Explanation:
Okay, so I need to find the greatest possible value of the slope of the line connecting points A and B. Point A is somewhere within or on the square with opposite corners at (0,0) and (2,2). Point B is within or on the square with opposite corners at (4,2) and (5,3).

First, let me visualize the squares. The first square has corners at (0,0) and (2,2). Since those are opposite corners, the square is from (0,0) to (2,2), so it's a square with side length 2, aligned with the axes. Similarly, the second square has opposite corners at (4,2) and (5,3). Wait, those aren't aligned with the axes. Let me check. The distance between (4,2) and (5,3) is sqrt[(5-4)^2 + (3-2)^2] = sqrt(1 + 1) = sqrt(2). So that's the diagonal of the square. Therefore, the square has side length 1, since the diagonal of a square with side length s is s√2. So if the diagonal is sqrt(2), then the side length is 1.

But where exactly is this square located? The square with opposite corners at (4,2) and (5,3). Let me figure out the other two corners. Since the square is axis-aligned? Wait, no, because the diagonal is from (4,2) to (5,3). If the square is not axis-aligned, then the other two corners can be found by rotating the diagonal. Alternatively, maybe the square is axis-aligned? Wait, if the square is axis-aligned, then the other two corners would be (4,3) and (5,2). Let me check. If the square is axis-aligned, then moving from (4,2) to (5,3) would require moving right 1 and up 1, but that's the diagonal. So the square would have sides of length 1. So the square would have corners at (4,2), (5,2), (5,3), and (4,3). Wait, but then the opposite corners would be (4,2) and (5,3), which is the diagonal. So yes, that square is axis-aligned. So the square for point B is from (4,2) to (5,3), so it's a unit square with sides parallel to the axes.

So point A is in the square from (0,0) to (2,2), and point B is in the square from (4,2) to (5,3).

Now, the question is to find the maximum possible slope of the line AB.

Slope is (y2 - y1)/(x2 - x1). To maximize the slope, we need to maximize the difference in y-coordinates while minimizing the difference in x-coordinates. However, since slope can be positive or negative, but we want the maximum slope, which would be a positive slope. So we need to maximize (yB - yA)/(xB - xA).

But since slope can be positive or negative, but maximum slope would be the largest positive value. So to maximize the slope, we need to maximize the numerator (yB - yA) and minimize the denominator (xB - xA). However, if the denominator is negative, then the slope would be negative. So to get the maximum slope, we need to have the numerator as large as possible and the denominator as small as possible (but positive).

Alternatively, if we can have a large positive numerator and a small positive denominator, that would give a large positive slope. Alternatively, if we have a large positive numerator and a negative denominator, that would give a negative slope. So to get the maximum slope, we need to maximize (yB - yA)/(xB - xA) where both numerator and denominator are positive.

Alternatively, maybe there's a case where the denominator is negative but the numerator is negative, leading to a positive slope? Wait, if both numerator and denominator are negative, then the slope is positive. So maybe we can have a large positive slope by having both yB - yA and xB - xA negative.

But to get the maximum slope, we need to maximize the absolute value of the slope, but considering the sign. So perhaps the maximum slope occurs when we have the maximum possible (yB - yA) divided by the minimum possible (xB - xA). But depending on the direction.

Alternatively, maybe the maximum slope occurs when we have the maximum difference in y over the minimum difference in x.

But let me think. Let me consider coordinates.

Point A is in the square from (0,0) to (2,2). So coordinates of A can be (x1, y1) where 0 ≤ x1 ≤ 2 and 0 ≤ y1 ≤ 2.

Point B is in the square from (4,2) to (5,3). So coordinates of B can be (x2, y2) where 4 ≤ x2 ≤ 5 and 2 ≤ y2 ≤ 3.

So the slope between A and B is (y2 - y1)/(x2 - x1).

To maximize this slope, we need to maximize the numerator and minimize the denominator. However, since x2 is between 4 and 5, and x1 is between 0 and 2, the difference x2 - x1 is between (4 - 2) = 2 and (5 - 0) = 5. Wait, but if we want to minimize the denominator, we need to minimize x2 - x1. Since x2 is at least 4 and x1 is at most 2, the minimum value of x2 - x1 is 4 - 2 = 2. Wait, but if x2 is as small as possible (4) and x1 is as large as possible (2), then x2 - x1 is 2. If x2 is larger, then x2 - x1 would be larger. Similarly, if x1 is smaller, then x2 - x1 would be larger. So the minimum value of x2 - x1 is 2.

Similarly, the maximum value of y2 - y1 is when y2 is as large as possible (3) and y1 is as small as possible (0). So that would be 3 - 0 = 3.

Therefore, if we take A at (2,0) and B at (4,3), then the slope would be (3 - 0)/(4 - 2) = 3/2 = 1.5.

But is this the maximum? Wait, maybe there's a way to get a larger slope.

Wait, suppose we take A at (0,0) and B at (4,3). Then the slope is (3 - 0)/(4 - 0) = 3/4 = 0.75, which is less than 3/2.

Alternatively, if we take A at (2,0) and B at (5,3). Then slope is (3 - 0)/(5 - 2) = 3/3 = 1. Which is less than 3/2.

Alternatively, if we take A at (2, something) and B at (4, something). Let me think.

But maybe there's another way. Suppose we take A at (x1, y1) and B at (x2, y2). To maximize (y2 - y1)/(x2 - x1).

Alternatively, maybe if we take A at (x1, y1) where y1 is as small as possible and B at (x2, y2) where y2 is as large as possible, but also x2 - x1 is as small as possible.

But if we take A at (2, 0) and B at (4, 3), then x2 - x1 is 2, and y2 - y1 is 3. So slope is 3/2.

Alternatively, if we take A at (something else) and B at (something else). For example, if we take A at (x1, y1) where x1 is as large as possible (so that x2 - x1 is minimized) and y1 as small as possible.

But maybe there's a case where the denominator is negative? Wait, if x2 - x1 is negative, then the slope would be negative if y2 - y1 is positive. But if we want the maximum slope, which is positive, then we need to have both numerator and denominator positive or both negative.

If both numerator and denominator are negative, then the slope is positive. So maybe if we take A at (x1, y1) where x1 is as large as possible (so x2 - x1 is negative if x2 is smaller than x1? Wait, but x2 is between 4 and 5, and x1 is between 0 and 2. So x2 is always larger than x1. Therefore, x2 - x1 is always positive. Therefore, the denominator is always positive. Therefore, to maximize the slope, we need to maximize the numerator (y2 - y1) and minimize the denominator (x2 - x1).

Therefore, the maximum slope is achieved when y2 - y1 is maximum and x2 - x1 is minimum.

So maximum y2 - y1 is 3 - 0 = 3.

Minimum x2 - x1 is 4 - 2 = 2.

Therefore, slope is 3/2.

But wait, is there a way to get a larger slope? Let me think.

Suppose instead of taking A at (2, 0), maybe taking A at some other point where y1 is smaller? But y1 can't be smaller than 0. So 0 is the minimum. Similarly, y2 can't be larger than 3. So 3 is the maximum.

But maybe if we take A not at (2, 0) but somewhere else? For example, if we take A at (x1, y1) where x1 is less than 2, but then x2 - x1 would be larger, so denominator would be larger, which would decrease the slope.

Alternatively, if we take A at (x1, y1) where x1 is 2 and y1 is 0, then x2 - x1 is minimized (since x2 is minimized at 4, so 4 - 2 = 2).

