The maximum area of a triangle whose one vertex is at (0, 0) and the other two vertices lie on the curve y = –2x² + 54 at points (x, y) and (–x, y), where y > 0, is

The correct answer is 108.

We are given a triangle with one vertex at the origin (0, 0) and the other two vertices on the parabola y = -2x² + 54, specifically at points (x, y) and (-x, y), with y > 0.

Because the two vertices on the parabola are symmetric about the y-axis (one at x and one at -x, both with the same y-coordinate), the triangle is isosceles. We can easily identify its base and height:

Base = horizontal distance between (x, y) and (-x, y) = 2x

Height = vertical distance from (0, 0) to the horizontal line y = constant = y

So the area of the triangle is:

A=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 2x\times y=xy

Now substitute the curve equation y = -2x² + 54 into the area formula:

A\left(x\right)=x(-2x^2+54)=-2x^3+54x

To find the maximum, differentiate with respect to x and set the derivative to zero:

\frac{\mathrm{dA}}{\mathrm{dx}}=-6x^2+54=0

6x^2=54\Rightarrow x^2=9\Rightarrow x=3

(We take x = 3 since x must be positive for a valid triangle.)

Verify it's a maximum using the second derivative:

\frac{d^2}{A}=-12x

At x = 3: -12\left(3\right)=-36<0 ✓ (confirms a maximum)

Calculate the maximum area:

First find y at x = 3:
y = -2(3)² + 54 = -18 + 54 = 36

A_{\max }=x\times y=3\times 36=108

Therefore, the maximum area of the triangle is 108.