Explanation:
Okay, so I need to find the greatest possible value of the slope of the line connecting points A and B, where point A is inside or on the square with opposite corners at (0,0) and (2,2), and point B is inside or on the square with opposite corners at (4,2) and (5,3). Alright, let me break this down step by step.

First, let me visualize the squares. The first square has corners at (0,0) and (2,2). Since it's a square, the other two corners should be at (0,2) and (2,0), right? So that square spans from x=0 to x=2 and y=0 to y=2. Then the second square has opposite corners at (4,2) and (5,3). Let me figure out the other two corners. The square can be a bit tricky here because the diagonal isn't necessarily aligned with the axes. Wait, actually, since it's a square, the other two corners can be found by moving perpendicularly from the diagonal.

But maybe I don't need to find the exact coordinates. Let me think. The square with opposite corners at (4,2) and (5,3) – since the sides are equal and the angles are 90 degrees. The vector from (4,2) to (5,3) is (1,1). To get the other diagonal, we need a vector perpendicular to (1,1). The perpendicular vectors would be (-1,1) and (1,-1). But since it's a square, the length of the diagonal is the same as the length of the vector (1,1), which is √(1² + 1²) = √2. Therefore, the other diagonal should also have length √2, but in the perpendicular direction. Wait, no. Wait, in a square, the diagonals are equal in length. However, actually, in a square, the diagonals are equal in length and they intersect at 90 degrees? No, in a square, the diagonals bisect each other at 90 degrees? Wait, no, in a square the diagonals bisect each other and are equal in length, but they intersect at 90 degrees? Wait, no. In a square, the diagonals are equal in length and bisect each other at 90 degrees? Wait, actually, in a square, the diagonals are equal in length and they bisect each other at 90 degrees? Wait, no. In a square, the diagonals are not perpendicular unless it's a rhombus. Wait, no. Wait, in a square, the diagonals are equal and they intersect at 90 degrees? Wait, let me recall. In a square, the diagonals are equal in length and they bisect each other, and they intersect at 90 degrees. So they are perpendicular. So, if one diagonal is from (4,2) to (5,3), then the other diagonal should be perpendicular to that.

So the vector along the first diagonal is (1,1). The perpendicular vector would be (-1,1) or (1,-1). But the length of the diagonal is √2. Since the diagonals of a square are equal, the length of the other diagonal is also √2. Wait, but in a square, the diagonals are of equal length. Wait, but the length of the diagonal of the square is sqrt( (side length)^2 + (side length)^2 ) = side length * sqrt(2). So if the diagonal is from (4,2) to (5,3), which has length √2, then the side length of the square would be √2 / √2 = 1. Therefore, each side is length 1. Therefore, the other diagonal should also have length √2, but since we already have one diagonal, which is from (4,2) to (5,3), the other diagonal would be centered at the midpoint of this diagonal.

Midpoint of (4,2) and (5,3) is ((4+5)/2, (2+3)/2) = (4.5, 2.5). So the other diagonal is perpendicular to the first one, same midpoint, and same length. Let me calculate the other two corners.

Since the original diagonal is vector (1,1), the other diagonal should be in the direction of (-1,1) or (1,-1). However, since the length of the original diagonal is √2, the other diagonal must also have length √2. So, we need to find points such that moving from the midpoint (4.5, 2.5) in direction (-1,1) or (1,-1) by half the diagonal length.

Wait, the diagonal vector is (1,1). The half-diagonal vector is (0.5, 0.5). So the other diagonal is perpendicular. So the direction perpendicular to (1,1) is (-1,1) or (1,-1). Let me take (-1,1). So the half-diagonal vector is (-0.5, 0.5) in that direction? Wait, no. The length of the diagonal is √2, so half-diagonal is √2 / 2. The vector (-1,1) has length √(1 + 1) = √2, so half of that vector would be (-1/√2, 1/√2). Wait, but maybe there is a simpler way.

Alternatively, since the square has side length 1, the other diagonal would be from midpoint (4.5, 2.5) in direction perpendicular to (1,1). Let me parametrize. Let me find the other two corners.

Let me denote the midpoint as M(4.5, 2.5). Let me find two points P and Q such that M is the midpoint of both diagonals. So, given that one diagonal is between (4,2) and (5,3), then the other diagonal is between P and Q, with midpoint at M.

So if I take a vector perpendicular to (1,1), which is (-1,1) or (1,-1). Let me take vector (-1,1). Then, since the length of the diagonal is √2, the vector from midpoint M to P would be half of that vector. Wait, the entire diagonal is length √2, so half-diagonal is √2 / 2. The vector (-1,1) has length √2, so half of that would be (-1/2, 1/2). Therefore, point P would be M + (-1/2, 1/2) = (4.5 - 0.5, 2.5 + 0.5) = (4.0, 3.0). Similarly, point Q would be M + (1/2, -1/2) = (4.5 + 0.5, 2.5 - 0.5) = (5.0, 2.0).

Therefore, the square has corners at (4,2), (5,3), (4,3), and (5,2)? Wait, let me check:

Wait, original diagonal is from (4,2) to (5,3). The other diagonal is from (4,3) to (5,2). Let me check if those points form a square.

Looking at the points (4,2), (5,2), (5,3), (4,3). Wait, that's actually a square with side length 1, right? From (4,2) to (5,2) is 1 unit right, then up to (5,3), then left to (4,3), then down to (4,2). So yes, that's a square with side length 1, sitting between x=4 to 5 and y=2 to 3. So the square for point B is actually a unit square from (4,2) to (5,3). Therefore, point B lies within or on that square.

So now, point A is in the square with corners at (0,0), (2,0), (0,2), (2,2). So that's the square from (0,0) to (2,2). Point B is in the square from (4,2) to (5,3). So to maximize the slope of AB, we need to find points A and B such that the slope (y_B - y_A)/(x_B - x_A) is as large as possible.

Slope is rise over run. To maximize the slope, we need to maximize the rise and minimize the run, but also considering the direction. Wait, but slope is (y_B - y_A)/(x_B - x_A). So to maximize slope, which could be positive or negative. Since we want the maximum slope, which would be a positive number, or maybe even a very large negative? Wait, no, the maximum slope would be positive infinity? But since points are constrained, we can't have that. Wait, but given the positions of the squares, maybe the maximum slope occurs when A is as low as possible and B is as high as possible, but also considering the x-coordinates.

Wait, let's think. The slope of AB is (y_B - y_A)/(x_B - x_A). If we want to maximize this slope, we need to maximize the numerator (y_B - y_A) and minimize the denominator (x_B - x_A). However, if x_B - x_A is negative, then the slope becomes negative. But if we want the maximum slope, which is the maximum value, it's possible that we can have a very large positive slope if we can have a large positive numerator and a small positive denominator.

Alternatively, if we can have a large positive numerator and a very small positive denominator, that would give a large positive slope. But if the denominator is negative, then the slope would be negative. Alternatively, if the denominator is positive and numerator is positive, slope is positive. If numerator is positive and denominator negative, slope is negative. If numerator is negative and denominator negative, slope is positive. So maximum slope could be achieved in different ways.

