360 cubes of 1cm are arranged to make a cuboid. The largest base area of the cuboid is a perfect square. Find the smallest possible height of the cuboid. Given 4k + 3r2 = 10, make r2 the subject of the formula.

Since 360 cubes of 1 cm are used, the volume of each cube is $1 \text{ cm} \times 1 \text{ cm} \times 1 \text{ cm} = 1 \text{ cm}^3$.
The total volume of the cuboid is $360 \times 1 \text{ cm}^3 = 360 \text{ cm}^3$.

The volume of a cuboid is given by $V = \text{Base Area} \times \text{Height}$.
So, $\text{Height} = \frac{\text{Volume}}{\text{Base Area}}$.
To find the smallest possible height, we need to find the largest possible base area.

The volume is $360 \text{ cm}^3$. We need to find the largest perfect square that divides 360.
Let's list the perfect square factors of 360:

• $1^2 = 1$ (360 is divisible by 1)
• $2^2 = 4$ (360 is divisible by 4, $360 \div 4 = 90$)
• $3^2 = 9$ (360 is divisible by 9, $360 \div 9 = 40$)
• $4^2 = 16$ (360 is not divisible by 16)
• $5^2 = 25$ (360 is not divisible by 25)
• $6^2 = 36$ (360 is divisible by 36, $360 \div 36 = 10$) The next perfect square is $7^2 = 49$, which is not a factor of 360. The largest perfect square factor of 360 is 36. Therefore, the largest base area is $36 \text{ cm}^2$.

The smallest possible height of the cuboid is $\boxed{10 \text{ cm}}$.

The formula with $r^2$ as the subject is $\boxed{r^2 = \frac{10 - 4k}{3}}$.