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:

• A. 88
• B. 108
• C. 92
• D. 122

Answer 1

The Correct Option is B

To determine the maximum area of a triangle with vertices at (0, 0), (x, y), and (−x, y), given the relation \( y = -2x^2 + 54 \), we follow these steps:
The triangle's base is the distance between (x, y) and (−x, y), which equals 2x.
The triangle's height is y, representing the perpendicular distance from the origin (0, 0) to the line segment connecting (x, y) and (−x, y).
The area \( \Delta \) of the triangle is calculated as:
\[\Delta = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2x \times y = x \times y\]

Substituting \( y = -2x^2 + 54 \) into the area formula:
\[\Delta = x \times (-2x^2 + 54) = -2x^3 + 54x\]

To find the maximum \( \Delta \), we differentiate with respect to \( x \) and set the derivative to zero:
\[\frac{d\Delta}{dx} = -6x^2 + 54 = 0\]
Solving for x: \( -6x^2 = -54 \Rightarrow x^2 = 9 \Rightarrow x = 3 \) (assuming \( y>0 \)).

Now, substitute \( x = 3 \) into the equation for \( y \):
\[ y = -2(3)^2 + 54 = -18 + 54 = 36 \]

The maximum area is therefore:
\[ \Delta = x \times y = 3 \times 36 = 108 \]