But given the positions of the squares, let's analyze.

Point A is in the square [0,2] x [0,2].

Point B is in the square [4,5] x [2,3].

So, the x-coordinate of B is between 4 and 5, while the x-coordinate of A is between 0 and 2. Therefore, x_B - x_A is between 4 - 2 = 2 and 5 - 0 = 5. So all denominators are positive. Therefore, the slope would be positive if y_B - y_A is positive, which would be the case if y_B > y_A. If y_B < y_A, the slope is negative. Therefore, to maximize the slope, we need to maximize (y_B - y_A) / (x_B - x_A).

Therefore, to get the maximum slope, we need to maximize the numerator and minimize the denominator. However, since x_B - x_A is minimized when x_B is as small as possible and x_A is as large as possible. Similarly, y_B - y_A is maximized when y_B is as large as possible and y_A is as small as possible.

Therefore, to maximize the slope, we should take:

- y_B maximum: 3 (since B is in [4,5] x [2,3])

- y_A minimum: 0 (since A is in [0,2] x [0,2])

- x_B minimum: 4 (since B is in [4,5] x [2,3])

- x_A maximum: 2 (since A is in [0,2] x [0,2])

So then, the maximum slope would be (3 - 0)/(4 - 2) = 3/2.

But wait, is that the maximum? Let me confirm.

Alternatively, is there a case where even though the denominator isn't minimized, the numerator is larger? For example, if we can increase the numerator more by moving y_B up and y_A down, while keeping the denominator small. But since y_B can't be more than 3 and y_A can't be less than 0, that gives the maximum possible numerator. So (3 - 0) is 3. Then denominator is (4 - 2) = 2, so 3/2. Is that the maximum?

Wait, but let me check if there's a case where the slope is higher if we take different positions. For example, suppose point A is not at (2,0) but somewhere else, and point B is not at (4,3). Maybe if we take point A at (2,0) and point B at (4,3), the slope is 3/2. But what if point B is at (5,3) and point A is at (2,0)? Then the slope is (3 - 0)/(5 - 2) = 3/3 = 1, which is less than 3/2. So worse. If point A is at (2,0) and point B is at (4,2), the slope is (2 - 0)/(4 - 2) = 1, also less. If point A is at (2,0) and point B is at (4,3), slope is 3/2. If point A is at (something else) and point B is at (something else), can we get a higher slope?

Wait, suppose we take point A near (2,0) and point B near (4,3). Then slope is approaching 3/2. Suppose we take point A slightly to the left of (2,0), say (2 - ε, 0), and point B at (4, 3). Then slope is (3 - 0)/(4 - (2 - ε)) = 3/(2 + ε), which approaches 3/2 as ε approaches 0. So that seems maximum.

But is there a case where the slope is higher? For example, if we take point A at (0, something) and point B at (4, something else). Wait, but then the denominator would be 4 - 0 = 4, and the numerator would be y_B - y_A. If y_B is 3 and y_A is 0, then 3/4, which is less than 3/2.

Alternatively, if we take point A at (x, y) where x is small and y is small, and point B at (x', y') where x' is small and y' is large. Wait, but x' is between 4 and 5, so even if x is 0, x' - x is between 4 and 5. So the denominator is larger.

Alternatively, suppose we take point A at (something) and point B at (something) where the line AB has a steeper slope. Wait, but since the denominator is x_B - x_A, which is between 2 and 5. The maximum numerator is 3 - 0 = 3, so maximum slope is 3/2. Wait, but is there a possibility that the numerator is larger? Let me check.

Wait, for point B, the maximum y-coordinate is 3, and for point A, the minimum y-coordinate is 0. So the maximum possible difference in y is 3 - 0 = 3. So unless there is a case where y_B - y_A is larger than 3? No, since y_B can't be more than 3 and y_A can't be less than 0. Therefore, that's maximum.

Therefore, slope is 3 / 2. But wait, let me check the coordinates again. Suppose point A is at (2, 0) and point B is at (4, 3). Then the slope is (3 - 0)/(4 - 2) = 3/2. Is this the maximum?

But wait, maybe there's another combination where even though the difference in y is less, the difference in x is smaller, leading to a higher slope. For example, suppose point A is at (x_A, 0) and point B is at (x_B, 3), where x_A is as large as possible (2) and x_B is as small as possible (4). Then the slope is 3 / (4 - 2) = 3/2. But what if point A is not on the bottom edge?

Suppose point A is somewhere inside the square, and point B is somewhere inside the square. For example, suppose point A is (x, y) where y is less than 0? No, because A is within or on the square, so y can't be less than 0. Similarly, point B can't be above y=3. Therefore, the maximum difference in y is 3 - 0 = 3.

But perhaps if we take point A such that y_A is slightly less than 0? But it can't be, since the square is from y=0 to y=2. Similarly, point B can't be above y=3.

Alternatively, perhaps there's another orientation where slope is higher. Wait, slope can also be calculated if the line goes from a high point of A to a low point of B, but since slope is (y_B - y_A)/(x_B - x_A), if y_B - y_A is negative and x_B - x_A is negative, that would be positive. But since x_B is always greater than x_A (as x_B is between 4 and 5, and x_A is between 0 and 2), so x_B - x_A is always positive, so the denominator is positive. Therefore, to have positive slope, we need y_B - y_A positive, which is y_B > y_A. So, the maximum slope occurs when y_B is maximum and y_A is minimum, with x_B - x_A as small as possible. Hence, that would be when point A is at (2,0) and point B is at (4,3), giving slope 3/2.

But let me check another possibility. What if point A is not at (2,0) but somewhere else, and point B is not at (4,3), but somewhere else, such that the slope is higher?

For example, suppose we take point A at (2, something) and point B at (4, 3). Let me take A at (2, 0) gives slope 3/2. If I take point A at (2, ε) where ε > 0, then slope is (3 - ε)/(4 - 2) = (3 - ε)/2, which is less than 3/2. Similarly, if I take point A at (something less than 2, 0), say x_A = 2 - δ, then slope is (3 - 0)/(4 - (2 - δ)) = 3/(2 + δ), which is less than 3/2.

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

Alternatively, suppose point A is at (1, 0) and point B is at (4, 3). Then slope is 3/3 = 1, which is less than 3/2.

Alternatively, suppose point A is at (x, 0) and point B is at (x + 2, 3). Then the slope would be (3 - 0)/2 = 3/2. Wait, but x + 2 has to be ≤5, so x ≤ 3, but since x is between 0 and 2, that's okay. So for example, if x is 2, then x + 2 is 4, so that's point B at (4,3). If x is 1, then x + 2 is 3, but point B's x-coordinate is between 4 and 5, so 3 is not allowed. Therefore, the minimal x_B is 4, so x_A must be 4 - 2 = 2. Therefore, the maximum slope is achieved at x_A = 2, x_B = 4, giving slope 3/2.

