A triangle is formed by vertices (0, 0), (x, y), (-x, y) on xy-plane. If the point (x, y) and (-x, y) lies on `y =-x^2 + 54`, then maximum area of triangle is

Options

A. `18 sqrt 2`
B. `108 sqrt 2`
C. `36 sqrt 2`
D. `54 sqrt 2`

Answer

To find the maximum area of the triangle formed by the vertices (0, 0), (x, y), and (-x, y), we will follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices of the triangle are:
- A(0, 0)
- B(x, y)
- C(-x, y)

### Step 2: Calculate the base and height of the triangle
The base of the triangle (BC) can be calculated as:
\[ \text{Base} = |x - (-x)| = |x + x| = 2x \]

The height of the triangle is simply the y-coordinate of points B and C, which is:
\[ \text{Height} = y \]

### Step 3: Write the formula for the area of the triangle
The area (A) of the triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{Base} \times \text{Height} \]
Substituting the values we found:
\[ A = \frac{1}{2} \times (2x) \times y = xy \]

### Step 4: Substitute y in terms of x
According to the problem, points (x, y) and (-x, y) lie on the parabola given by:
\[ y = -x^2 + 54 \]
Substituting this into the area formula:
\[ A = x(-x^2 + 54) = -x^3 + 54x \]

### Step 5: Find the critical points to maximize the area
To find the maximum area, we need to take the derivative of A with respect to x and set it to zero:
\[ \frac{dA}{dx} = -3x^2 + 54 \]
Setting the derivative equal to zero:
\[ -3x^2 + 54 = 0 \]
\[ 3x^2 = 54 \]
\[ x^2 = 18 \]
\[ x = \pm 3\sqrt{2} \]

### Step 6: Determine whether this critical point is a maximum
To confirm that this critical point is a maximum, we can take the second derivative:
\[ \frac{d^2A}{dx^2} = -6x \]
Evaluating the second derivative at \( x = 3\sqrt{2} \):
\[ \frac{d^2A}{dx^2} = -6(3\sqrt{2}) < 0 \]
Since the second derivative is negative, this indicates a maximum.

### Step 7: Calculate the maximum area
Now we can substitute \( x = 3\sqrt{2} \) back into the area formula to find the maximum area:
\[ y = -x^2 + 54 = -18 + 54 = 36 \]
Thus, the maximum area is:
\[ A = x \cdot y = (3\sqrt{2}) \cdot 36 = 108\sqrt{2} \]

### Final Answer
The maximum area of the triangle is:
\[ \boxed{108\sqrt{2}} \]