Alternatively, if we take B at (4, 3) and A at (2, 0), slope is 3/2.

But maybe there's another combination where the slope is higher?

Wait, suppose we take A at (something) and B at (something else). For example, if we take A at (0, 0) and B at (4, 3), slope is 3/4. If we take A at (0, 0) and B at (5, 3), slope is 3/5. If we take A at (2, 0) and B at (5, 3), slope is 3/3 = 1.

Alternatively, if we take A at (2, 0) and B at (4, 3), slope is 3/2.

Alternatively, if we take A at (something else). Let me think. Suppose A is at (x1, y1) and B is at (x2, y2). Let me think of the slope as (y2 - y1)/(x2 - x1).

To maximize this, we can think of it as maximizing (y2 - y1) while minimizing (x2 - x1).

But maybe there's a case where even though (y2 - y1) is not maximum, but (x2 - x1) is smaller? Wait, but (x2 - x1) can't be smaller than 2, since x2 is at least 4 and x1 is at most 2. So 4 - 2 = 2 is the minimum.

Therefore, if we can achieve the maximum (y2 - y1) with that minimum (x2 - x1), then that would be the maximum slope.

But is there a case where (y2 - y1) is larger? For example, if A is at (something) where y1 is negative? But no, because A is within the square from (0,0) to (2,2), so y1 can't be less than 0. Similarly, y2 can't be more than 3.

Therefore, the maximum difference in y is 3 - 0 = 3.

Therefore, the maximum slope is 3/2.

But wait, let me check if there's another way. Suppose we take A at (x1, y1) and B at (x2, y2) where the line AB has a steeper slope. For example, if we take A at (0, 0) and B at (4, 3), slope is 3/4. If we take A at (0, 0) and B at (5, 3), slope is 3/5. If we take A at (2, 0) and B at (4, 3), slope is 3/2.

Alternatively, if we take A at (2, 0) and B at (4, 3), slope is 3/2.

But maybe if we take A at (something else) where y1 is not 0? For example, suppose A is at (2, y1) and B is at (4, 3). Then the slope is (3 - y1)/(4 - 2) = (3 - y1)/2. To maximize this, we need to minimize y1. The minimum y1 is 0, so slope is 3/2.

Similarly, if we take A at (x1, 0) and B at (4, 3), then slope is (3 - 0)/(4 - x1) = 3/(4 - x1). To maximize this, we need to minimize (4 - x1), which is equivalent to maximizing x1. The maximum x1 is 2, so slope is 3/(4 - 2) = 3/2.

So same result.

Alternatively, if we take A at (x1, y1) and B at (x2, y2) where x2 is 4 and y2 is 3, then slope is (3 - y1)/(4 - x1). To maximize this, we need to maximize (3 - y1) and minimize (4 - x1). Which is again achieved when y1 is minimized (0) and x1 is maximized (2).

Alternatively, if we take B at (5, 3) and A at (something). Then slope is (3 - y1)/(5 - x1). To maximize this, we need to maximize (3 - y1) and minimize (5 - x1). So again, maximize (3 - y1) by taking y1 = 0, and minimize (5 - x1) by taking x1 as large as possible, which is 2. Then slope is 3/(5 - 2) = 1. Which is less than 3/2.

Alternatively, if we take B at (4, 3) and A at (something else).

Alternatively, if we take B at (4, 2) and A at (something). Then slope is (2 - y1)/(4 - x1). To maximize this, we need to maximize (2 - y1) and minimize (4 - x1). So maximum (2 - y1) is when y1 is minimized (0), so 2 - 0 = 2. Minimum (4 - x1) is when x1 is maximized (2), so 4 - 2 = 2. Then slope is 2/2 = 1. Which is less than 3/2.

Alternatively, if we take B at (5, 2) and A at (something). Then slope is (2 - y1)/(5 - x1). To maximize, maximize (2 - y1) and minimize (5 - x1). So again, 2 - 0 = 2, and 5 - 2 = 3, so slope is 2/3.

So that's worse.

Alternatively, if we take B at (4, 3) and A at (something else).

Wait, maybe if we take A not on the edge? For example, suppose A is somewhere inside the square. But since we are trying to maximize the slope, we need to maximize (y2 - y1) and minimize (x2 - x1). Since both y2 and x2 are at their maximums and x1 and y1 are at their minimums, then the maximum slope is achieved when A is at (2, 0) and B is at (4, 3).

But let me check another possibility. Suppose we take A at (x1, y1) where x1 is less than 2, but y1 is also less than 0? No, because A is within the square, so y1 can't be less than 0.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is something else. For example, if A is at (2, 0), slope is 3/2. If A is at (2, 1), then slope is (3 - 1)/2 = 1. Which is less.

Alternatively, if we take A at (something else). Wait, maybe if we take B not at (4, 3) but somewhere else? For example, if B is at (5, 3), but then the denominator is larger.

Alternatively, if we take B at (4, 3) and A at (2, 0), slope is 3/2.

But let me think if there's another combination where the slope is higher. For example, suppose we take A at (0, 0) and B at (4, 3). Slope is 3/4. If we take A at (0, 0) and B at (5, 3), slope is 3/5. If we take A at (0, 0) and B at (4, 2), slope is 2/4 = 1/2.

Alternatively, if we take A at (something) where y1 is negative? No, since A is in the square from (0,0) to (2,2).

Wait, but what if we take A at (something) where x1 is less than 2 and y1 is less than 0? But no, since the square is from (0,0) to (2,2), so coordinates can't be less than 0.

Alternatively, maybe if we take A at (something) where x1 is 2 and y1 is 0, and B at (4, 3).

Alternatively, maybe there's a case where the line AB passes through other points? Wait, but the maximum slope is achieved when the line is as steep as possible.

Wait, another thought: maybe if we take A at (x1, y1) and B at (x2, y2) such that the line AB is tangent to some edge of the squares? Or maybe not.

Alternatively, think of the problem as optimizing the slope. Let me consider variables.

Let me denote A as (x1, y1) with 0 ≤ x1 ≤ 2, 0 ≤ y1 ≤ 2.

B as (x2, y2) with 4 ≤ x2 ≤ 5, 2 ≤ y2 ≤ 3.

We need to maximize (y2 - y1)/(x2 - x1).

Let me consider this as a function f(x1, y1, x2, y2) = (y2 - y1)/(x2 - x1).

To maximize this, we can think of it as maximizing the numerator while minimizing the denominator.

But since x2 - x1 is always positive, as discussed earlier, we can think of it as maximizing (y2 - y1) and minimizing (x2 - x1).

But maybe there's a case where even though (y2 - y1) is not maximum, but (x2 - x1) is smaller? But since (x2 - x1) can't be smaller than 2, as x2 ≥ 4 and x1 ≤ 2, then 4 - 2 = 2 is the minimum.

Therefore, the maximum slope is achieved when (y2 - y1) is maximum and (x2 - x1) is minimum.

Therefore, maximum (y2 - y1) is 3 - 0 = 3, and minimum (x2 - x1) is 4 - 2 = 2. Therefore, slope is 3/2.

But let me check if there's another combination where (y2 - y1)/(x2 - x1) is larger.

Suppose we take A at (x1, y1) where y1 is not 0, but x1 is such that x2 - x1 is smaller? Wait, but x2 - x1 can't be smaller than 2.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is something, and B at (x2, y2) where x2 is 4 and y2 is something.