Wait, but maybe if we take point B not on the top edge? Suppose point B is somewhere else. For example, suppose point B is at (4, 3) which is the top corner, and point A is at (2, 0), which is the bottom right corner of the first square. Then slope is 3/2. If we take point A at (2, 0) and point B at (4, something less than 3), then the slope would be less. Similarly, if we take point A at (something less than 2, 0) and point B at (4, 3), then the slope is 3/(4 - x_A). If x_A is less than 2, 4 - x_A is larger, so the slope is smaller. Therefore, that case gives less slope.

Alternatively, suppose we take point A not on the bottom edge but somewhere else, and point B not on the top edge, but such that (y_B - y_A)/(x_B - x_A) is greater than 3/2. Let me think. For example, suppose we take point A at (x, y) and point B at (x', y'), where y' - y is more than 3, but since y' can't exceed 3 and y can't be less than 0, that's impossible. So maximum y' - y is 3.

Therefore, 3 divided by minimal x' - x, which is 4 - 2 = 2, gives 3/2.

But wait, is there a case where the denominator x_B - x_A is smaller? For example, if x_B is 4 and x_A is 2, then x_B - x_A = 2. If we can have x_B - x_A less than 2? Well, since x_A is at least 0 and x_B is at least 4, then x_B - x_A is at least 4 - 2 = 2. So minimal value of denominator is 2, hence maximum slope is 3/2. Therefore, that seems to be the maximum.

But let me think again. Is there a possibility that if point A is not on the bottom edge and point B is not on the top edge, but the line AB has a steeper slope?

For example, suppose point A is at (x, y) and point B is at (x', y') where the slope is (y' - y)/(x' - x). Let's consider a line that goes from somewhere in the lower left of square A to somewhere in the upper right of square B. Wait, but square B is already to the right of square A. So the line from point A to point B would go from the left square to the right square. The maximum slope would be achieved when going from the lowest point of A to the highest point of B, but with the smallest horizontal difference. Which would be from the rightmost point of A (x=2, y=0) to the leftmost point of B (x=4, y=3). Which is exactly the case we considered before, slope 3/2.

But let me confirm with coordinates. If we take A(2, 0) and B(4, 3), slope is (3 - 0)/(4 - 2) = 3/2. If we take A(2, 0) and B(4, 3), that's valid because A is on the square (since (2,0) is a corner of the square from (0,0) to (2,2)), and B is on the square (since (4,3) is a corner of the square from (4,2) to (5,3)).

But let me check if there is a line that goes from some other point A and B with a steeper slope. For example, if point A is at (2, 0) and point B is at (4.5, 3), which is the upper right corner of square B? Wait, square B has corners at (4,2), (5,2), (5,3), (4,3). So (4.5, 3) is not a corner, but is inside the square. Wait, (4.5, 3) is outside the square because the square's y-coordinate only goes up to 3. Wait, actually, since the square's corners are (4,2), (5,2), (5,3), (4,3), then the square spans x from 4 to 5 and y from 2 to 3. So (4.5, 3) is on the top edge. So yes, it is inside the square. Then, if we take point B as (4.5, 3), then slope from A(2, 0) is (3 - 0)/(4.5 - 2) = 3 / 2.5 = 6/5 = 1.2, which is less than 3/2 = 1.5. Therefore, less steep.

Alternatively, what if point B is at (4, 3) and point A is at (2, 0). That gives 3/2. If we take point A at (something) and point B at (something else) such that the line AB is steeper?

Alternatively, suppose we take point A at (x, y) where y is as low as possible and point B at (x', y') where y' is as high as possible, but x' - x is smaller. Wait, but x' can't be less than 4 and x can't be more than 2. So the minimal x' - x is 4 - 2 = 2. Therefore, that is the minimal possible denominator.

Alternatively, could there be a case where the line passes through both squares in such a way that the slope is steeper? For example, maybe a line that starts at a lower y in square A and ends at a higher y in square B, but with a smaller horizontal distance?

Wait, but if you take point A at (2, 0) and point B at (4, 3), the horizontal distance is 2. If you take point A slightly to the left of 2, like (2 - ε, 0) and point B slightly to the right of 4, like (4 + ε, 3), then the horizontal distance is (4 + ε) - (2 - ε) = 2 + 2ε, which is larger. So the denominator increases, making the slope smaller. Therefore, that would decrease the slope.

Alternatively, suppose we take point A not at y=0 but somewhere else, but point B at some position where y' - y is larger? Wait, but y' cannot be larger than 3 and y cannot be smaller than 0, so that difference can't be larger than 3.

Wait, but perhaps if point A is at a lower y and point B is at a higher y relative to some other point? Wait, but no, since the maximum y for B is 3 and minimum y for A is 0, so that difference is fixed.

Alternatively, is there a case where the slope is not (y_B - y_A)/(x_B - x_A) but something else? No, that is the formula for slope between two points.

Wait, but maybe I need to consider the possibility that the line AB is tangent to both squares or something? But since we can choose any points within the squares, the maximum slope should be achieved at some corner points.

In general, the maximum or minimum slopes between two convex sets are achieved at the extreme points, i.e., the corners. So maybe we can check all combinations of corners of the two squares and find the maximum slope.

So let's try that approach. Let me list all the corners of square A and square B.

Square A has corners at (0,0), (2,0), (2,2), (0,2).

Square B has corners at (4,2), (5,2), (5,3), (4,3).

So there are 4 corners for A and 4 for B, so 16 possible combinations of points. Let me calculate the slope for each pair and see which is the maximum.

Starting with point A(0,0):

- To B(4,2): slope = (2 - 0)/(4 - 0) = 2/4 = 1/2

- To B(5,2): (2 - 0)/(5 - 0) = 2/5 = 0.4

- To B(4,3): (3 - 0)/(4 - 0) = 3/4 = 0.75

- To B(5,3): (3 - 0)/(5 - 0) = 3/5 = 0.6

So the maximum here is 3/4.

Next, point A(2,0):

- To B(4,2): (2 - 0)/(4 - 2) = 2/2 = 1

- To B(5,2): (2 - 0)/(5 - 2) = 2/3 ≈ 0.666...

- To B(4,3): (3 - 0)/(4 - 2) = 3/2 = 1.5

- To B(5,3): (3 - 0)/(5 - 2) = 3/3 = 1

So the maximum here is 3/2.

Next, point A(2,2):

- To B(4,2): (2 - 2)/(4 - 2) = 0/2 = 0

- To B(5,2): (2 - 2)/(5 - 2) = 0

- To B(4,3): (3 - 2)/(4 - 2) = 1/2 = 0.5

- To B(5,3): (3 - 2)/(5 - 2) = 1/3 ≈ 0.333...

Maximum here is 1/2.

Next, point A(0,2):

- To B(4,2): (2 - 2)/(4 - 0) = 0/4 = 0

- To B(5,2): same as above, 0

- To B(4,3): (3 - 2)/(4 - 0) = 1/4 = 0.25

- To B(5,3): (3 - 2)/(5 - 0) = 1/5 = 0.2

Maximum here is 1/4.

So among all the corner pairs, the maximum slope is 3/2, achieved when A is (2,0) and B is (4,3). Therefore, that seems to be the maximum. However, just to be thorough, maybe there are non-corner points that could give a higher slope? Let me think.

