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

Solution: To find the maximum area of the triangle with vertices at (0, 0), (x, y), and (−x, y) where \( y = -2x^2 + 54 \), we can proceed as follows:
The base of the triangle is the distance between (x, y) and (−x, y), which is 2x.
The height of the triangle is y, which is the distance from the origin (0, 0) to the line joining (x, y) and (−x, y).
Thus, the area \( \Delta \) of the triangle is:
\[\Delta = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2x \times y = x \times y\]

Since \( y = -2x^2 + 54 \), we can substitute this into the area expression:
\[\Delta = x \times (-2x^2 + 54) = -2x^3 + 54x\]

To maximize \( \Delta \), we take the derivative with respect to \( x \) and set it to zero:
\[\frac{d\Delta}{dx} = -6x^2 + 54 = 0\]
\[-6x^2 = -54 \Rightarrow x^2 = 9 \Rightarrow x = 3 \quad (\text{since } y > 0) \]

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

Thus, the maximum area is:
\[ \Delta = x \times y = 3 \times 36 = 108 \]