If we take A at (2, y1) and B at (4, y2), then slope is (y2 - y1)/2. To maximize this, we need to maximize (y2 - y1). Since y2 can be up to 3 and y1 can be as low as 0, so maximum is 3 - 0 = 3, so slope is 3/2.

Alternatively, if we take A at (x1, y1) and B at (x2, y2) where x2 is 4 and y2 is 3, and x1 is 2 and y1 is 0.

Alternatively, if we take A at (x1, y1) where x1 is less than 2, but then x2 - x1 would be larger.

Alternatively, if we take B at (x2, y2) where x2 is 4 and y2 is 3, and A at (x1, y1) where x1 is 2 and y1 is 0.

So slope is 3/2.

But let me think again. Suppose we take A at (something) where y1 is not 0, but maybe the slope is higher? For example, if A is at (2, 0) and B is at (4, 3), slope is 3/2. If A is at (2, 0.5) and B is at (4, 3), slope is (3 - 0.5)/2 = 2.5/2 = 1.25, which is less than 1.5.

Alternatively, if A is at (2, 0) and B is at (4, 3), slope is 3/2.

Alternatively, if we take A at (something else). Suppose A is at (x1, y1) where x1 is 2 and y1 is 0, and B is at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, if we take A at (x1, y1) where x1 is 2 and y1 is 0, and B is at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, if we take A at (x1, y1) where x1 is 2 and y1 is 0, and B is at (x2, y2) where x2 is 4 and y2 is 3.

So that's the same as before.

Wait, but what if we take A at (x1, y1) where x1 is less than 2, but y1 is also less than 0? But no, since A is in the square from (0,0) to (2,2).

Alternatively, maybe there's a case where the line AB goes from a point on the edge of the first square to a point on the edge of the second square, but not necessarily the corners.

Wait, for example, suppose A is at (2, 0) and B is at (4, 3). That's a corner to corner.

Alternatively, suppose A is at (2, 0) and B is at (4, 3).

Alternatively, suppose A is at (2, 0) and B is at (4, 3).

But maybe if we take A at (something else) where the line AB is steeper?

Wait, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Wait, I keep getting the same point.

Alternatively, maybe there's a case where the slope is higher if we take A not at (2, 0) but somewhere else?

Wait, suppose we take A at (x1, y1) where x1 is 2 and y1 is something, and B at (x2, y2) where x2 is 4 and y2 is something.

If we take A at (2, y1) and B at (4, y2), then slope is (y2 - y1)/2. To maximize this, we need to maximize y2 - y1. Since y2 can be up to 3 and y1 can be as low as 0, so maximum is 3 - 0 = 3, so slope is 3/2.

Alternatively, if we take A at (x1, y1) where x1 is less than 2, but then x2 - x1 is larger.

Alternatively, if we take B not at (4, 3) but somewhere else. For example, if B is at (4, 3) and A is at (2, 0), slope is 3/2. If B is at (5, 3) and A is at (something), slope is (3 - y1)/(5 - x1). To maximize this, we need to maximize (3 - y1) and minimize (5 - x1). So maximum (3 - y1) is 3 - 0 = 3, and minimum (5 - x1) is 5 - 2 = 3, so slope is 3/3 = 1. Which is less than 3/2.

Alternatively, if B is at (4, 3) and A is at (something else).

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (4, 3).

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (4, 3).

Wait, maybe there's another way. Let me think of parametric equations.

Suppose we want to maximize (y2 - y1)/(x2 - x1). Let me think of this as a linear function.

But since A and B are constrained within their squares, maybe we can use calculus or some optimization technique.

But since it's a problem with constraints, maybe we can consider the maximum occurs at the boundary.

In optimization problems, the extrema often occur at the boundaries. So, for the maximum slope, it's likely achieved when A and B are at the corners of their respective squares.

So let's check all possible combinations of corners.

First square (A's square) has corners at (0,0), (0,2), (2,0), (2,2).

Second square (B's square) has corners at (4,2), (4,3), (5,2), (5,3).

So there are 4 corners for A and 4 corners for B, so 16 combinations. Let me check the slopes for all these combinations.

1. A=(0,0), B=(4,2): slope = (2 - 0)/(4 - 0) = 2/4 = 1/2
2. A=(0,0), B=(4,3): slope = (3 - 0)/4 = 3/4
3. A=(0,0), B=(5,2): slope = (2 - 0)/5 = 2/5
4. A=(0,0), B=(5,3): slope = 3/5
5. A=(0,2), B=(4,2): slope = (2 - 2)/4 = 0
6. A=(0,2), B=(4,3): slope = (3 - 2)/4 = 1/4
7. A=(0,2), B=(5,2): slope = 0
8. A=(0,2), B=(5,3): slope = (3 - 2)/5 = 1/5
9. A=(2,0), B=(4,2): slope = (2 - 0)/2 = 1
10. A=(2,0), B=(4,3): slope = (3 - 0)/2 = 3/2
11. A=(2,0), B=(5,2): slope = (2 - 0)/3 = 2/3
12. A=(2,0), B=(5,3): slope = (3 - 0)/3 = 1
13. A=(2,2), B=(4,2): slope = (2 - 2)/2 = 0
14. A=(2,2), B=(4,3): slope = (3 - 2)/2 = 1/2
15. A=(2,2), B=(5,2): slope = 0
16. A=(2,2), B=(5,3): slope = (3 - 2)/3 = 1/3

So looking at all these slopes, the maximum is 3/2, which occurs at case 10: A=(2,0) and B=(4,3).

Therefore, the maximum slope is 3/2.

But wait, let me check if there are any other points on the edges that could give a higher slope. For example, suppose A is not at a corner but somewhere on the edge.

For example, suppose A is on the right edge of the first square, which is x=2, y from 0 to 2. Let me parametrize A as (2, y1) where 0 ≤ y1 ≤ 2. Then B is on the left edge of the second square, which is x=4, y from 2 to 3. Let me parametrize B as (4, y2) where 2 ≤ y2 ≤ 3. Then the slope is (y2 - y1)/(4 - 2) = (y2 - y1)/2. To maximize this, we need to maximize y2 - y1. Since y2 can be up to 3 and y1 can be as low as 0, the maximum is 3 - 0 = 3, so slope is 3/2. So same as before.

Alternatively, if A is on the bottom edge of the first square, x from 0 to 2, y=0. Let me parametrize A as (x1, 0) where 0 ≤ x1 ≤ 2. Then B is on the top edge of the second square, x from 4 to 5, y=3. Let me parametrize B as (x2, 3) where 4 ≤ x2 ≤ 5. Then slope is (3 - 0)/(x2 - x1) = 3/(x2 - x1). To maximize this, we need to minimize x2 - x1. The minimum x2 - x1 is 4 - 2 = 2, so slope is 3/2. Again same result.

Alternatively, if A is on the left edge of the first square, x=0, y from 0 to 2. Then B is on the top edge of the second square, x from 4 to 5, y=3. Then slope is (3 - y1)/(x2 - 0) = (3 - y1)/x2. To maximize this, we need to maximize (3 - y1) and minimize x2. So maximum (3 - y1) is 3 - 0 = 3, and minimum x2 is 4. So slope is 3/4. Which is less than 3/2.

