This was the question we answered in my calculus class.
First I showed this applet and had students play with it
Next I asked what do we need to know?
Students asked for the width of the hallways which are 0.8 m and 0.9 m wide.
Next we realized that actually to determine the largest TV we actually need to MINIMIZE the length of the line. As the smallest line will be the line that can fit around the entire corner.
Calling the, angle between the TV and the 0.8 m wall, theta you get the equation of the TV length at any angle to be
Substituting this back into the equation gives us a TV (or any rigid object) with a length of 2.4 m or 94.45 in across.
We then did have a discussion around what assumptions are we making? Some are...
- The TV has no depth at all
- The TV will scrap across the wall
- The TV is out of the box