math
posted by saman on .
Ex. 120 m of fencing is to be used to form three sides of a rectangular enclosure , the fourth side being an existing wall . Find the maximum possible area of the enclosure

L + 2 W = 120 so L = 120  2 W
A = L W
A = (1202W)W = 120 W  2 W^2
so
2 W^2 120 W =  A
W^2  60 W = A/2
W^2  60 W + 900 = A/2 + 900
(W30)^2 = (1/2)(A1800)
vertax of parabola at (30,1800)
so
W = 30
L = 120  60 = 60
and A = 1800 m^2