For example, suppose point A is somewhere along the bottom edge of square A, (x, 0), where x is between 0 and 2, and point B is somewhere along the top edge of square B, (4, y), where y is between 2 and 3. Then the slope would be (y - 0)/(4 - x) = y/(4 - x). To maximize this, we need to maximize y and minimize (4 - x). Since y can be up to 3 and (4 - x) can be as small as 4 - 2 = 2, giving 3 / 2 = 1.5, which is the same as the corner case. If we take y = 3 and x = 2, then we get 3/2. If we take y less than 3, then even with x = 2, the slope would be less. If we take x less than 2, then (4 - x) would be larger, leading to smaller slope.

Similarly, suppose point B is along the right edge of square B, (5, y), where y is between 2 and 3, and point A is along the bottom edge of square A, (x, 0). Then slope is (y - 0)/(5 - x). To maximize this, we need y as large as possible (3) and x as large as possible (2). So slope is 3/(5 - 2) = 1, which is less than 3/2.

If we take point B on the top edge and point A on the right edge? So A(2, y) and B(4, 3). Then slope is (3 - y)/(4 - 2) = (3 - y)/2. To maximize this, minimize y. y can be as low as 0, but since A is in square A, y can be 0. Then slope is (3 - 0)/2 = 3/2, same as before.

Alternatively, if point A is somewhere else. Suppose point A is at (x, y) and point B is at (x', y'), not restricted to edges. But since we already considered the maximum difference in y and minimal difference in x, and that case is achieved at the corners, I think that 3/2 is indeed the maximum.

But just to check another possibility: suppose we have a line that starts from (2, 0) and goes to some point inside square B, but not at the corner. For example, suppose point B is at (4 + ε, 3). Then slope is (3 - 0)/(4 + ε - 2) = 3/(2 + ε). As ε increases, this slope decreases. So maximum is when ε = 0, which is 3/2. Similarly, if point B is at (4, 3 - ε), then slope is (3 - ε - 0)/(4 - 2) = (3 - ε)/2, which is less than 3/2.

Alternatively, if we take point A not on the bottom edge. For example, suppose point A is at (2, ε) and point B is at (4, 3). Then slope is (3 - ε)/2. To maximize this, set ε as small as possible, which is 0. So again slope is 3/2.

Alternatively, suppose we take point A somewhere in the square and point B somewhere in the square such that the line has a steeper slope. Wait, but since we can't get a larger numerator or smaller denominator than 3 and 2, respectively, it's not possible. Therefore, 3/2 seems to be the maximum.

But just to be thorough, let me consider another approach. The maximum slope between two regions can be found by considering the extremal points. Since both squares are convex, the maximum slope should be achieved either at two corners or at a corner and a boundary point, or something like that. However, we already checked all the corner combinations and found 3/2 as maximum. So unless there is a case where a non-corner point gives a higher slope, but I can't think of any.

Wait, but let's think about the line equations. Suppose we have a line that goes through square A and square B. What's the maximum slope of such a line?

The line needs to pass through some point A in square A and some point B in square B. The slope is (y_B - y_A)/(x_B - x_A). Since x_B ≥ 4 and x_A ≤ 2, x_B - x_A ≥ 2. Also, y_B ≤ 3 and y_A ≥ 0, so y_B - y_A ≤ 3. So maximum slope is 3/2, as before. Therefore, 3/2 is indeed the maximum.

But let me think again. Is there a way to have y_B - y_A larger than 3? For example, if point A is below the square? But no, A is within the square from (0,0) to (2,2), so y_A ≥ 0. Similarly, point B is within the square from (4,2) to (5,3), so y_B ≤ 3. So maximum difference is 3 - 0 = 3.

Therefore, I think that 3/2 is indeed the correct answer.

Wait, but let me check with another method. Suppose we parametrize the points. Let me suppose point A is (a, b) where 0 ≤ a ≤ 2, 0 ≤ b ≤ 2. Point B is (c, d) where 4 ≤ c ≤ 5, 2 ≤ d ≤ 3. Then slope is (d - b)/(c - a). To maximize this, we need to maximize d - b and minimize c - a. The maximum of d - b is 3 - 0 = 3. The minimum of c - a is 4 - 2 = 2. Hence, maximum slope is 3/2.

Yes, that confirms again that 3/2 is the maximum. Therefore, the answer is 3/2.

But just to be 100% sure, let me think if there is any other way. For example, suppose we take a line that goes through both squares and has a steeper slope. But given the positions of the squares, the steepest slope would be achieved by going from the lowest point of A to the highest point of B, as we did. So I think that is correct.

Alternatively, suppose someone argues that if point A is at (0,0) and point B is at (4,3), the slope is 3/4, which is less than 3/2. So that's not helpful. Similarly, other combinations. Therefore, I think 3/2 is correct.

Therefore, the greatest possible value of the slope is 3/2, which is 3 over 2, so expressed as a common fraction, it's \boxed{\dfrac{3}{2}}.

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

To determine the **greatest possible value of the slope** of the line connecting points $ A $ and $ B $, where:

- Point $ A $ lies within or on the square with opposite corners at $ (0, 0) $ and $ (2, 2) $
- Point $ B $ lies within or on the square with opposite corners at $ (4, 2) $ and $ (5, 3) $,

we analyze the geometry of the two regions.

---

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

**For Point A**:
The square from (0, 0) to (2, 2) is a square with side length 2, and its boundaries are:
- $ x \in [0, 2] $
- $ y \in [0, 2] $

**For Point B**:
The square has corners at $ (4, 2) $, $ (5, 2) $, $ (5, 3) $, and $ (4, 3) $. This is a unit square with side length 1, centered at $ (4.5, 2.5) $, and its coordinates satisfy:
- $ x \in [4, 5] $
- $ y \in [2, 3] $

Since $ x_B > x_A $ and $ y_B \ge y_A $, the slope of the line $ AB $ is non-negative.

---

### **Step 2: Maximize the Slope**

The slope between points $ A = (x_1, y_1) $ and $ B = (x_2, y_2) $ is:

$$
\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}
$$

To **maximize this**, we want:

- **Maximize the numerator**: Choose $ y_2 = y_{\max} = 3 $ and $ y_1 = y_{\min} = 0 $
- **Minimize the denominator**: Choose $ x_2 = x_{\min} = 4 $ and $ x_1 = x_{\max} = 2 $

So the maximum possible slope is:

$$
\text{slope}_{\max} = \frac{3 - 0}{4 - 2} = \frac{3}{2}
$$

---

### **Step 3: Verify with Corners**

The squares have 4 corners each. Evaluating all combinations of corners for both squares confirms that the **maximum slope** occurs when:

- $ A = (2, 0) $ (bottom-right corner of square A)
- $ B = (4, 3) $ (top-left corner of square B)

This gives:

$$
\text{slope} = \frac{3 - 0}{4 - 2} = \frac{3}{2}
$$

All other combinations of points from the squares result in a smaller slope. Thus, no point within the squares produces a larger slope than this value.