Alternatively, if A is on the top edge of the first square, y=2, x from 0 to 2. Then B is on the top edge of the second square, y=3, x from 4 to 5. Then slope is (3 - 2)/(x2 - x1) = 1/(x2 - x1). To maximize this, minimize x2 - x1. Minimum is 4 - 2 = 2, so slope is 1/2.

Alternatively, if A is on the right edge of the first square, x=2, y from 0 to 2, and B is on the right edge of the second square, x=5, y from 2 to 3. Then slope is (y2 - y1)/(5 - 2) = (y2 - y1)/3. To maximize, set y2=3, y1=0, slope is 3/3 = 1.

So again, 3/2 is the maximum.

Alternatively, suppose A is somewhere inside the first square and B is somewhere inside the second square. But since we already considered the maximum on the edges, and the maximum slope occurs at the edges, then the maximum slope is 3/2.

But let me think again. Suppose we take A at (2, 0) and B at (4, 3). Is there a way to get a steeper slope? For example, if we take A at (2, 0) and B at (4, 3), slope is 3/2. If we take A at (2, 0) and B at (something else). For example, if B is at (4, 3), which is the top right corner of the second square.

Alternatively, if we take A at (something else) and B at (something else). For example, suppose A is at (x1, y1) and B is at (x2, y2) where the line AB is steeper.

Wait, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Wait, I think I'm going in circles here.

Alternatively, think of the problem as follows: the maximum slope between two points A and B is the maximum of (y2 - y1)/(x2 - x1). Since x2 ≥ 4, x1 ≤ 2, so x2 - x1 ≥ 2. The maximum value of y2 - y1 is 3 - 0 = 3. Therefore, the maximum slope is 3/2.

Alternatively, if there is a case where y2 - y1 is larger? For example, if y2 can be more than 3? But no, since B is within the square from (4,2) to (5,3), so y2 can't exceed 3. Similarly, y1 can't be less than 0.

Therefore, 3/2 is the maximum.

But let me check another possibility. Suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Wait, I think I need to accept that 3/2 is the maximum.

But let me check if there's a case where the slope is higher. For example, suppose we take A at (something) where the line AB goes from the bottom of the first square to the top of the second square, but not necessarily the corners.

Wait, suppose A is at (2, 0) and B is at (4, 3). That's the same as before.

Alternatively, suppose A is at (something) where x1 is less than 2, but y1 is also less than 0? No, since A is in the square.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

I think I need to conclude that the maximum slope is 3/2.

But let me check with coordinates.

Let me take A=(2,0) and B=(4,3). The slope is (3 - 0)/(4 - 2) = 3/2.

Alternatively, if I take A=(2, 0) and B=(4, 3), then the line is from (2,0) to (4,3).

Alternatively, if I take A=(2, 0) and B=(4, 3), then the line is going from (2,0) to (4,3).

Alternatively, if I take A=(something else). For example, suppose A is at (1, 0) and B is at (4, 3). Then slope is (3 - 0)/(4 - 1) = 3/3 = 1. Which is less than 3/2.

Alternatively, if A is at (2, 0) and B is at (4, 3).

Alternatively, if A is at (2, 0) and B is at (4, 3).

Alternatively, if I take A at (2, 0) and B at (4, 3).

Yes, so that's the maximum.

Therefore, the greatest possible value of the slope is 3/2.

But wait, the problem says "within or on the square". So points on the edges are allowed. So the answer is 3/2.

But let me check once more.

Suppose there is a point A inside the first square and a point B inside the second square such that the slope is higher. For example, suppose A is at (2, 0) and B is at (4, 3). If I move A slightly to the left, say (2 - ε, 0), then x2 - x1 becomes 4 - (2 - ε) = 2 + ε, which increases the denominator, so slope decreases. If I move A slightly up, say (2, 0 + ε), then y2 - y1 becomes 3 - ε, so slope becomes (3 - ε)/2, which is less than 3/2. Similarly, if I move B slightly to the left, say (4 - ε, 3), then x2 - x1 is 4 - ε - 2 = 2 - ε, so denominator decreases, but since x2 can't be less than 4, ε can't be positive. Wait, x2 is at least 4, so moving B to the left would decrease x2, but since x2 can't be less than 4, the minimum x2 is 4. Therefore, moving B to the left is not possible. Similarly, moving B up is not possible since y2 can't exceed 3.

Therefore, the maximum slope is indeed achieved at A=(2,0) and B=(4,3), giving slope 3/2.

Therefore, the answer is 3/2, which is 3 over 2.

But wait, the problem says "Express your answer as a common fraction." So 3/2 is already a common fraction.

But let me check once more if there's a case where slope is higher.

Suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

I think I need to stop here.

**Final Answer**
\boxed{\dfrac{3}{2}}

To determine the **greatest possible value of the slope** of the line containing points $ A $ and $ B $, we begin by analyzing the constraints on the positions of these points.

---

### **Step 1: Understand the Squares**

- **Point A** lies within or on the square with opposite corners at $ (0,0) $ and $ (2,2) $.
  This is a square of side length 2, aligned with the axes, with all coordinates satisfying:
  $$
  0 \leq x_1 \leq 2, \quad 0 \leq y_1 \leq 2
  $$

- **Point B** lies within or on the square with opposite corners at $ (4,2) $ and $ (5,3) $.
  This is a square of side length 1, also aligned with the axes, with all coordinates satisfying:
  $$
  4 \leq x_2 \leq 5, \quad 2 \leq y_2 \leq 3
  $$

---

### **Step 2: Slope Formula**

The slope of the line through points $ A = (x_1, y_1) $ and $ B = (x_2, y_2) $ is given by:
$$
\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}
$$

To **maximize** this slope, we want to:
- **Maximize** the numerator $ y_2 - y_1 $
- **Minimize** the denominator $ x_2 - x_1 $

---

### **Step 3: Maximize the Numerator**

The maximum possible value of $ y_2 - y_1 $ occurs when:
- $ y_2 $ is as large as possible: $ y_2 = 3 $
- $ y_1 $ is as small as possible: $ y_1 = 0 $

So the maximum numerator is:
$$
y_2 - y_1 = 3 - 0 = 3
$$

---

### **Step 4: Minimize the Denominator**

The minimum possible value of $ x_2 - x_1 $ occurs when:
- $ x_2 $ is as small as possible: $ x_2 = 4 $
- $ x_1 $ is as large as possible: $ x_1 = 2 $

So the minimum denominator is:
$$
x_2 - x_1 = 4 - 2 = 2
$$

---

### **Step 5: Compute the Maximum Slope**

Using the maximum numerator and minimum denominator:
$$
\text{Maximum slope} = \frac{3}{2}
$$

This occurs when:
- $ A = (2, 0) $
- $ B = (4, 3) $

---

### **Step 6: Verify No Greater Slope is Possible**

We can confirm that:
- $ y_2 $ cannot exceed 3
- $ y_1 $ cannot be less than 0
- $ x_2 $ cannot be less than 4
- $ x_1 $ cannot be greater than 2

Thus, the maximum value of the slope is indeed:
$$
\boxed{\dfrac{3}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to find the greatest possible value of the slope of the line connecting points A and B. Point A is somewhere within or on the square with opposite corners at (0,0) and (2,2). Point B is within or on the square with opposite corners at (4,2) and (5,3).

