, 2019

Solution:

The greatest possible value for the product is $3 \cdot 4 \cdot 4 \cdot 6 \cdot 8 = 2304$, achieved when the rectangles are $3 \times 1$, $1 \times 4$, $2 \times 2$, $2 \times 3$, $4 \times 2$. To see that this is possible, orient these rectangles so that the first number is the horizontal dimension and the second number is the vertical dimension. Then, place the bottom-left corners of these rectangles at $(2,4)$, $(4,0)$, $(2,2)$, $(0,2)$, $(0,0)$ respectively on the grid.

We will now prove that no larger product can be achieved. Suppose that there is at least one rectangle of area at most $2$. Then the product is at most $2 \cdot 5.75^{4} = 2 \cdot 33.0625^{2} < 2 \cdot 1100 = 2200$ by AM-GM. Now suppose that there is at least one rectangle of area at least $9$. Then the product is at most $9 \cdot 4^{4} = 2304$ by AM-GM. (Neither of these is tight, since you cannot have non-integer areas, nor can you have four rectangles all of area $4$.)

Now consider the last possibility that is not covered by any of the above: that there are no rectangles of size at most $2$ and no rectangles of area at least $9$. There can be at most one rectangle of area $3, 5, 6, 8$ each, at most two rectangles of area $4$, and no rectangles of area $7$. The only way to achieve a sum of $25$ with these constraints is $3, 4, 4, 6, 8$, which produces a product of $2304$. We have shown through the earlier cases that a larger product cannot be achieved, so this is indeed the maximum.