
Let, area of triangle ABC be denoted by P.
⇒P=21∣0x−x0yy111∣
⇒P=21∣0−0+2xy∣
⇒P=∣xy∣
⇒P=∣x(−2x2+54)∣
⇒P=∣−2x3+54x∣
⇒dxdP=∣−6x2+54∣
For critical points, dxdP=0
⇒∣−6x2+54∣=0
⇒x=±3
So, the maximum area will be,
A=3(−2×9+54)
⇒A=108
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 :
Held on 30 Jan 2024 · Verified 6 Jul 2026.
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