Answer
a yes
b largest possible sail area is 3291 ft$^{2}$
Work Step by Step
a) substitute 60 for L , 3400 for A and 650 for L in the equation $0.3*L+.38\sqrt {A}-3\sqrt[3]{V}$
you get $0.3*60+.38*\sqrt {3400}-3\sqrt[3]{650}=14.17$
Since $14.17\leq16$ the boat qualify for the race
b) when substitute the known value in the equation $0.3*L+.38\sqrt {A}-3\sqrt[3]{V}$ and we get
$16=.3*65+.38*\sqrt A -3\sqrt[3]{600}$
We solve for a by subtract $.3*65-3\sqrt[3]{600}$ from both sides and get
$16-( .3*65-3\sqrt[3]{600})=.38*\sqrt A$
we simplify the right side
$21.8=.38*\sqrt A$
divide both sides by .38
$57.37=\sqrt A$
Square both sides
3291=A
