The maximum area of a triangle whose one vertex is at (0,0) and the other two vertices lie on the curve y=-2x2+54 at points (x, y) and (-x, y), where y>0, is :
- 108
- 122
- 88
- 92
Answer
(1) 108
Solution
[figure]
Area = 12 = 12(2x)(y)=xy
A(x)=x (54-2x2 )=54x-2x3
dAdx=54-6x2=0 x=3; d2Adx2=-12x<0 (maximum)
A =3(54-18)=108
Answer: (1)