---

### ✅ **Final Answer**

$$
\boxed{\frac{3}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to find the greatest possible value of the slope of the line connecting points A and B, where point A is inside or on the square with opposite corners at (0,0) and (2,2), and point B is inside or on the square with opposite corners at (4,2) and (5,3). Alright, let me break this down step by step.

First, let me visualize the squares. The first square has corners at (0,0) and (2,2). Since it's a square, the other two corners should be at (0,2) and (2,0), right? So that square spans from x=0 to x=2 and y=0 to y=2. Then the second square has opposite corners at (4,2) and (5,3). Let me figure out the other two corners. The square can be a bit tricky here because the diagonal isn't necessarily aligned with the axes. Wait, actually, since it's a square, the other two corners can be found by moving perpendicularly from the diagonal.

But maybe I don't need to find the exact coordinates. Let me think. The square with opposite corners at (4,2) and (5,3) – since the sides are equal and the angles are 90 degrees. The vector from (4,2) to (5,3) is (1,1). To get the other diagonal, we need a vector perpendicular to (1,1). The perpendicular vectors would be (-1,1) and (1,-1). But since it's a square, the length of the diagonal is the same as the length of the vector (1,1), which is √(1² + 1²) = √2. Therefore, the other diagonal should also have length √2, but in the perpendicular direction. Wait, no. Wait, in a square, the diagonals are equal in length. However, actually, in a square, the diagonals are equal in length and they intersect at 90 degrees? No, in a square, the diagonals bisect each other at 90 degrees? Wait, no, in a square the diagonals bisect each other and are equal in length, but they intersect at 90 degrees? Wait, no. In a square, the diagonals are equal in length and bisect each other at 90 degrees? Wait, actually, in a square, the diagonals are equal in length and they bisect each other at 90 degrees? Wait, no. In a square, the diagonals are not perpendicular unless it's a rhombus. Wait, no. Wait, in a square, the diagonals are equal and they intersect at 90 degrees? Wait, let me recall. In a square, the diagonals are equal in length and they bisect each other, and they intersect at 90 degrees. So they are perpendicular. So, if one diagonal is from (4,2) to (5,3), then the other diagonal should be perpendicular to that.

So the vector along the first diagonal is (1,1). The perpendicular vector would be (-1,1) or (1,-1). But the length of the diagonal is √2. Since the diagonals of a square are equal, the length of the other diagonal is also √2. Wait, but in a square, the diagonals are of equal length. Wait, but the length of the diagonal of the square is sqrt( (side length)^2 + (side length)^2 ) = side length * sqrt(2). So if the diagonal is from (4,2) to (5,3), which has length √2, then the side length of the square would be √2 / √2 = 1. Therefore, each side is length 1. Therefore, the other diagonal should also have length √2, but since we already have one diagonal, which is from (4,2) to (5,3), the other diagonal would be centered at the midpoint of this diagonal.

Midpoint of (4,2) and (5,3) is ((4+5)/2, (2+3)/2) = (4.5, 2.5). So the other diagonal is perpendicular to the first one, same midpoint, and same length. Let me calculate the other two corners.

Since the original diagonal is vector (1,1), the other diagonal should be in the direction of (-1,1) or (1,-1). However, since the length of the original diagonal is √2, the other diagonal must also have length √2. So, we need to find points such that moving from the midpoint (4.5, 2.5) in direction (-1,1) or (1,-1) by half the diagonal length.

Wait, the diagonal vector is (1,1). The half-diagonal vector is (0.5, 0.5). So the other diagonal is perpendicular. So the direction perpendicular to (1,1) is (-1,1) or (1,-1). Let me take (-1,1). So the half-diagonal vector is (-0.5, 0.5) in that direction? Wait, no. The length of the diagonal is √2, so half-diagonal is √2 / 2. The vector (-1,1) has length √(1 + 1) = √2, so half of that vector would be (-1/√2, 1/√2). Wait, but maybe there is a simpler way.

Alternatively, since the square has side length 1, the other diagonal would be from midpoint (4.5, 2.5) in direction perpendicular to (1,1). Let me parametrize. Let me find the other two corners.

Let me denote the midpoint as M(4.5, 2.5). Let me find two points P and Q such that M is the midpoint of both diagonals. So, given that one diagonal is between (4,2) and (5,3), then the other diagonal is between P and Q, with midpoint at M.

So if I take a vector perpendicular to (1,1), which is (-1,1) or (1,-1). Let me take vector (-1,1). Then, since the length of the diagonal is √2, the vector from midpoint M to P would be half of that vector. Wait, the entire diagonal is length √2, so half-diagonal is √2 / 2. The vector (-1,1) has length √2, so half of that would be (-1/2, 1/2). Therefore, point P would be M + (-1/2, 1/2) = (4.5 - 0.5, 2.5 + 0.5) = (4.0, 3.0). Similarly, point Q would be M + (1/2, -1/2) = (4.5 + 0.5, 2.5 - 0.5) = (5.0, 2.0).

Therefore, the square has corners at (4,2), (5,3), (4,3), and (5,2)? Wait, let me check:

Wait, original diagonal is from (4,2) to (5,3). The other diagonal is from (4,3) to (5,2). Let me check if those points form a square.

Looking at the points (4,2), (5,2), (5,3), (4,3). Wait, that's actually a square with side length 1, right? From (4,2) to (5,2) is 1 unit right, then up to (5,3), then left to (4,3), then down to (4,2). So yes, that's a square with side length 1, sitting between x=4 to 5 and y=2 to 3. So the square for point B is actually a unit square from (4,2) to (5,3). Therefore, point B lies within or on that square.

So now, point A is in the square with corners at (0,0), (2,0), (0,2), (2,2). So that's the square from (0,0) to (2,2). Point B is in the square from (4,2) to (5,3). So to maximize the slope of AB, we need to find points A and B such that the slope (y_B - y_A)/(x_B - x_A) is as large as possible.

Slope is rise over run. To maximize the slope, we need to maximize the rise and minimize the run, but also considering the direction. Wait, but slope is (y_B - y_A)/(x_B - x_A). So to maximize slope, which could be positive or negative. Since we want the maximum slope, which would be a positive number, or maybe even a very large negative? Wait, no, the maximum slope would be positive infinity? But since points are constrained, we can't have that. Wait, but given the positions of the squares, maybe the maximum slope occurs when A is as low as possible and B is as high as possible, but also considering the x-coordinates.

Wait, let's think. The slope of AB is (y_B - y_A)/(x_B - x_A). If we want to maximize this slope, we need to maximize the numerator (y_B - y_A) and minimize the denominator (x_B - x_A). However, if x_B - x_A is negative, then the slope becomes negative. But if we want the maximum slope, which is the maximum value, it's possible that we can have a very large positive slope if we can have a large positive numerator and a small positive denominator.

Alternatively, if we can have a large positive numerator and a very small positive denominator, that would give a large positive slope. But if the denominator is negative, then the slope would be negative. Alternatively, if the denominator is positive and numerator is positive, slope is positive. If numerator is positive and denominator negative, slope is negative. If numerator is negative and denominator negative, slope is positive. So maximum slope could be achieved in different ways.

But given the positions of the squares, let's analyze.

