Maximize the area of a triangle inscribed in a semicircle. http://i.imgur.com/Q5gjaSG.png
Consider the semicircle with radius 1, the diameter is AB. Let C be a point on the semicircle and D the projection of C onto AB. Maximize the area of the triangle BDC.
My attempt so far, I'm new at these problems and have only done a few so far. Thanks for any hints. I feel I'm not setting up the correct equation. 
$x^{2}$+$y^2$=$r^2$ 
$y=\sqrt{r^2 - x^2}=\sqrt{1-x^2}\;(\text{since}\:r=1).$
The area is A=$\dfrac{1}{2}$($x+1$)($y$)  
A'=$\frac{2x^2-x-1}{2\sqrt{1-x^2}}$
A'=$0$ When $x$ = -   $\dfrac{1}{2}$
$y$=$\dfrac{\sqrt{3}}{2}$
Plugging everything back into the original area equation I get $\dfrac{ \sqrt{3}}{8}$
 A: Almost: Since the area is
$$
A = \frac{(1+x)\sqrt{1-x^2}}{2}
$$
we want to find the zero of the derivative
$$
\frac{dA}{dx} = \frac{1-x-2x^2}{2\sqrt{1-x^2}}
$$
Since $-1 < x < 1$ (do you see why?), we can just focus on the numerator, and then
$$
2x^2+x-1 = 0
$$
or $x = -1$ or $x = 1/2$.  The first solution is discarded, and we use the second solution: $x = 1/2$, so $y = \sqrt{3}/2$, and $A = (1+x)y/2 = 3\sqrt{3}/8 \doteq 0.64952$.
You just lost a minus sign, is all, I think.  But it is useful to evaluate your answer to see if it makes sense.  $\sqrt{3}/8 \doteq 0.21651$ is too small for something that must be larger than half the unit square.
A: Another approach is to use $\angle CBD=\phi\in(0,\frac\pi2)$:

\begin{align}
|BC|&=|AB|\cos\phi=2\cos\phi
\\
|BD|&=|BC|\cos\phi=2\cos^2\phi
\\
S_{\triangle BCD}&=
\tfrac12|BD|\cdot|BC|\sin\phi
\\
&=
2\sin\phi\cos^3\phi
\\
S'(\phi)
&=
2\cos^2\phi(\cos^2\phi-3\sin^2\phi)
\\
&=
2\cos^2\phi(1-4\sin^2\phi)
\end{align}
The only suitable solution to $S'(\phi)=0$
is $\phi=\arcsin\tfrac12=\tfrac\pi6$,
hence 
\begin{align}
S_{\max}
&=
2\cdot\tfrac12\cdot\left(\tfrac{\sqrt3}{2}\right)^3
=\tfrac{3\sqrt3}{8}.
\end{align}
