2006 AMC 10B Problem 5

This asks for the smallest square, so there are two halves: show a size-5 square works, then show a size-4 square cannot.

Part 1: a $5 \times 5$ square is enough.

1. Stand the $3 \times 4$ rectangle up so it is $3$ wide and $4$ tall.
2. Put the $2 \times 3$ rectangle right beside it, standing $2$ wide and $3$ tall.
3. Together they are $3 + 2 = 5$ wide, and the taller one is $4$ tall. A block $5$ wide and $4$ tall fits inside a $5 \times 5$ square, with room to spare.

Part 2: a $4 \times 4$ square is too small.

1. The rectangles have areas $2 \cdot 3 = 6$ and $3 \cdot 4 = 12$, so together they need $18$ square units of space.
2. A $4 \times 4$ square has only $16$ square units. Since they cannot overlap, they do not fit. Any smaller square is even worse.

Among the answer choices, $16$ is too small and $25$ works, so the smallest area is $5^2 = 25$.

The answer is $\boxed{\textbf{(B)}\ 25}$.

"Smallest possible" needs both an example that works and a reason nothing smaller works. Comparing areas is the fastest way to rule out small squares. To build the example, line the rectangles up so their short sides add together, keeping the overall shape as close to square as possible.

Lower bound: the rectangles have areas $6$ and $12$, so the square must have area at least $18$. A $4 \times 4$ square has area $16 < 18$, so side $4$ is impossible, and any smaller side is too.

Construction for side $5$: stand the $3 \times 4$ rectangle with its $3$-side horizontal and $4$-side vertical, and stand the $2 \times 3$ rectangle next to it with its $2$-side horizontal and $3$-side vertical. Together they span width $3 + 2 = 5$ and height $\max(4, 3) = 4$, which fits inside a $5 \times 5$ square.

So the smallest square has area $5^2 = 25$.

The answer is $\boxed{\textbf{(B)}\ 25}$.