Answer
(a) $L=2x+\frac{200}{x}$
(b) minimum $L=40m$ when $x=10m$ and $y=10m$
Work Step by Step
(a) Assume the length of fencing is $x$ and the width is $y$, we have $xy=100$ and $y=\frac{100}{x}$
The total length of fencing $L=2x+2y=2x+\frac{200}{x}$
(b) Graph the above function as shown in the figure. A minimum can be found $L=40m$ when $x=10m$
which gives $y=10m$