Key Concepts and Formulas

• Area of a Triangle: Given vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, the area is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. If one vertex is at the origin, the area simplifies to $\frac{1}{2} |x_2y_3 - x_3y_2|$.
• Optimization using Derivatives: To find the maximum or minimum of a function $f(x)$, find the critical points by solving $f'(x) = 0$. Use the second derivative test: if $f''(x) < 0$, it's a local maximum; if $f''(x) > 0$, it's a local minimum.

Step-by-Step Solution

Step 1: Define the Vertices We are given the vertices of the triangle as $(0,0)$, $(x, y)$, and $(-x, y)$, where the points $(x, y)$ and $(-x, y)$ lie on the curve $y = -2x^2 + 54$ and $y > 0$.

Step 2: Calculate the Area of the Triangle Using the formula for the area of a triangle with one vertex at the origin: $\text{Area} = \frac{1}{2} |x_2 y_3 - x_3 y_2|$ Substituting the coordinates $(x_2, y_2) = (x, y)$ and $(x_3, y_3) = (-x, y)$: $\text{Area} = \frac{1}{2} |x \cdot y - (-x) \cdot y|$ $\text{Area} = \frac{1}{2} |xy + xy|$ $\text{Area} = \frac{1}{2} |2xy|$ $\text{Area} = |xy|$ Since $y > 0$, we have $\text{Area} = |x|y$.

Step 3: Express the Area as a Function of x We are given $y = -2x^2 + 54$. Substitute this into the area equation: $\text{Area}(x) = |x|(-2x^2 + 54)$ Also, we know $y > 0$, so $-2x^2 + 54 > 0$. This implies: $2x^2 < 54$ $x^2 < 27$ $-\sqrt{27} < x < \sqrt{27}$ Since the function is symmetric (even), we can consider only the case when $x > 0$. Then, $0 < x < \sqrt{27}$, and $|x| = x$. Therefore, $\text{Area}(x) = x(-2x^2 + 54) = -2x^3 + 54x$.

Step 4: Find the Critical Points To find the maximum area, we differentiate $\text{Area}(x)$ with respect to $x$ and set the result equal to zero: $\text{Area}'(x) = \frac{d}{dx}(-2x^3 + 54x) = -6x^2 + 54$ Setting $\text{Area}'(x) = 0$: $-6x^2 + 54 = 0$ $6x^2 = 54$ $x^2 = 9$ $x = \pm 3$ Since we are considering the case where $x > 0$, we take $x = 3$.

Step 5: Verify Maximum Using the Second Derivative Test Calculate the second derivative: $\text{Area}''(x) = \frac{d}{dx}(-6x^2 + 54) = -12x$ Evaluate the second derivative at $x = 3$: $\text{Area}''(3) = -12(3) = -36$ Since $\text{Area}''(3) < 0$, the area function has a local maximum at $x = 3$.

Step 6: Calculate the Maximum Area Substitute $x = 3$ into the area function: $\text{Area}(3) = -2(3)^3 + 54(3) = -2(27) + 162 = -54 + 162 = 108$

Common Mistakes & Tips

• Remember to check the domain of the function based on the constraint $y > 0$.
• When dealing with absolute values, consider different cases to simplify the problem. In this case, leveraging the symmetry of the problem simplified the calculations.
• Always verify that the critical point corresponds to a maximum (or minimum) using the second derivative test.

Summary We found the maximum area of the triangle by expressing the area as a function of x, taking its derivative, finding the critical points, and verifying that the critical point corresponds to a maximum. By substituting the x-value of the critical point into the area equation, we calculated the maximum area to be 108.

Final Answer The final answer is \boxed{108}, which corresponds to option (A).