The lengths of the sides of an isosceles triangle are `9 + x^(2), 9 + x^(2)` and `18 - 2x^(2)` units. Calculate the area of the triangle in terms of x and find the value of x which makes the area maximum.

Answer

To solve the problem, we need to calculate the area of the isosceles triangle in terms of \( x \) and then find the value of \( x \) that maximizes this area.

### Step 1: Identify the sides of the triangle
The sides of the isosceles triangle are given as:
- \( a = 9 + x^2 \)
- \( b = 9 + x^2 \)
- \( c = 18 - 2x^2 \)

### Step 2: Calculate the semi-perimeter \( S \)
The semi-perimeter \( S \) is calculated using the formula:
\[ S = \frac{a + b + c}{2} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ S = \frac{(9 + x^2) + (9 + x^2) + (18 - 2x^2)}{2} \]
\[ S = \frac{36}{2} = 18 \]

### Step 3: Apply Heron's formula to find the area \( A \)
Heron's formula states:
\[ A = \sqrt{S(S - a)(S - b)(S - c)} \]
Substituting \( S = 18 \):
\[ A = \sqrt{18(18 - (9 + x^2))(18 - (9 + x^2))(18 - (18 - 2x^2))} \]
Calculating each term:
\[ A = \sqrt{18(9 - x^2)(9 - x^2)(2x^2)} \]
\[ A = \sqrt{18 \cdot 2x^2 \cdot (9 - x^2)^2} \]
\[ A = \sqrt{36x^2(9 - x^2)^2} \]
\[ A = 6x(9 - x^2) \]

### Step 4: Find the value of \( x \) that maximizes the area
To maximize the area, we find the derivative \( \frac{dA}{dx} \) and set it to zero:
\[ A = 6x(9 - x^2) \]
Using the product rule:
\[ \frac{dA}{dx} = 6(9 - x^2) + 6x(-2x) = 6(9 - x^2 - 2x^2) = 6(9 - 3x^2) \]
Setting the derivative to zero:
\[ 6(9 - 3x^2) = 0 \]
\[ 9 - 3x^2 = 0 \]
\[ 3x^2 = 9 \]
\[ x^2 = 3 \]
\[ x = \sqrt{3} \]

### Step 5: Calculate the maximum area
Substituting \( x = \sqrt{3} \) back into the area formula:
\[ A = 6(\sqrt{3})(9 - 3) = 6\sqrt{3} \cdot 6 = 36\sqrt{3} \]

### Final Answer
The area of the triangle in terms of \( x \) is \( 6x(9 - x^2) \) and the maximum area occurs at \( x = \sqrt{3} \) with a maximum area of \( 36\sqrt{3} \) square units.