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

Answer: (B)

108

Answer 2

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:
$\Delta = x \times y = 3 \times 36 = 108$