The maximum area of a triangle whose one vertex is at (0,0) and the other two vertices lie…

Question

The maximum area of a triangle whose one vertex is at $(0,0)$ and the other two vertices lie on the curve $y=-2 x^2+54$ at points $(x, y)$ and $(-x, y)$, where $y>0$, is :

Accepted Answer

Area of $$\Delta$$

$$\begin{aligned}
& =\frac{1}{2}\left|\[\begin{array}{ccc} 0 & 0 & 1 \ x & y & 1 \ -x & y & 1 \end{array}\end{aligned}$$\right| \
\& \Rightarrow\left|\frac{1}{2}(x y+x y)\right|=|x y| \
\& \operatorname{Area}(\Delta)=|x y|=\left|x\left(-2 x\^2+54\right)\right| \
\& \frac{d(\Delta)}{d x}=\left|\left(-6 x\^2+54\right)\right| \Rightarrow \frac{d \Delta}{d x}=0 \text { at } x=3 \
\& \text { Area }=3(-2 \times 9+54)=108
\]

Solutions · 1