First, let me visualize the squares. The first square has corners at (0,0) and (2,2). Since those are opposite corners, the square is from (0,0) to (2,2), so it's a square with side length 2, aligned with the axes. Similarly, the second square has opposite corners at (4,2) and (5,3). Wait, those aren't aligned with the axes. Let me check. The distance between (4,2) and (5,3) is sqrt[(5-4)^2 + (3-2)^2] = sqrt(1 + 1) = sqrt(2). So that's the diagonal of the square. Therefore, the square has side length 1, since the diagonal of a square with side length s is s√2. So if the diagonal is sqrt(2), then the side length is 1.

But where exactly is this square located? The square with opposite corners at (4,2) and (5,3). Let me figure out the other two corners. Since the square is axis-aligned? Wait, no, because the diagonal is from (4,2) to (5,3). If the square is not axis-aligned, then the other two corners can be found by rotating the diagonal. Alternatively, maybe the square is axis-aligned? Wait, if the square is axis-aligned, then the other two corners would be (4,3) and (5,2). Let me check. If the square is axis-aligned, then moving from (4,2) to (5,3) would require moving right 1 and up 1, but that's the diagonal. So the square would have sides of length 1. So the square would have corners at (4,2), (5,2), (5,3), and (4,3). Wait, but then the opposite corners would be (4,2) and (5,3), which is the diagonal. So yes, that square is axis-aligned. So the square for point B is from (4,2) to (5,3), so it's a unit square with sides parallel to the axes.

So point A is in the square from (0,0) to (2,2), and point B is in the square from (4,2) to (5,3).

Now, the question is to find the maximum possible slope of the line AB.

Slope is (y2 - y1)/(x2 - x1). To maximize the slope, we need to maximize the difference in y-coordinates while minimizing the difference in x-coordinates. However, since slope can be positive or negative, but we want the maximum slope, which would be a positive slope. So we need to maximize (yB - yA)/(xB - xA).

But since slope can be positive or negative, but maximum slope would be the largest positive value. So to maximize the slope, we need to maximize the numerator (yB - yA) and minimize the denominator (xB - xA). However, if the denominator is negative, then the slope would be negative. So to get the maximum slope, we need to have the numerator as large as possible and the denominator as small as possible (but positive).

Alternatively, if we can have a large positive numerator and a small positive denominator, that would give a large positive slope. Alternatively, if we have a large positive numerator and a negative denominator, that would give a negative slope. So to get the maximum slope, we need to maximize (yB - yA)/(xB - xA) where both numerator and denominator are positive.

Alternatively, maybe there's a case where the denominator is negative but the numerator is negative, leading to a positive slope? Wait, if both numerator and denominator are negative, then the slope is positive. So maybe we can have a large positive slope by having both yB - yA and xB - xA negative.

But to get the maximum slope, we need to maximize the absolute value of the slope, but considering the sign. So perhaps the maximum slope occurs when we have the maximum possible (yB - yA) divided by the minimum possible (xB - xA). But depending on the direction.

Alternatively, maybe the maximum slope occurs when we have the maximum difference in y over the minimum difference in x.

But let me think. Let me consider coordinates.

Point A is in the square from (0,0) to (2,2). So coordinates of A can be (x1, y1) where 0 ≤ x1 ≤ 2 and 0 ≤ y1 ≤ 2.

Point B is in the square from (4,2) to (5,3). So coordinates of B can be (x2, y2) where 4 ≤ x2 ≤ 5 and 2 ≤ y2 ≤ 3.

So the slope between A and B is (y2 - y1)/(x2 - x1).

To maximize this slope, we need to maximize the numerator and minimize the denominator. However, since x2 is between 4 and 5, and x1 is between 0 and 2, the difference x2 - x1 is between (4 - 2) = 2 and (5 - 0) = 5. Wait, but if we want to minimize the denominator, we need to minimize x2 - x1. Since x2 is at least 4 and x1 is at most 2, the minimum value of x2 - x1 is 4 - 2 = 2. Wait, but if x2 is as small as possible (4) and x1 is as large as possible (2), then x2 - x1 is 2. If x2 is larger, then x2 - x1 would be larger. Similarly, if x1 is smaller, then x2 - x1 would be larger. So the minimum value of x2 - x1 is 2.

Similarly, the maximum value of y2 - y1 is when y2 is as large as possible (3) and y1 is as small as possible (0). So that would be 3 - 0 = 3.

Therefore, if we take A at (2,0) and B at (4,3), then the slope would be (3 - 0)/(4 - 2) = 3/2 = 1.5.

But is this the maximum? Wait, maybe there's a way to get a larger slope.

Wait, suppose we take A at (0,0) and B at (4,3). Then the slope is (3 - 0)/(4 - 0) = 3/4 = 0.75, which is less than 3/2.

Alternatively, if we take A at (2,0) and B at (5,3). Then slope is (3 - 0)/(5 - 2) = 3/3 = 1. Which is less than 3/2.

Alternatively, if we take A at (2, something) and B at (4, something). Let me think.

But maybe there's another way. Suppose we take A at (x1, y1) and B at (x2, y2). To maximize (y2 - y1)/(x2 - x1).

Alternatively, maybe if we take A at (x1, y1) where y1 is as small as possible and B at (x2, y2) where y2 is as large as possible, but also x2 - x1 is as small as possible.

But if we take A at (2, 0) and B at (4, 3), then x2 - x1 is 2, and y2 - y1 is 3. So slope is 3/2.

Alternatively, if we take A at (something else) and B at (something else). For example, if we take A at (x1, y1) where x1 is as large as possible (so that x2 - x1 is minimized) and y1 as small as possible.

But maybe there's a case where the denominator is negative? Wait, if x2 - x1 is negative, then the slope would be negative if y2 - y1 is positive. But if we want the maximum slope, which is positive, then we need to have both numerator and denominator positive or both negative.

If both numerator and denominator are negative, then the slope is positive. So maybe if we take A at (x1, y1) where x1 is as large as possible (so x2 - x1 is negative if x2 is smaller than x1? Wait, but x2 is between 4 and 5, and x1 is between 0 and 2. So x2 is always larger than x1. Therefore, x2 - x1 is always positive. Therefore, the denominator is always positive. Therefore, to maximize the slope, we need to maximize the numerator (y2 - y1) and minimize the denominator (x2 - x1).

Therefore, the maximum slope is achieved when y2 - y1 is maximum and x2 - x1 is minimum.

So maximum y2 - y1 is 3 - 0 = 3.

Minimum x2 - x1 is 4 - 2 = 2.

Therefore, slope is 3/2.

But wait, is there a way to get a larger slope? Let me think.

Suppose instead of taking A at (2, 0), maybe taking A at some other point where y1 is smaller? But y1 can't be smaller than 0. So 0 is the minimum. Similarly, y2 can't be larger than 3. So 3 is the maximum.

But maybe if we take A not at (2, 0) but somewhere else? For example, if we take A at (x1, y1) where x1 is less than 2, but then x2 - x1 would be larger, so denominator would be larger, which would decrease the slope.

Alternatively, if we take A at (x1, y1) where x1 is 2 and y1 is 0, then x2 - x1 is minimized (since x2 is minimized at 4, so 4 - 2 = 2).

Alternatively, if we take B at (4, 3) and A at (2, 0), slope is 3/2.

But maybe there's another combination where the slope is higher?

Wait, suppose we take A at (something) and B at (something else). For example, if we take A at (0, 0) and B at (4, 3), slope is 3/4. If we take A at (0, 0) and B at (5, 3), slope is 3/5. If we take A at (2, 0) and B at (5, 3), slope is 3/3 = 1.

