Trigonometric functions, find the value of x that maximizes the area of quadrilateral. I've been stuck trying to solve this problem for the whole day. Also, I'm trying to translate the problem as good as I can, as my English skills aren't the greatest; sorry for that.
Problem is as follows: Points OBDE form a quadrilateral. Points B and D are on the line x=1. Find the value of x that maximizes the area of quadrilateral in a unit circle. Here is a picture of the problem: https://i.imgur.com/LuXgMa4.png
What I have figure out, is that the quadrilateral consists of right-angled triangle and a rectangle, so I'm assuming that the function that I'm supposed to form should be $A\left(x\right)=\frac{1}{2}ah+ah$
I thought that the correct function would be $A(x)=\frac{1}{2}sin\left(x\right)cos\left(x\right)+sin\left(x\right)cos\left(x\right)$, I also derivated it, but it didn't give the right answer. I've also been thinking about using tangent, since the line $x=1$ is tangent to the circle, but I'm not sure how and what kind of function I'm supposed to form using the tangent.
I'm not really looking for a straight-up answer, I could go on if anyone could at least give me the function that I'm supposed to form. I know how to derivate trigonometric functions and how to find their min/max.
 A: If you take $\theta$ as being measured from the y- axis, then the area of the quadrilateral should be $sin(\theta)*1+\cfrac{1}{2}sin(\theta)cos(\theta)$, since the width of the rectangular part is $1$, and the triangle is $sin(\theta)$ by $cos(\theta)$, so you should differentiate that instead
A: Use the formula of a trapezoid: $A=\frac{1}{2}(a+b)(h)$.
We know that $E$ is at $(\cos x, \sin x)$, so $a = 1 - \cos x$, $b = 1$, and $h = |\sin x|$ (since $h$ is never negative). Putting it all together, we have:
$$A = \frac{1}{2}\left((2-\cos x)(|\sin x| \right)$$
$$= |\sin x| - \frac{1}{2} \sin x \cos x$$
$$= |\sin x| - \frac{1}{4} \sin(2x),$$
and now you can differentiate.
A: 
... I'm assuming that the function that I'm supposed to form should be $A\left(x\right)=\frac{1}{2}ah+ah$

Notice that you used the same symbols, $a$ and $h$, in both parts of your formula.
Supposing that $h$ represents height (measured parallel to the $y$ axis),
it's fine to use $h$ twice like this, since both the rectangle and the triangle have the same height.
But look at the diagram and you can clearly see that in general the rectangle and triangle do not have the same width. By using the symbol $a$ twice, your formula assumes the triangle will always have exactly half the width of the rectangle, whereas it actually never will.
A better way is to work out the area of the rectangle as a function of your input variable,
then work out the area of the triangle as a function of your input variable,
and then add the two areas.
In that way you will not have to come up with too many symbols to represent the height and the different widths of these figures,
and you will not be tempted to mistakenly use the same symbol for two different things.
Another way your question uses one symbol for two purposes is by putting the points $B$ and $D$ on the line $x = 1$ (implying that $x$ is a Cartesian coordinate) and by using the same symbol, $x,$ to represent an angle.
It would be better if the angle were represented by some other symbol such ast $\theta.$
If you do that, then $\theta,$ not $x,$ will be the "input variable" in the previous paragraph.
Doing all this correctly, you will end up with something like Seth's answer.