Point A is in the square [0,2] x [0,2].

Point B is in the square [4,5] x [2,3].

So, the x-coordinate of B is between 4 and 5, while the x-coordinate of A is between 0 and 2. Therefore, x_B - x_A is between 4 - 2 = 2 and 5 - 0 = 5. So all denominators are positive. Therefore, the slope would be positive if y_B - y_A is positive, which would be the case if y_B > y_A. If y_B < y_A, the slope is negative. Therefore, to maximize the slope, we need to maximize (y_B - y_A) / (x_B - x_A).

Therefore, to get the maximum slope, we need to maximize the numerator and minimize the denominator. However, since x_B - x_A is minimized when x_B is as small as possible and x_A is as large as possible. Similarly, y_B - y_A is maximized when y_B is as large as possible and y_A is as small as possible.

Therefore, to maximize the slope, we should take:

- y_B maximum: 3 (since B is in [4,5] x [2,3])

- y_A minimum: 0 (since A is in [0,2] x [0,2])

- x_B minimum: 4 (since B is in [4,5] x [2,3])

- x_A maximum: 2 (since A is in [0,2] x [0,2])

So then, the maximum slope would be (3 - 0)/(4 - 2) = 3/2.

But wait, is that the maximum? Let me confirm.

Alternatively, is there a case where even though the denominator isn't minimized, the numerator is larger? For example, if we can increase the numerator more by moving y_B up and y_A down, while keeping the denominator small. But since y_B can't be more than 3 and y_A can't be less than 0, that gives the maximum possible numerator. So (3 - 0) is 3. Then denominator is (4 - 2) = 2, so 3/2. Is that the maximum?

Wait, but let me check if there's a case where the slope is higher if we take different positions. For example, suppose point A is not at (2,0) but somewhere else, and point B is not at (4,3). Maybe if we take point A at (2,0) and point B at (4,3), the slope is 3/2. But what if point B is at (5,3) and point A is at (2,0)? Then the slope is (3 - 0)/(5 - 2) = 3/3 = 1, which is less than 3/2. So worse. If point A is at (2,0) and point B is at (4,2), the slope is (2 - 0)/(4 - 2) = 1, also less. If point A is at (2,0) and point B is at (4,3), slope is 3/2. If point A is at (something else) and point B is at (something else), can we get a higher slope?

Wait, suppose we take point A near (2,0) and point B near (4,3). Then slope is approaching 3/2. Suppose we take point A slightly to the left of (2,0), say (2 - ε, 0), and point B at (4, 3). Then slope is (3 - 0)/(4 - (2 - ε)) = 3/(2 + ε), which approaches 3/2 as ε approaches 0. So that seems maximum.

But is there a case where the slope is higher? For example, if we take point A at (0, something) and point B at (4, something else). Wait, but then the denominator would be 4 - 0 = 4, and the numerator would be y_B - y_A. If y_B is 3 and y_A is 0, then 3/4, which is less than 3/2.

Alternatively, if we take point A at (x, y) where x is small and y is small, and point B at (x', y') where x' is small and y' is large. Wait, but x' is between 4 and 5, so even if x is 0, x' - x is between 4 and 5. So the denominator is larger.

Alternatively, suppose we take point A at (something) and point B at (something) where the line AB has a steeper slope. Wait, but since the denominator is x_B - x_A, which is between 2 and 5. The maximum numerator is 3 - 0 = 3, so maximum slope is 3/2. Wait, but is there a possibility that the numerator is larger? Let me check.

Wait, for point B, the maximum y-coordinate is 3, and for point A, the minimum y-coordinate is 0. So the maximum possible difference in y is 3 - 0 = 3. So unless there is a case where y_B - y_A is larger than 3? No, since y_B can't be more than 3 and y_A can't be less than 0. Therefore, that's maximum.

Therefore, slope is 3 / 2. But wait, let me check the coordinates again. Suppose point A is at (2, 0) and point B is at (4, 3). Then the slope is (3 - 0)/(4 - 2) = 3/2. Is this the maximum?

But wait, maybe there's another combination where even though the difference in y is less, the difference in x is smaller, leading to a higher slope. For example, suppose point A is at (x_A, 0) and point B is at (x_B, 3), where x_A is as large as possible (2) and x_B is as small as possible (4). Then the slope is 3 / (4 - 2) = 3/2. But what if point A is not on the bottom edge?

Suppose point A is somewhere inside the square, and point B is somewhere inside the square. For example, suppose point A is (x, y) where y is less than 0? No, because A is within or on the square, so y can't be less than 0. Similarly, point B can't be above y=3. Therefore, the maximum difference in y is 3 - 0 = 3.

But perhaps if we take point A such that y_A is slightly less than 0? But it can't be, since the square is from y=0 to y=2. Similarly, point B can't be above y=3.

Alternatively, perhaps there's another orientation where slope is higher. Wait, slope can also be calculated if the line goes from a high point of A to a low point of B, but since slope is (y_B - y_A)/(x_B - x_A), if y_B - y_A is negative and x_B - x_A is negative, that would be positive. But since x_B is always greater than x_A (as x_B is between 4 and 5, and x_A is between 0 and 2), so x_B - x_A is always positive, so the denominator is positive. Therefore, to have positive slope, we need y_B - y_A positive, which is y_B > y_A. So, the maximum slope occurs when y_B is maximum and y_A is minimum, with x_B - x_A as small as possible. Hence, that would be when point A is at (2,0) and point B is at (4,3), giving slope 3/2.

But let me check another possibility. What if point A is not at (2,0) but somewhere else, and point B is not at (4,3), but somewhere else, such that the slope is higher?

For example, suppose we take point A at (2, something) and point B at (4, 3). Let me take A at (2, 0) gives slope 3/2. If I take point A at (2, ε) where ε > 0, then slope is (3 - ε)/(4 - 2) = (3 - ε)/2, which is less than 3/2. Similarly, if I take point A at (something less than 2, 0), say x_A = 2 - δ, then slope is (3 - 0)/(4 - (2 - δ)) = 3/(2 + δ), which is less than 3/2.

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

Alternatively, suppose point A is at (1, 0) and point B is at (4, 3). Then slope is 3/3 = 1, which is less than 3/2.

Alternatively, suppose point A is at (x, 0) and point B is at (x + 2, 3). Then the slope would be (3 - 0)/2 = 3/2. Wait, but x + 2 has to be ≤5, so x ≤ 3, but since x is between 0 and 2, that's okay. So for example, if x is 2, then x + 2 is 4, so that's point B at (4,3). If x is 1, then x + 2 is 3, but point B's x-coordinate is between 4 and 5, so 3 is not allowed. Therefore, the minimal x_B is 4, so x_A must be 4 - 2 = 2. Therefore, the maximum slope is achieved at x_A = 2, x_B = 4, giving slope 3/2.

