Determining the max area of a trapezoid with no known sides A trapezoid is drawn inside a semi circle cross section with the upper base length, being the length of the circle diameter, $d$, and a lower base, $x$, touching the lower sides of the semi circle.
What is the maximum length of the lower base, $x$, of the trapezoid if its area is to be a maximum?
What is the maximum are of the trapezoidal cross section in terms of $d$? 
So far I have determined from $A=(\frac{x+d}2)h$, that $h$ is equal to $\sqrt{(\frac{d}2)^2 - (\frac{x}2)^2}$
And therefore $A$ in terms of $x$ & $d$.   
$$A=(\frac{x+d}2)(\sqrt{(\frac{d}2)^2 - (\frac{x}2)^2}$$
$$A=(\frac{x+d}2)(\sqrt{\frac{d^2}4 - \frac{x^2}4})$$
$$A=(\frac{x+d}2)(\frac{\sqrt{d^2 - x^2}}2)$$
$$A= \frac14 (x+d) (\sqrt{d^2 - x^2})$$
this is where I got stuck... help?
 A: Answer:
I really do not know what the problem is when you got the steps absolutely right.
If differentiation is the problem:
$$\frac{dA}{dx} = \sqrt{d^2-x^2} + (x+d) \frac{-2x}{2\sqrt{d^2-x^2}}$$
$$\frac{dA}{dx} = 0$$
Assuming two functions $U = x+d$ and $V = \sqrt{d^2-x^2}$
$$\frac{dA}{dx} = V\frac{dU}{dx}+U\frac{dV}{dx}$$
$\frac{dU}{dx} = 1$  and $\frac{dV}{dx} = \frac{-2x}{2\sqrt{d^2-x^2}}$
$$\frac{dA}{dx} =  \sqrt{d^2-x^2}+(x+d)\frac{-2x}{2\sqrt{d^2-x^2}} = d^2-x^2 -x(x+d) $$ $$= d^2 - 2x^2-xd = 2x^2 +xd-d^2=0$$  This a quadratic with x as the variable and two roots of x in terms of d is what follows.
the two roots of a quadratic$ax^2+bx+c = 0$ is as follows:
$$ x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$.
In our case $a = 2$, $b = d$ and $c = -d^2$
\begin{align}
 x &= \frac{-d\pm\sqrt{d^2 + 8d^2}}{4}\\
 &= \frac{-d\pm\sqrt{9d^2}}{4}\\  &= \frac{-d\pm3d}{4}\\
  &= \frac{d}{2} \text{ or } \frac{-2d}{2}\text{ (rej.)}\\
\end{align}.
Now substitute the value of d in the area and find A in terms of d.
Goodluck
Thanks
Satish
A: You just have to continue differentiating the product by Chain Rule.
If $v \cdot v $ = const, then $ \dfrac{u}{v} = -\dfrac {u^{'}}{v^{'} } $
$ \dfrac{x+d}{\sqrt {d^2-x^2 }}$ =  $ -\dfrac{-\sqrt {d^2-x^2 }}{x} $
Cross multiply and simplify,
$ (x-d/2) (x +d ) = 0, $ from which the first root is taken and the second one rejected.
Incidentally note that solution $ x = ( d/2= r ) $ is giving you bottom half of a regular hexagon, a solution adopted by civil engg. in canal digging for maximum water flow section area.
A: It is not necessary to assume the quadrilateral is a trapezoid. Connect the two free vertices to the center with radii, thus dividing the area into three isosceles triangles. Let the three vertex angles at the circle center be ${\theta}_1,{\theta}_2$ and ${\theta}_3$. Let the radius of the circle be one. Then the area of the quadrilateral is $\frac12(\sin{\theta}_1+\sin{\theta}_2+\sin{\theta}_3)$, together with constraint that the sum of the three angles is 180 degrees. The sine function is concave downward over the range from zero to 180 degrees, so we can apply Jensen's inequality. This gives the maximum area when the three angles are equal.
