What is the area of the largest trapezoid that can be inscribed in a semi-circle with radius $r=1$? Steps I took:
I drew out a circle with a radius of 1 and drew a trapezoid inscribed in the top portion of it. I outlined the rectangle within the trapezoid and the two right triangles within it. This allowed me to come to the conclusion, using the pythagorean theorem, that the height of the trapezoid is $h=\sqrt { 1-\frac { x^{ 2 } }{ 4 }  } $
The formula for the area of a trapezoid is $A=\frac { a+b }{ 2 } (h)$
I am basically solving for the $a$ here and so far I have: 
$A=\frac { x+2 }{ 2 } (\sqrt { 1-\frac { x^{ 2 } }{ 4 }  } )$
I simplified this in order to be able to easily take the derivative of it as such:
$$A=\frac { x+2 }{ 2 } (\sqrt { \frac { 4-x^{ 2 } }{ 4 }  } )$$
$$A=\frac { x+2 }{ 2 } (\frac { \sqrt { 4-x^{ 2 } }  }{ 2 } )$$
$$A=(\frac { 1 }{ 4 } )(x+2)(\sqrt { 4-x^{ 2 } } )$$
Then I took the derivative:
$$A'=\frac { 1 }{ 4 } [(1)(\sqrt { 4-x^{ 2 } } )+(x+2)((\frac { 1 }{ 2 } )(4-x^{ 2 })^{ -1/2 }(-2x)]$$
This all simplified to:
$$A'=\frac { 1 }{ 4 } [\sqrt { 4-x^{ 2 } } -\frac { 2x^{ 2 }-4x }{ \sqrt { 4-x^{ 2 } }  } ]$$
Next, I set the derivative equal to zero in order to find the maximum point (I know that I can prove that it is actually a max point by taking the second derivative later)
$$\frac { 1 }{ 4 } [\sqrt { 4-x^{ 2 } } -\frac { 2x^{ 2 }-4x }{ \sqrt { 4-x^{ 2 } }  } ]=0$$
$$\sqrt { 4-x^{ 2 } } -\frac { 2x^{ 2 }-4x }{ \sqrt { 4-x^{ 2 } }  } =0$$
$$\frac { 4-x^ 2-2x^ 2-4x }{ \sqrt { 4-x^ 2 }  } =0$$
$$4-x^{ 2 }-2x^{ 2 }-4x=0$$
$$-3x^{ 2 }-4x+4=0$$
so I got:
 $$x=-2\quad or\quad x=\frac { 2 }{ 3 }$$
$x=-2$ wouldn't make sense so I chose $x=\frac { 2 }{ 3 }$ and plugged it back into my original formula for the area of this trapezoid but my answer doesn't seem to match any of the multiple choice solutions. Where did I go wrong? 
 A: When you went from $$\sqrt{4-x^2} - {{2x^2 - 4x} \over \sqrt{4-x^2}} = 0$$
to $$ {{4 - x^2 - 2x^2 - 4x} \over \sqrt{4-x^2}} = 0$$ you slipped a sign. The last term in the numerator should be $+4x$ instead of $-4x$.
Never mind. That error just reverses a previous erroneous sign. Your real error is in your first simplification of $A'$. You have factors of $1 \over 2$ and $-2x$. The $2$s should have cancelled, but you instead left out the $1 \over 2$.
A: You may even ask

What is the largest quadrilateral inscribed in a semicircle, given that one side is the diameter?

If this maximal quadrilateral happens to be a trapezoid (which will be the case), this also answers the original question.
Complete the figure by refletcing it along the diameter of the semicircle. The question then becomes:

What is the largest hexagon inscribed in a circle, given that at least one long diagonal is a diameter?

and may ge generalized again to 

What is the largest hexagon inscribed in a circle?

In this formulation, it may already strike your eye that almost by symmetry alone the maximal hexagon is the regular one.
Actually,even more generally: 

The maximal $n$-gon inscribed in a circle is the regular $n$-gon

Indeed, any non-regular $n$-gon can be improved by adjusting a vertex that is adjacent to two different edges.
Now finally the area of the inscribed regular hexagon is $6\cdot \frac{\sqrt3}{4}$, so the answer to the original question is $$\frac{3\sqrt3}{4}.$$
A: I also found the area equation for the trapezoid, but found $h$ in a different way.
The area of a trapezoid is $A=\dfrac{(b_1+b_2)H}{2}$. Base $1$ becomes $2$ units, Base $2$ is $2x$, and using the Pythagorean theorem you can find that $H=\sqrt{1-x^2}$. You then plug these values into the Area formula and find its derivative. 
$A=(1+x)\sqrt{1-x^2}$, and thus $A'=\sqrt{1-x^2}+ (-x-x^2)(\sqrt{1-x^2})^{-1}$
You can then set the derivative equal to zero and solve for $x$.
$\sqrt{1-x^2} + (-x-x^2)(\sqrt{1-x^2})^{-1}=0$
Move $\sqrt{1-x^2}$ to the other side of the equation.
$(-x-x^2)(\sqrt{1-x^2})^{-1}=-\sqrt{1-x^2}$
Multiply both sides by $\sqrt{1-x^2}$.
$(-x-x^2)=-(1-x^2)$
$-x-x^2=x^2-1$
$x^2+x-1=0$
Use quadratic formula or other method to deduce that $x = -1, 0.5$. $x$ can't be negative as you can't have a negative measurement, so $x = 0.5$ is the only answer.
Plugging that back into the Area formula you get $\dfrac{3\sqrt{3}}{4}$ or $1.299 un^2$ as your biggest area.
