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?
Best Answer
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 $1 \over 2$.