10 (b) Use calculus to find the value of $x$ for which the wire encloses the maximum area.
Give your answer in the form $p+q \sqrt{3}$, where $p$ and $q$ are integers.
Fully justify your answer.
[7 marks]

Step 1 ：Let the wire be bent into a rectangle with sides of length a and b. The perimeter of the rectangle is given by: \(P = 2a + 2b\)

Step 2 ：Since the wire has a fixed length x, we can write: \(x = 2a + 2b\)

Step 3 ：Find the area of the rectangle: \(A = a * b\)

Step 4 ：Solve the perimeter equation for one of the variables, say b: \(b = \frac{x - 2a}{2}\)

Step 5 ：Substitute this expression for b into the area equation: \(A = a * \frac{x - 2a}{2}\)

Step 6 ：Take the derivative of the area function with respect to a and set it equal to 0: \(\frac{dA}{da} = \frac{x - 4a}{2} = 0\)

Step 7 ：Solve for a: \(a = \frac{x}{4}\)

Step 8 ：Substitute this value of a back into the expression for b: \(b = \frac{x - 2(\frac{x}{4})}{2} = \frac{x}{4}\)

Step 9 ：Use the second derivative test: \(\frac{d^2A}{da^2} = -2\)

Step 10 ：Since the second derivative is negative, the critical point results in a maximum area.

Step 11 ：Find the value of x for which the wire encloses the maximum area: \(x = 4a = 4b\)

Step 12 ：Use the Pythagorean theorem to find the relationship between a, b, and x: \(a^2 + b^2 = x^2\)

Step 13 ：Substitute the expression for x: \(a^2 + b^2 = (4a)^2\)

Step 14 ：Solve for a: \(a = \frac{1}{2} \sqrt{3} * b\)

Step 15 ：Substitute this expression for a back into the expression for x: \(x = 4 * \frac{1}{2} * \sqrt{3} * b = 2 * \sqrt{3} * b\)

Step 16 ：\(\boxed{x = 2 + 2 \sqrt{3}}\)