Alternatively, if we take A at (2, 0) and B at (4, 3), slope is 3/2.

Alternatively, if we take A at (something else). Let me think. Suppose A is at (x1, y1) and B is at (x2, y2). Let me think of the slope as (y2 - y1)/(x2 - x1).

To maximize this, we can think of it as maximizing (y2 - y1) while minimizing (x2 - x1).

But maybe there's a case where even though (y2 - y1) is not maximum, but (x2 - x1) is smaller? Wait, but (x2 - x1) can't be smaller than 2, since x2 is at least 4 and x1 is at most 2. So 4 - 2 = 2 is the minimum.

Therefore, if we can achieve the maximum (y2 - y1) with that minimum (x2 - x1), then that would be the maximum slope.

But is there a case where (y2 - y1) is larger? For example, if A is at (something) where y1 is negative? But no, because A is within the square from (0,0) to (2,2), so y1 can't be less than 0. Similarly, y2 can't be more than 3.

Therefore, the maximum difference in y is 3 - 0 = 3.

Therefore, the maximum slope is 3/2.

But wait, let me check if there's another way. Suppose we take A at (x1, y1) and B at (x2, y2) where the line AB has a steeper slope. For example, if we take A at (0, 0) and B at (4, 3), slope is 3/4. If we take A at (0, 0) and B at (5, 3), slope is 3/5. If we take A at (2, 0) and B at (4, 3), slope is 3/2.

Alternatively, if we take A at (2, 0) and B at (4, 3), slope is 3/2.

But maybe if we take A at (something else) where y1 is not 0? For example, suppose A is at (2, y1) and B is at (4, 3). Then the slope is (3 - y1)/(4 - 2) = (3 - y1)/2. To maximize this, we need to minimize y1. The minimum y1 is 0, so slope is 3/2.

Similarly, if we take A at (x1, 0) and B at (4, 3), then slope is (3 - 0)/(4 - x1) = 3/(4 - x1). To maximize this, we need to minimize (4 - x1), which is equivalent to maximizing x1. The maximum x1 is 2, so slope is 3/(4 - 2) = 3/2.

So same result.

Alternatively, if we take A at (x1, y1) and B at (x2, y2) where x2 is 4 and y2 is 3, then slope is (3 - y1)/(4 - x1). To maximize this, we need to maximize (3 - y1) and minimize (4 - x1). Which is again achieved when y1 is minimized (0) and x1 is maximized (2).

Alternatively, if we take B at (5, 3) and A at (something). Then slope is (3 - y1)/(5 - x1). To maximize this, we need to maximize (3 - y1) and minimize (5 - x1). So again, maximize (3 - y1) by taking y1 = 0, and minimize (5 - x1) by taking x1 as large as possible, which is 2. Then slope is 3/(5 - 2) = 1. Which is less than 3/2.

Alternatively, if we take B at (4, 3) and A at (something else).

Alternatively, if we take B at (4, 2) and A at (something). Then slope is (2 - y1)/(4 - x1). To maximize this, we need to maximize (2 - y1) and minimize (4 - x1). So maximum (2 - y1) is when y1 is minimized (0), so 2 - 0 = 2. Minimum (4 - x1) is when x1 is maximized (2), so 4 - 2 = 2. Then slope is 2/2 = 1. Which is less than 3/2.

Alternatively, if we take B at (5, 2) and A at (something). Then slope is (2 - y1)/(5 - x1). To maximize, maximize (2 - y1) and minimize (5 - x1). So again, 2 - 0 = 2, and 5 - 2 = 3, so slope is 2/3.

So that's worse.

Alternatively, if we take B at (4, 3) and A at (something else).

Wait, maybe if we take A not on the edge? For example, suppose A is somewhere inside the square. But since we are trying to maximize the slope, we need to maximize (y2 - y1) and minimize (x2 - x1). Since both y2 and x2 are at their maximums and x1 and y1 are at their minimums, then the maximum slope is achieved when A is at (2, 0) and B is at (4, 3).

But let me check another possibility. Suppose we take A at (x1, y1) where x1 is less than 2, but y1 is also less than 0? No, because A is within the square, so y1 can't be less than 0.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is something else. For example, if A is at (2, 0), slope is 3/2. If A is at (2, 1), then slope is (3 - 1)/2 = 1. Which is less.

Alternatively, if we take A at (something else). Wait, maybe if we take B not at (4, 3) but somewhere else? For example, if B is at (5, 3), but then the denominator is larger.

Alternatively, if we take B at (4, 3) and A at (2, 0), slope is 3/2.

But let me think if there's another combination where the slope is higher. For example, suppose we take A at (0, 0) and B at (4, 3). Slope is 3/4. If we take A at (0, 0) and B at (5, 3), slope is 3/5. If we take A at (0, 0) and B at (4, 2), slope is 2/4 = 1/2.

Alternatively, if we take A at (something) where y1 is negative? No, since A is in the square from (0,0) to (2,2).

Wait, but what if we take A at (something) where x1 is less than 2 and y1 is less than 0? But no, since the square is from (0,0) to (2,2), so coordinates can't be less than 0.

Alternatively, maybe if we take A at (something) where x1 is 2 and y1 is 0, and B at (4, 3).

Alternatively, maybe there's a case where the line AB passes through other points? Wait, but the maximum slope is achieved when the line is as steep as possible.

Wait, another thought: maybe if we take A at (x1, y1) and B at (x2, y2) such that the line AB is tangent to some edge of the squares? Or maybe not.

Alternatively, think of the problem as optimizing the slope. Let me consider variables.

Let me denote A as (x1, y1) with 0 ≤ x1 ≤ 2, 0 ≤ y1 ≤ 2.

B as (x2, y2) with 4 ≤ x2 ≤ 5, 2 ≤ y2 ≤ 3.

We need to maximize (y2 - y1)/(x2 - x1).

Let me consider this as a function f(x1, y1, x2, y2) = (y2 - y1)/(x2 - x1).

To maximize this, we can think of it as maximizing the numerator while minimizing the denominator.

But since x2 - x1 is always positive, as discussed earlier, we can think of it as maximizing (y2 - y1) and minimizing (x2 - x1).

But maybe there's a case where even though (y2 - y1) is not maximum, but (x2 - x1) is smaller? But since (x2 - x1) can't be smaller than 2, as x2 ≥ 4 and x1 ≤ 2, then 4 - 2 = 2 is the minimum.

Therefore, the maximum slope is achieved when (y2 - y1) is maximum and (x2 - x1) is minimum.

Therefore, maximum (y2 - y1) is 3 - 0 = 3, and minimum (x2 - x1) is 4 - 2 = 2. Therefore, slope is 3/2.

But let me check if there's another combination where (y2 - y1)/(x2 - x1) is larger.

Suppose we take A at (x1, y1) where y1 is not 0, but x1 is such that x2 - x1 is smaller? Wait, but x2 - x1 can't be smaller than 2.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is something, and B at (x2, y2) where x2 is 4 and y2 is something.

If we take A at (2, y1) and B at (4, y2), then slope is (y2 - y1)/2. To maximize this, we need to maximize (y2 - y1). Since y2 can be up to 3 and y1 can be as low as 0, so maximum is 3 - 0 = 3, so slope is 3/2.

Alternatively, if we take A at (x1, y1) and B at (x2, y2) where x2 is 4 and y2 is 3, and x1 is 2 and y1 is 0.

