Leaning Ladder

Let the horizontal distance from $A$ to the box be $x$ and the vertical distance from $B$ to the box be $y$ . Then by the similarity of the two blue triangles framed by the ladder, box and wall we see that

$\dfrac{2}{x} = \dfrac{y}{2} \Longrightarrow y = \dfrac{4}{x}$ .

Next, by Pythagoras we have that $(3\sqrt{5})^{2} = (2 + x)^{2} + (2 + y)^{2}$

$\Longrightarrow 45 = 4 + 4x + x^{2} + 4 + 4y + y^{2} = 8 + 4x + x^{2} + \dfrac{16}{x} + \dfrac{16}{x^{2}} = \left(x^{2} + \dfrac{16}{x^{2}} + 8\right) + \left(4x + \dfrac{16}{x}\right) = \left(x + \dfrac{4}{x}\right)^{2} + 4\left(x + \dfrac{4}{x}\right)$ .

Letting $y = x + \dfrac{4}{x}$ we have that $y^{2} + 4y - 45 = 0 \Longrightarrow (y + 9)(y - 5) = 0 \Longrightarrow y = 5$ as we must have $y \gt 0$ .

So $y = x + \dfrac{4}{x} = 5 \Longrightarrow x^{2} - 5x + 4 = 0 \Longrightarrow (x - 4)(x - 1) = 0$ , so either $(x,y) = (4,1)$ or $(x,y) = (1,4)$ .

Finally, given the assumption that $y \gt x$ we see that $y = 4$ and thus the right of the ladder is $y + 2 = 4 + 2 = \boxed{6}$ m.