Question 335398
Typically, running tracks are a rectangle with two half-circles attached at each end.

For the two sides of the rectangle that do not have the half-circles on the end, 200 meters of the fence will be used. This leaves 430 meters to be used for the circles.

Since the width of the track is 81 m, the radius of each half-circle is 81/2 m. However, if the gap is x meters in width, the result will be a circle with radius 81/2 + x.

To maximize the gap, we will need to use all of the fence. Thus, we have:

2π(81/2 + x) = 430 (the circumference of each half-circle is C = πr)
==> π(81 + 2x) = 430
==> 2x + 81 = 430/π
==> 2x = 430/π - 81
==> x = 215/π - 81/2 ≈ 28 m.