Wait, but maybe if we take point B not on the top edge? Suppose point B is somewhere else. For example, suppose point B is at (4, 3) which is the top corner, and point A is at (2, 0), which is the bottom right corner of the first square. Then slope is 3/2. If we take point A at (2, 0) and point B at (4, something less than 3), then the slope would be less. Similarly, if we take point A at (something less than 2, 0) and point B at (4, 3), then the slope is 3/(4 - x_A). If x_A is less than 2, 4 - x_A is larger, so the slope is smaller. Therefore, that case gives less slope.

Alternatively, suppose we take point A not on the bottom edge but somewhere else, and point B not on the top edge, but such that (y_B - y_A)/(x_B - x_A) is greater than 3/2. Let me think. For example, suppose we take point A at (x, y) and point B at (x', y'), where y' - y is more than 3, but since y' can't exceed 3 and y can't be less than 0, that's impossible. So maximum y' - y is 3.

Therefore, 3 divided by minimal x' - x, which is 4 - 2 = 2, gives 3/2.

But wait, is there a case where the denominator x_B - x_A is smaller? For example, if x_B is 4 and x_A is 2, then x_B - x_A = 2. If we can have x_B - x_A less than 2? Well, since x_A is at least 0 and x_B is at least 4, then x_B - x_A is at least 4 - 2 = 2. So minimal value of denominator is 2, hence maximum slope is 3/2. Therefore, that seems to be the maximum.

But let me think again. Is there a possibility that if point A is not on the bottom edge and point B is not on the top edge, but the line AB has a steeper slope?

For example, suppose point A is at (x, y) and point B is at (x', y') where the slope is (y' - y)/(x' - x). Let's consider a line that goes from somewhere in the lower left of square A to somewhere in the upper right of square B. Wait, but square B is already to the right of square A. So the line from point A to point B would go from the left square to the right square. The maximum slope would be achieved when going from the lowest point of A to the highest point of B, but with the smallest horizontal difference. Which would be from the rightmost point of A (x=2, y=0) to the leftmost point of B (x=4, y=3). Which is exactly the case we considered before, slope 3/2.

But let me confirm with coordinates. If we take A(2, 0) and B(4, 3), slope is (3 - 0)/(4 - 2) = 3/2. If we take A(2, 0) and B(4, 3), that's valid because A is on the square (since (2,0) is a corner of the square from (0,0) to (2,2)), and B is on the square (since (4,3) is a corner of the square from (4,2) to (5,3)).

But let me check if there is a line that goes from some other point A and B with a steeper slope. For example, if point A is at (2, 0) and point B is at (4.5, 3), which is the upper right corner of square B? Wait, square B has corners at (4,2), (5,2), (5,3), (4,3). So (4.5, 3) is not a corner, but is inside the square. Wait, (4.5, 3) is outside the square because the square's y-coordinate only goes up to 3. Wait, actually, since the square's corners are (4,2), (5,2), (5,3), (4,3), then the square spans x from 4 to 5 and y from 2 to 3. So (4.5, 3) is on the top edge. So yes, it is inside the square. Then, if we take point B as (4.5, 3), then slope from A(2, 0) is (3 - 0)/(4.5 - 2) = 3 / 2.5 = 6/5 = 1.2, which is less than 3/2 = 1.5. Therefore, less steep.

Alternatively, what if point B is at (4, 3) and point A is at (2, 0). That gives 3/2. If we take point A at (something) and point B at (something else) such that the line AB is steeper?

Alternatively, suppose we take point A at (x, y) where y is as low as possible and point B at (x', y') where y' is as high as possible, but x' - x is smaller. Wait, but x' can't be less than 4 and x can't be more than 2. So the minimal x' - x is 4 - 2 = 2. Therefore, that is the minimal possible denominator.

Alternatively, could there be a case where the line passes through both squares in such a way that the slope is steeper? For example, maybe a line that starts at a lower y in square A and ends at a higher y in square B, but with a smaller horizontal distance?

Wait, but if you take point A at (2, 0) and point B at (4, 3), the horizontal distance is 2. If you take point A slightly to the left of 2, like (2 - ε, 0) and point B slightly to the right of 4, like (4 + ε, 3), then the horizontal distance is (4 + ε) - (2 - ε) = 2 + 2ε, which is larger. So the denominator increases, making the slope smaller. Therefore, that would decrease the slope.

Alternatively, suppose we take point A not at y=0 but somewhere else, but point B at some position where y' - y is larger? Wait, but y' cannot be larger than 3 and y cannot be smaller than 0, so that difference can't be larger than 3.

Wait, but perhaps if point A is at a lower y and point B is at a higher y relative to some other point? Wait, but no, since the maximum y for B is 3 and minimum y for A is 0, so that difference is fixed.

Alternatively, is there a case where the slope is not (y_B - y_A)/(x_B - x_A) but something else? No, that is the formula for slope between two points.

Wait, but maybe I need to consider the possibility that the line AB is tangent to both squares or something? But since we can choose any points within the squares, the maximum slope should be achieved at some corner points.

In general, the maximum or minimum slopes between two convex sets are achieved at the extreme points, i.e., the corners. So maybe we can check all combinations of corners of the two squares and find the maximum slope.

So let's try that approach. Let me list all the corners of square A and square B.

Square A has corners at (0,0), (2,0), (2,2), (0,2).

Square B has corners at (4,2), (5,2), (5,3), (4,3).

So there are 4 corners for A and 4 for B, so 16 possible combinations of points. Let me calculate the slope for each pair and see which is the maximum.

Starting with point A(0,0):

- To B(4,2): slope = (2 - 0)/(4 - 0) = 2/4 = 1/2

- To B(5,2): (2 - 0)/(5 - 0) = 2/5 = 0.4

- To B(4,3): (3 - 0)/(4 - 0) = 3/4 = 0.75

- To B(5,3): (3 - 0)/(5 - 0) = 3/5 = 0.6

So the maximum here is 3/4.

Next, point A(2,0):

- To B(4,2): (2 - 0)/(4 - 2) = 2/2 = 1

- To B(5,2): (2 - 0)/(5 - 2) = 2/3 ≈ 0.666...

- To B(4,3): (3 - 0)/(4 - 2) = 3/2 = 1.5

- To B(5,3): (3 - 0)/(5 - 2) = 3/3 = 1

So the maximum here is 3/2.

Next, point A(2,2):

- To B(4,2): (2 - 2)/(4 - 2) = 0/2 = 0

- To B(5,2): (2 - 2)/(5 - 2) = 0

- To B(4,3): (3 - 2)/(4 - 2) = 1/2 = 0.5

- To B(5,3): (3 - 2)/(5 - 2) = 1/3 ≈ 0.333...

Maximum here is 1/2.

Next, point A(0,2):

- To B(4,2): (2 - 2)/(4 - 0) = 0/4 = 0

- To B(5,2): same as above, 0

- To B(4,3): (3 - 2)/(4 - 0) = 1/4 = 0.25

- To B(5,3): (3 - 2)/(5 - 0) = 1/5 = 0.2

Maximum here is 1/4.

So among all the corner pairs, the maximum slope is 3/2, achieved when A is (2,0) and B is (4,3). Therefore, that seems to be the maximum. However, just to be thorough, maybe there are non-corner points that could give a higher slope? Let me think.

