Parabola
Maximum Area of Triangle with Vertex at Origin and Two Points on Parabola
nta_pyq_2024_jan
Grade 11
Question:
The maximum area of a triangle whose one vertex is at $(0,0)$ and the other two vertices lie on the curve $y=-2x^2+54$ at points $(x,y)$ and $(-x,y)$ where $y>0$ is:
Step-by-Step Solution
Key Concept: For vertices $(0,0)$, $(x,y)$, $(-x,y)$: area $=\frac{1}{2}|2x|\cdot y=|xy|$. Maximize $|xy|=|x(-2x^2+54)|=|x||54-2x^2|$ subject to $y>0$ (so $x^2<27$). Differentiate: $\frac{d}{dx}(54x-2x^3)=54-6x^2=0\Rightarrow x=3$.
Area $=xy=x(-2x^2+54)$. Max at $x=3$: Area $=3(54-18)=108$.
Correct Answer: 4