3 - 1)/(4 - 1) = 2/3 ≈ 0.666. Still less than 3/2.

Alternatively, if A is at (2,2) and B is at (4,3), the slope is (3 - 2)/(4 - 2) = 1/2. Still less. If A is at (2,0) and B is at (5,3), the slope is (3 - 0)/(5 - 2) = 1. If A is at (2,0) and B is at (4,3), the slope is 3/2. So, that seems to be the maximum.

Wait, but what if we take A not at the corner, but somewhere else? For example, if A is at (2,0) and B is at (4,3), slope is 3/2. If A is at (2,0) and B is at (4.5,3), but wait, the second square only goes up to x=5. So, B can't be at (4.5,3), that's outside the second square. So, the maximum x for B is 5. But if we take B at (5,3), then the slope is (3 - 0)/(5 - 2) = 1. So, that's less than 3/2. Therefore, the maximum slope occurs when B is at (4,3) and A is at (2,0). So, 3/2 is the answer.

But let me check another approach. Let's parametrize the points. Let point A be (x1, y1) where 0 ≤ x1 ≤ 2 and 0 ≤ y1 ≤ 2. Point B is (x2, y2) where 4 ≤ x2 ≤ 5 and 2 ≤ y2 ≤ 3. We need to maximize (y2 - y1)/(x2 - x1). To maximize this, we need to maximize y2 and minimize y1, and also minimize x2 - x1. So, the maximum y2 is 3, the minimum y1 is 0, and the minimum x2 is 4, the maximum x1 is 2. Therefore, x2 - x1 is minimized when x2 is 4 and x1 is 2, giving 4 - 2 = 2. Therefore, the maximum slope is (3 - 0)/2 = 3/2. So, that's 1.5.

Therefore, the answer is 3/2. But let me confirm once again.

Wait, suppose we take A at (2,0) and B at (4,3). The line between them has a slope of (3 - 0)/(4 - 2) = 3/2. If we take A at (2,0) and B at any other point in the second square, say (4,2), the slope would be (2 - 0)/(4 - 2) = 1. If we take B at (4,3), that's the highest y in the second square, so that gives the maximum rise. Similarly, if we take A at (2,0), the lowest y in the first square, then the difference in y is maximized. So, combining these two extremes gives the maximum slope.

Alternatively, could there be a point A in the first square with a lower y than 0? No, because the first square starts at y=0. Similarly, could there be a point B with a higher y than 3? No, because the second square ends at y=3. So, those are the extremes.

Therefore, the greatest possible slope is 3/2.

**Final Answer**
The greatest possible value of the slope is \boxed{\dfrac{3}{2}}.
Therefore, the maximum slope of the line containing points $A$ and $B$ is $\boxed{\frac{3}{2}}$.
</think>