PartialMathematician

Ok then... :P

We also do not know if the 5m wire was cut into 2 equal length wires or 2 random wires. 

Actually, since the wire is cut into two pieces (that are equal, I assume), the perimeter of the square is 5/2 meters or 250 cm. Each side of the square would then be 250/4 cm, which equals 62.5 cm. \(62.5^2\)  is 3906. The total area of the square would then be 3906.25 \(cm^2\) or 0.390625 \(m^2\). 

The minimum area would be 0...just don't place the wire.

I think you meant to state the question differently...

Hope this sort of helps,

- PartialMathematician

Ok then...

Please do not ask or post offensive or disrespectful questions on this website. 

You can put that into the "LaTeX" box to make the GCD part look nicer. 

In the proof, 0 is restricted because the denominator of a fraction cannot be 0 (cannot divide by 0). If we set x to 1, then the part \(3f(\dfrac{x-1}{x})\) cancels out and turns into 0 because \(3f(0) = 0\). Also, if you look around in the proof, you will find \(x\)  and \(1-x\) many times as the denominator, which cannot equal 0. Simply stated, 0 and 1 are restricted, AKA not in the domain of the function. 

Hope this helps,

- PartialMathematician

Ok, Guest, I'll see if I can help you. 

Ok, thanks, CoolStuffYouTube. 

You probably saved a grade...