a. A=LW where L=length of rectangular area, W=width.
b. 150=2L+W.
c. W=150-2L. But the true perimeter of the area is P=2L+2W, so W=(1/2)(P-2L).
d. A=L(P-2L)/2=LP/2-L^2.
e. f(L)=-L^2+LP/2-A.
f. -b/2a=(-P/2)/(-2)=P/4. This is the vertex of f(L), and L=P/4 is the value for the maximum area.
g. From (c) W=(1/2)(P-2L)=P/4, so W=L=P/4. This makes the area a square of area P^2/16.
From (b) 150=2L+W=P/2+P/4=3P/4, making P=4/3*150=200, so W=L=200/4=50 yd.
h. Maximum area is 50^2=2500 sq yd.