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 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:
\[ \Delta = x \times y = 3 \times 36 = 108 \]

Answer 2

The problem involves finding the maximum area of a triangle with a vertex at the origin \((0, 0)\) and the other two vertices on the parabola defined by \(y = -2x^2 + 54\). The symmetry in the problem arises because the vertices are \((x, y)\) and \((-x, y)\), meaning the triangle's base is along the x-axis, and its height is perpendicular to it.

1. Given the curve \(y = -2x^2 + 54\), substitute the coordinates of the two vertices into this equation to maintain their alignment on the curve.
2. The area \(A\) of the triangle with a vertex at the origin and a base along the x-axis will be given by: \(A = \frac{1}{2} \times \text{base} \times \text{height}\).
3. Here, the base is the horizontal distance between \((x, 0)\) and \((-x, 0)\), which is \(2x\), and the height is \(y\). \(A = \frac{1}{2} \times 2x \times y = x \times y\).
4. Substitute for \(y\): \(y = -2x^2 + 54\), giving: \(A = x(-2x^2 + 54) = -2x^3 + 54x\).
5. To find the maximum area, differentiate \(A\) with respect to \(x\) and set it to zero: \(\frac{dA}{dx} = -6x^2 + 54 = 0\). Solving gives: \(-6x^2 = -54 \Rightarrow x^2 = 9 \Rightarrow x = 3 \text{ or } x = -3\).
6. Choose \(x = 3\) (since symmetry and maximum areas are positive). Compute \(y\) at this point using: \(y = -2(3)^2 + 54 = -18 + 54 = 36\).
7. The area \(A\) at \(x = 3\) is: \(A = 3 \cdot 36 = 108\).

Thus, the maximum area of the triangle is 108 square units.