For example, suppose point A is somewhere along the bottom edge of square A, (x, 0), where x is between 0 and 2, and point B is somewhere along the top edge of square B, (4, y), where y is between 2 and 3. Then the slope would be (y - 0)/(4 - x) = y/(4 - x). To maximize this, we need to maximize y and minimize (4 - x). Since y can be up to 3 and (4 - x) can be as small as 4 - 2 = 2, giving 3 / 2 = 1.5, which is the same as the corner case. If we take y = 3 and x = 2, then we get 3/2. If we take y less than 3, then even with x = 2, the slope would be less. If we take x less than 2, then (4 - x) would be larger, leading to smaller slope.

Similarly, suppose point B is along the right edge of square B, (5, y), where y is between 2 and 3, and point A is along the bottom edge of square A, (x, 0). Then slope is (y - 0)/(5 - x). To maximize this, we need y as large as possible (3) and x as large as possible (2). So slope is 3/(5 - 2) = 1, which is less than 3/2.

If we take point B on the top edge and point A on the right edge? So A(2, y) and B(4, 3). Then slope is (3 - y)/(4 - 2) = (3 - y)/2. To maximize this, minimize y. y can be as low as 0, but since A is in square A, y can be 0. Then slope is (3 - 0)/2 = 3/2, same as before.

Alternatively, if point A is somewhere else. Suppose point A is at (x, y) and point B is at (x', y'), not restricted to edges. But since we already considered the maximum difference in y and minimal difference in x, and that case is achieved at the corners, I think that 3/2 is indeed the maximum.

But just to check another possibility: suppose we have a line that starts from (2, 0) and goes to some point inside square B, but not at the corner. For example, suppose point B is at (4 + ε, 3). Then slope is (3 - 0)/(4 + ε - 2) = 3/(2 + ε). As ε increases, this slope decreases. So maximum is when ε = 0, which is 3/2. Similarly, if point B is at (4, 3 - ε), then slope is (3 - ε - 0)/(4 - 2) = (3 - ε)/2, which is less than 3/2.

Alternatively, if we take point A not on the bottom edge. For example, suppose point A is at (2, ε) and point B is at (4, 3). Then slope is (3 - ε)/2. To maximize this, set ε as small as possible, which is 0. So again slope is 3/2.

Alternatively, suppose we take point A somewhere in the square and point B somewhere in the square such that the line has a steeper slope. Wait, but since we can't get a larger numerator or smaller denominator than 3 and 2, respectively, it's not possible. Therefore, 3/2 seems to be the maximum.

But just to be thorough, let me consider another approach. The maximum slope between two regions can be found by considering the extremal points. Since both squares are convex, the maximum slope should be achieved either at two corners or at a corner and a boundary point, or something like that. However, we already checked all the corner combinations and found 3/2 as maximum. So unless there is a case where a non-corner point gives a higher slope, but I can't think of any.

Wait, but let's think about the line equations. Suppose we have a line that goes through square A and square B. What's the maximum slope of such a line?

The line needs to pass through some point A in square A and some point B in square B. The slope is (y_B - y_A)/(x_B - x_A). Since x_B ≥ 4 and x_A ≤ 2, x_B - x_A ≥ 2. Also, y_B ≤ 3 and y_A ≥ 0, so y_B - y_A ≤ 3. So maximum slope is 3/2, as before. Therefore, 3/2 is indeed the maximum.

But let me think again. Is there a way to have y_B - y_A larger than 3? For example, if point A is below the square? But no, A is within the square from (0,0) to (2,2), so y_A ≥ 0. Similarly, point B is within the square from (4,2) to (5,3), so y_B ≤ 3. So maximum difference is 3 - 0 = 3.

Therefore, I think that 3/2 is indeed the correct answer.

Wait, but let me check with another method. Suppose we parametrize the points. Let me suppose point A is (a, b) where 0 ≤ a ≤ 2, 0 ≤ b ≤ 2. Point B is (c, d) where 4 ≤ c ≤ 5, 2 ≤ d ≤ 3. Then slope is (d - b)/(c - a). To maximize this, we need to maximize d - b and minimize c - a. The maximum of d - b is 3 - 0 = 3. The minimum of c - a is 4 - 2 = 2. Hence, maximum slope is 3/2.

Yes, that confirms again that 3/2 is the maximum. Therefore, the answer is 3/2.

But just to be 100% sure, let me think if there is any other way. For example, suppose we take a line that goes through both squares and has a steeper slope. But given the positions of the squares, the steepest slope would be achieved by going from the lowest point of A to the highest point of B, as we did. So I think that is correct.

Alternatively, suppose someone argues that if point A is at (0,0) and point B is at (4,3), the slope is 3/4, which is less than 3/2. So that's not helpful. Similarly, other combinations. Therefore, I think 3/2 is correct.

Therefore, the greatest possible value of the slope is 3/2, which is 3 over 2, so expressed as a common fraction, it's \boxed{\dfrac{3}{2}}.

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

To determine the **greatest possible value of the slope** of the line connecting points $ A $ and $ B $, where:

- Point $ A $ lies within or on the square with opposite corners at $ (0, 0) $ and $ (2, 2) $
- Point $ B $ lies within or on the square with opposite corners at $ (4, 2) $ and $ (5, 3) $,

we analyze the geometry of the two regions.

---

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

**For Point A**:
The square from (0, 0) to (2, 2) is a square with side length 2, and its boundaries are:
- $ x \in [0, 2] $
- $ y \in [0, 2] $

**For Point B**:
The square has corners at $ (4, 2) $, $ (5, 2) $, $ (5, 3) $, and $ (4, 3) $. This is a unit square with side length 1, centered at $ (4.5, 2.5) $, and its coordinates satisfy:
- $ x \in [4, 5] $
- $ y \in [2, 3] $

Since $ x_B > x_A $ and $ y_B \ge y_A $, the slope of the line $ AB $ is non-negative.

---

### **Step 2: Maximize the Slope**

The slope between points $ A = (x_1, y_1) $ and $ B = (x_2, y_2) $ is:

$$
\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}
$$

To **maximize this**, we want:

- **Maximize the numerator**: Choose $ y_2 = y_{\max} = 3 $ and $ y_1 = y_{\min} = 0 $
- **Minimize the denominator**: Choose $ x_2 = x_{\min} = 4 $ and $ x_1 = x_{\max} = 2 $

So the maximum possible slope is:

$$
\text{slope}_{\max} = \frac{3 - 0}{4 - 2} = \frac{3}{2}
$$

---

### **Step 3: Verify with Corners**

The squares have 4 corners each. Evaluating all combinations of corners for both squares confirms that the **maximum slope** occurs when:

- $ A = (2, 0) $ (bottom-right corner of square A)
- $ B = (4, 3) $ (top-left corner of square B)

This gives:

$$
\text{slope} = \frac{3 - 0}{4 - 2} = \frac{3}{2}
$$

All other combinations of points from the squares result in a smaller slope. Thus, no point within the squares produces a larger slope than this value.

---

### ✅ **Final Answer**

$$
\boxed{\frac{3}{2}}
$$

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 $\boxed{\frac{3}{2}}$.