Solution
Maximize the area. The largest figure would be a circle. However, let's check a rectangle.
\(A_c = \pi() r^2\)
\(C = 2\pi()r\)
\(20 = 2 \pi() r\)
\(r = 3.18309886184\)
\(A_c = \pi() (3.18309886184)^2\)
\(A_c = 31.83 m^2\)
Let's Check a rectangle.
\(P=2l+2w\)
\(l = 10-w\)
\(A = lw\)
\(A = 10w -w^2\)
\(0=10-2w\)
\(w= 5\)
\(A = 10 (5) - 5^2 = 25 sq.m.\). This value is lower than the area of the circle.
\( x = 31.83 m^2 \)