Alternatively, if we take A at (x1, y1) where x1 is less than 2, but then x2 - x1 would be larger.

Alternatively, if we take B at (x2, y2) where x2 is 4 and y2 is 3, and A at (x1, y1) where x1 is 2 and y1 is 0.

So slope is 3/2.

But let me think again. Suppose we take A at (something) where y1 is not 0, but maybe the slope is higher? For example, if A is at (2, 0) and B is at (4, 3), slope is 3/2. If A is at (2, 0.5) and B is at (4, 3), slope is (3 - 0.5)/2 = 2.5/2 = 1.25, which is less than 1.5.

Alternatively, if A is at (2, 0) and B is at (4, 3), slope is 3/2.

Alternatively, if we take A at (something else). Suppose A is at (x1, y1) where x1 is 2 and y1 is 0, and B is at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, if we take A at (x1, y1) where x1 is 2 and y1 is 0, and B is at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, if we take A at (x1, y1) where x1 is 2 and y1 is 0, and B is at (x2, y2) where x2 is 4 and y2 is 3.

So that's the same as before.

Wait, but what if we take A at (x1, y1) where x1 is less than 2, but y1 is also less than 0? But no, since A is in the square from (0,0) to (2,2).

Alternatively, maybe there's a case where the line AB goes from a point on the edge of the first square to a point on the edge of the second square, but not necessarily the corners.

Wait, for example, suppose A is at (2, 0) and B is at (4, 3). That's a corner to corner.

Alternatively, suppose A is at (2, 0) and B is at (4, 3).

Alternatively, suppose A is at (2, 0) and B is at (4, 3).

But maybe if we take A at (something else) where the line AB is steeper?

Wait, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Wait, I keep getting the same point.

Alternatively, maybe there's a case where the slope is higher if we take A not at (2, 0) but somewhere else?

Wait, suppose we take A at (x1, y1) where x1 is 2 and y1 is something, and B at (x2, y2) where x2 is 4 and y2 is something.

If we take A at (2, y1) and B at (4, y2), then slope is (y2 - y1)/2. To maximize this, we need to maximize y2 - y1. Since y2 can be up to 3 and y1 can be as low as 0, so maximum is 3 - 0 = 3, so slope is 3/2.

Alternatively, if we take A at (x1, y1) where x1 is less than 2, but then x2 - x1 is larger.

Alternatively, if we take B not at (4, 3) but somewhere else. For example, if B is at (4, 3) and A is at (2, 0), slope is 3/2. If B is at (5, 3) and A is at (something), slope is (3 - y1)/(5 - x1). To maximize this, we need to maximize (3 - y1) and minimize (5 - x1). So maximum (3 - y1) is 3 - 0 = 3, and minimum (5 - x1) is 5 - 2 = 3, so slope is 3/3 = 1. Which is less than 3/2.

Alternatively, if B is at (4, 3) and A is at (something else).

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (4, 3).

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (4, 3).

Wait, maybe there's another way. Let me think of parametric equations.

Suppose we want to maximize (y2 - y1)/(x2 - x1). Let me think of this as a linear function.

But since A and B are constrained within their squares, maybe we can use calculus or some optimization technique.

But since it's a problem with constraints, maybe we can consider the maximum occurs at the boundary.

In optimization problems, the extrema often occur at the boundaries. So, for the maximum slope, it's likely achieved when A and B are at the corners of their respective squares.

So let's check all possible combinations of corners.

First square (A's square) has corners at (0,0), (0,2), (2,0), (2,2).

Second square (B's square) has corners at (4,2), (4,3), (5,2), (5,3).

So there are 4 corners for A and 4 corners for B, so 16 combinations. Let me check the slopes for all these combinations.

1. A=(0,0), B=(4,2): slope = (2 - 0)/(4 - 0) = 2/4 = 1/2
2. A=(0,0), B=(4,3): slope = (3 - 0)/4 = 3/4
3. A=(0,0), B=(5,2): slope = (2 - 0)/5 = 2/5
4. A=(0,0), B=(5,3): slope = 3/5
5. A=(0,2), B=(4,2): slope = (2 - 2)/4 = 0
6. A=(0,2), B=(4,3): slope = (3 - 2)/4 = 1/4
7. A=(0,2), B=(5,2): slope = 0
8. A=(0,2), B=(5,3): slope = (3 - 2)/5 = 1/5
9. A=(2,0), B=(4,2): slope = (2 - 0)/2 = 1
10. A=(2,0), B=(4,3): slope = (3 - 0)/2 = 3/2
11. A=(2,0), B=(5,2): slope = (2 - 0)/3 = 2/3
12. A=(2,0), B=(5,3): slope = (3 - 0)/3 = 1
13. A=(2,2), B=(4,2): slope = (2 - 2)/2 = 0
14. A=(2,2), B=(4,3): slope = (3 - 2)/2 = 1/2
15. A=(2,2), B=(5,2): slope = 0
16. A=(2,2), B=(5,3): slope = (3 - 2)/3 = 1/3

So looking at all these slopes, the maximum is 3/2, which occurs at case 10: A=(2,0) and B=(4,3).

Therefore, the maximum slope is 3/2.

But wait, let me check if there are any other points on the edges that could give a higher slope. For example, suppose A is not at a corner but somewhere on the edge.

For example, suppose A is on the right edge of the first square, which is x=2, y from 0 to 2. Let me parametrize A as (2, y1) where 0 ≤ y1 ≤ 2. Then B is on the left edge of the second square, which is x=4, y from 2 to 3. Let me parametrize B as (4, y2) where 2 ≤ y2 ≤ 3. Then the slope is (y2 - y1)/(4 - 2) = (y2 - y1)/2. To maximize this, we need to maximize y2 - y1. Since y2 can be up to 3 and y1 can be as low as 0, the maximum is 3 - 0 = 3, so slope is 3/2. So same as before.

Alternatively, if A is on the bottom edge of the first square, x from 0 to 2, y=0. Let me parametrize A as (x1, 0) where 0 ≤ x1 ≤ 2. Then B is on the top edge of the second square, x from 4 to 5, y=3. Let me parametrize B as (x2, 3) where 4 ≤ x2 ≤ 5. Then slope is (3 - 0)/(x2 - x1) = 3/(x2 - x1). To maximize this, we need to minimize x2 - x1. The minimum x2 - x1 is 4 - 2 = 2, so slope is 3/2. Again same result.

Alternatively, if A is on the left edge of the first square, x=0, y from 0 to 2. Then B is on the top edge of the second square, x from 4 to 5, y=3. Then slope is (3 - y1)/(x2 - 0) = (3 - y1)/x2. To maximize this, we need to maximize (3 - y1) and minimize x2. So maximum (3 - y1) is 3 - 0 = 3, and minimum x2 is 4. So slope is 3/4. Which is less than 3/2.

Alternatively, if A is on the top edge of the first square, y=2, x from 0 to 2. Then B is on the top edge of the second square, y=3, x from 4 to 5. Then slope is (3 - 2)/(x2 - x1) = 1/(x2 - x1). To maximize this, minimize x2 - x1. Minimum is 4 - 2 = 2, so slope is 1/2.

Alternatively, if A is on the right edge of the first square, x=2, y from 0 to 2, and B is on the right edge of the second square, x=5, y from 2 to 3. Then slope is (y2 - y1)/(5 - 2) = (y2 - y1)/3. To maximize, set y2=3, y1=0, slope is 3/3 = 1.

