Express the rectangles area as a function of $x$.  

Translation: 3214 C A rectangle is drawn inside a semicircle as the figure shows. 
FIG (See picture)
a) Express the rectangles area as a function of $v$.
b) Express the rectangles area as a function of $x$.
c) Determine the largest value of the rectangles area. 


My solution:
a) c) 
$x=12\sin v$
$a=12\cos v$
$A=2ax=144\sin 2v$ where $v=\frac{\pi}{4}$
$A''=-576\sin 2v$
$A''\left(\frac{\pi}{4}\right)\lt0:\:max$
$A_{max}\left(\frac{\pi}{4}\right)=144\sin 2\frac{\pi}{4}=144\;\text{cm}^2$
Answer: $A_{max}=144\;\text{cm}^2$.
b)
One correct answer is $A=2x\sqrt{144-x^2}$, but I do not know how to get there. Any hints, tips or solution(s) is appreciated.
 A: Hint. By the Pythagorean theorem, the base of the inscribed rectangle is
$$2\sqrt{12^2-x^2}.$$
A: Let $y$ be the side adjacent to the angle $v$. Then, the area of the rectangle would be equal to $x\cdot 2y$.

a)
$$
\sin{v}=\frac{x}{12}\implies x=12\cdot \sin{v}\\
\cos{v}=\frac{y}{12}\implies y=12\cdot \cos{v}\
$$
Therefore, the area of the rectangle expressed as a function of $v$ is $A(v)=2\cdot 12^2\cdot \sin{v} \cdot \cos{v}$.
Answer: $A(v)=288\cdot \sin{v} \cdot \cos{v}$.

b) Notice that the sides marked $x$, $12$ and $y$ form a right triangle. Therefore:
$$
x^2+y^2=12^2\implies y=\sqrt{12^2-x^2}
$$
So, the area of the rectangle expressed as a function of $x$ is $A(x)=2x\sqrt{12^2-x^2}$.
Answer: $A(x)=2x\sqrt{12^2-x^2}$.

c) Given $A(x)=2x\sqrt{12^2-x^2}$ with the domain $x\in (0,12)$ (that's the interval on which $x$ makes geometrical sense for this problem), the largest area is where the given function reaches a maximum value and that can be found by setting the first derivative to zero and solving for $x$.
$$
A'(x)=\left(2x\sqrt{144-x^2}\right)'=(2x)'\sqrt{144-x^2}+2x\left(\sqrt{144-x^2}\right)'=\\
2\sqrt{144-x^2}-\frac{2x^2}{\sqrt{144-x^2}}
$$
Now, set the derivative to zero and solve for $x$:
$$
2\sqrt{144-x^2}-\frac{2x^2}{\sqrt{144-x^2}}=0\\
2(144-x^2)-2x^2=0\\
x^2=\frac{288}{4}\\
x=\pm\sqrt{72}\\
x=\pm6\sqrt{2}
$$
$x=-6\sqrt{2}$ goes out the window because it's not part of the domain. So, the largest area you can get is $A(6\sqrt{2})=2\cdot 6\sqrt{2}\sqrt{144-(6\sqrt{2})^2}=144$ $cm^2$.
Answer: $144$ $cm^2$.
