Maximizing the Area of an Inscribed Symmetric Triangle

Q: Why is the base equal to 2x?

The points are (x, y) and (-x, y), which have the same y-coordinate. The horizontal distance between them is |x - (-x)| = 2x for x > 0.

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:

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A88B122C92D108

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Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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Why is the base equal to 2x?

The points are $(x, y)$ and $(- x, y)$ , which have the same y-coordinate. The horizontal distance between them is $∣ x - (- x) ∣ = 2 x$ for $x > 0$ .

Answer

The maximum area of the triangle is 108 square units, achieved when $x = 3$ .

Why each option works or fails

A: 88 - Incorrectly computing the value of $A (3)$ after finding the critical point $x = 3$ , or taking $x = 2$ without differentiating correctly. Evaluate $A (x) = x (54 - 2 x^{2})$ precisely at the critical point $x = 3$ : $A (3) = 3 (54 - 18) = 3 (36) = 108$ .
B: 122 - Adding rather than subtracting the cubic term or miscalculating derivative critical values. Set $A^{'} (x) = 54 - 6 x^{2} = 0$ , giving $x^{2} = 9 ⟹ x = 3$ , leading to $A (3) = 108$ .
C: 92 - Making an arithmetic error in the final multiplication $3 \times 36$ or using an incorrect base-length expression. The base length between $(- x, y)$ and $(x, y)$ is $2 x$ , so $Area = \frac{1}{2} (2 x) (y) = x (54 - 2 x^{2})$ . At $x = 3$ , $3 \times 36 = 108$ .
D · correct: 108 - This is the correct maximum area. Express area as $A (x) = x (54 - 2 x^{2})$ , find the critical point $x = 3$ by setting $A^{'} (x) = 54 - 6 x^{2} = 0$ , and evaluate $A (3) = 108$ .

Step-by-step solution

Given: Vertices of the triangle are $O (0, 0)$ , $A (x, y)$ , and $B (- x, y)$ , with $y = - 2 x^{2} + 54$ and $y > 0$ . Assume $x > 0$ without loss of generality.
Goal: Find the maximum possible area of the triangle.
Approach: The base of the triangle lies along the horizontal line segment connecting $(- x, y)$ and $(x, y)$ , giving base length $b = 2 x$ . The altitude to the vertex at $(0, 0)$ is simply the y-coordinate $y$ . Thus, the area is $A (x) = \frac{1}{2} (2 x) (y) = x y = x (54 - 2 x^{2})$ . We maximize $A (x)$ with respect to $x$ using first derivative.
Execute: Set $\frac{d A}{d x} = 0$ : $A (x) = 54 x - 2 x^{3} ⟹ \frac{d A}{d x} = 54 - 6 x^{2} = 0 ⟹ x^{2} = 9 ⟹ x = 3$ . Evaluating the maximum area: $A (3) = 3 (54 - 2 (3^{2})) = 3 (54 - 18) = 108$ .
Verify: Check second derivative: $\frac{d^{2} A}{d x^{2}} = - 12 x$ . At $x = 3$ , $\frac{d^{2} A}{d x^{2}} = - 36 < 0$ , confirming a local maximum. At $x = 3$ , $y = 36 > 0$ , satisfying the problem constraint.

Maximizing the Area of an Inscribed Symmetric Triangle

What feels right?

The maximum area of the triangle is 108 square units, achieved when $x = 3$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

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