Answer
(a) A solution is possible
(b) Refer to the picture below
(c) $x ≈ 15.49 ft.$
(d) $L ≈ 408.67 ft$
Work Step by Step
(a) The length of a track with straightaways of length $L$ and semicircles of radius $r$ is $P = (2)L + (2)(πr) ft$. Let $L = 360$ and $r = 80$ to get $P = 720 + 160π ≈ 1222.65 ft$. Since this is less than $1320 ft$ (a quarter-mile), a solution is possible.
(b) $P = 2L + 2πr = 1320$ and $2r = 2x + 160$, so $L = (1320 − 2πr)/2 = (1320 − 2π(80 + x))/2 = 660 − 80π − πx.$
Now the expression for $L$ helps us to plot the graph of $L$ versus $x$ as shown in the graph (refer to the graph below)
(c) Using the graph below, the shortest straightaway is $L = 360$, so we solve the equation $360 = 660 − 80π − πx$ to obtain $x = 300 π − 80 ≈ 15.49 ft.$
(d) Using the graph below, the longest straightaway occurs when $x = 0$, so $L = 660 − 80π ≈ 408.67 ft$
