If Paul is using a piece of fencing 80 meters long to build a rectangular enclosure for his dog, what is the greatest possible area that can be enclosed?

**Long way:**

Say the sides are *x* and *y* long. The perimeter of the fence must be 80, so 2*x* + 2*y* = 80. The area of the fence will be *xy*.

We can use the first equation to get the second equation in terms of only one variable!

2*x* + 2*y* = 80

*y* = 40 – *x*

Now you can say that the area of the rectangle will be *x*(40 – *x*). You can graph that if you want, or you can just recognize that it will be a downward facing parabola with roots at 0 and 40. That means it’ll have its maximum when *x* = 20. That means the greatest possible area will be when *x* = 20, when the area will equal 400 square meters.

**Shortcut: **

Just remember that in a case like this, the biggest rectangular area you can enclose is with a square. If you know the perimeter needs to be 80, then make a 20×20 square.