So again, 3/2 is the maximum.

Alternatively, suppose A is somewhere inside the first square and B is somewhere inside the second square. But since we already considered the maximum on the edges, and the maximum slope occurs at the edges, then the maximum slope is 3/2.

But let me think again. Suppose we take A at (2, 0) and B at (4, 3). Is there a way to get a steeper slope? For example, if we take A at (2, 0) and B at (4, 3), slope is 3/2. If we take A at (2, 0) and B at (something else). For example, if B is at (4, 3), which is the top right corner of the second square.

Alternatively, if we take A at (something else) and B at (something else). For example, suppose A is at (x1, y1) and B is at (x2, y2) where the line AB is steeper.

Wait, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Wait, I think I'm going in circles here.

Alternatively, think of the problem as follows: the maximum slope between two points A and B is the maximum of (y2 - y1)/(x2 - x1). Since x2 ≥ 4, x1 ≤ 2, so x2 - x1 ≥ 2. The maximum value of y2 - y1 is 3 - 0 = 3. Therefore, the maximum slope is 3/2.

Alternatively, if there is a case where y2 - y1 is larger? For example, if y2 can be more than 3? But no, since B is within the square from (4,2) to (5,3), so y2 can't exceed 3. Similarly, y1 can't be less than 0.

Therefore, 3/2 is the maximum.

But let me check another possibility. Suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Wait, I think I need to accept that 3/2 is the maximum.

But let me check if there's a case where the slope is higher. For example, suppose we take A at (something) where the line AB goes from the bottom of the first square to the top of the second square, but not necessarily the corners.

Wait, suppose A is at (2, 0) and B is at (4, 3). That's the same as before.

Alternatively, suppose A is at (something) where x1 is less than 2, but y1 is also less than 0? No, since A is in the square.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

I think I need to conclude that the maximum slope is 3/2.

But let me check with coordinates.

Let me take A=(2,0) and B=(4,3). The slope is (3 - 0)/(4 - 2) = 3/2.

Alternatively, if I take A=(2, 0) and B=(4, 3), then the line is from (2,0) to (4,3).

Alternatively, if I take A=(2, 0) and B=(4, 3), then the line is going from (2,0) to (4,3).

Alternatively, if I take A=(something else). For example, suppose A is at (1, 0) and B is at (4, 3). Then slope is (3 - 0)/(4 - 1) = 3/3 = 1. Which is less than 3/2.

Alternatively, if A is at (2, 0) and B is at (4, 3).

Alternatively, if A is at (2, 0) and B is at (4, 3).

Alternatively, if I take A at (2, 0) and B at (4, 3).

Yes, so that's the maximum.

Therefore, the greatest possible value of the slope is 3/2.

But wait, the problem says "within or on the square". So points on the edges are allowed. So the answer is 3/2.

But let me check once more.

Suppose there is a point A inside the first square and a point B inside the second square such that the slope is higher. For example, suppose A is at (2, 0) and B is at (4, 3). If I move A slightly to the left, say (2 - ε, 0), then x2 - x1 becomes 4 - (2 - ε) = 2 + ε, which increases the denominator, so slope decreases. If I move A slightly up, say (2, 0 + ε), then y2 - y1 becomes 3 - ε, so slope becomes (3 - ε)/2, which is less than 3/2. Similarly, if I move B slightly to the left, say (4 - ε, 3), then x2 - x1 is 4 - ε - 2 = 2 - ε, so denominator decreases, but since x2 can't be less than 4, ε can't be positive. Wait, x2 is at least 4, so moving B to the left would decrease x2, but since x2 can't be less than 4, the minimum x2 is 4. Therefore, moving B to the left is not possible. Similarly, moving B up is not possible since y2 can't exceed 3.

Therefore, the maximum slope is indeed achieved at A=(2,0) and B=(4,3), giving slope 3/2.

Therefore, the answer is 3/2, which is 3 over 2.

But wait, the problem says "Express your answer as a common fraction." So 3/2 is already a common fraction.

But let me check once more if there's a case where slope is higher.

Suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

Alternatively, suppose we take A at (x1, y1) where x1 is 2 and y1 is 0, and B at (x2, y2) where x2 is 4 and y2 is 3.

I think I need to stop here.

**Final Answer**
\boxed{\dfrac{3}{2}}

To determine the **greatest possible value of the slope** of the line containing points $ A $ and $ B $, we begin by analyzing the constraints on the positions of these points.

---

### **Step 1: Understand the Squares**

- **Point A** lies within or on the square with opposite corners at $ (0,0) $ and $ (2,2) $.
  This is a square of side length 2, aligned with the axes, with all coordinates satisfying:
  $$
  0 \leq x_1 \leq 2, \quad 0 \leq y_1 \leq 2
  $$

- **Point B** lies within or on the square with opposite corners at $ (4,2) $ and $ (5,3) $.
  This is a square of side length 1, also aligned with the axes, with all coordinates satisfying:
  $$
  4 \leq x_2 \leq 5, \quad 2 \leq y_2 \leq 3
  $$

---

### **Step 2: Slope Formula**

The slope of the line through points $ A = (x_1, y_1) $ and $ B = (x_2, y_2) $ is given by:
$$
\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}
$$

To **maximize** this slope, we want to:
- **Maximize** the numerator $ y_2 - y_1 $
- **Minimize** the denominator $ x_2 - x_1 $

---

### **Step 3: Maximize the Numerator**

The maximum possible value of $ y_2 - y_1 $ occurs when:
- $ y_2 $ is as large as possible: $ y_2 = 3 $
- $ y_1 $ is as small as possible: $ y_1 = 0 $

So the maximum numerator is:
$$
y_2 - y_1 = 3 - 0 = 3
$$

---

### **Step 4: Minimize the Denominator**

The minimum possible value of $ x_2 - x_1 $ occurs when:
- $ x_2 $ is as small as possible: $ x_2 = 4 $
- $ x_1 $ is as large as possible: $ x_1 = 2 $

So the minimum denominator is:
$$
x_2 - x_1 = 4 - 2 = 2
$$

---

### **Step 5: Compute the Maximum Slope**

Using the maximum numerator and minimum denominator:
$$
\text{Maximum slope} = \frac{3}{2}
$$

This occurs when:
- $ A = (2, 0) $
- $ B = (4, 3) $

---

### **Step 6: Verify No Greater Slope is Possible**

We can confirm that:
- $ y_2 $ cannot exceed 3
- $ y_1 $ cannot be less than 0
- $ x_2 $ cannot be less than 4
- $ x_1 $ cannot be greater than 2

Thus, the maximum value of the slope is indeed:
$$
\boxed{\dfrac{3}{2}}
$$

Solution: Since point $A$ is constrained to a rectangular region with sides parallel to the axes, its $x$ and $y$ coordinates can be chosen independently of one another.  The same is true of point $B$.  Therefore, the horizontal separation between $A$ and $B$ should be minimized and the vertical separation maximized.  The greatest possible $y$-coordinate for $B$ is 3 and the least possible $y$-coordinate for $A$ is 0.  The greatest possible $x$-coordinate for $A$ is 2 and the least possible $x$-coordinate for $B$ is 4. Therefore, the slope between $A$ and $B$ is maximized when $A$ has coordinates (2,0) and $B$ has coordinates (4,3).  The maximum slope is $\boxed{\frac{3}{2}}$.