1) Can the box be entirely cleaned?

2) If not, how much area will be missed?

3) If so, how many dips are required?

4) If just 2 dips are required, can they be done without having to turn the longer box upside down?

I currently just use a CAD software to find the answer through trial-and-error but academically I'd like to know a concise mathematical solution for it.

First, let's assume we can ignore depth and just focus on length and width.

I'm not sure how to calculate Ax_offset without trial-and-error. The angle theta is based on the offset, but the offset is based on the angle.

Then it seems like a long route to calculate the remaining items to start answering questions.

If Bx_top_in is greater than half of Bx, I think it can be dipped in 2 dips without having to flip the box upside down.

If Bcenter falls inside of the small box but Bx_top_in is less than half of Bx, then I think it can be dipped in 2 dips but the longer box must be flipped upside down on the 2nd dip.

I can't currently imagine a scenario where 4 dips would accomplish anything 2 dips (with a flip) wouldn't.

Can anyone help with some strategies for solving (long-hand... for fun) the original questions.

Thanks in advance,

Mike