Calculate the maximum possible area of the quadrilateral (functions and derivatives) The total area $A(x)$ in $m^2$ of a quadrilateral is given by:
$$A(x)= 2 \sqrt{ x^2 -16} +\frac 14\sqrt{68x^2-x^4-256},\;\;    (4 < x < 8)$$
How do I calculate the maximum possible area of the quadrilateral.
As i have understood I need to find the derivative of A(x) then find the stationary points but I am struggling to first find the derivative and then find the stationary points.
Thanks in advance.
 A: It looks like your derivative you posted in the comment to the other hint is correct.
Hint: Find the common denominator and add the fractions. The stationary points will occur when the derivative is equal to zero, and the derivative will equal zero only when the numerator is equal to zero. What values of $x$ that are within your range make the numerator zero?
$$\dfrac{2x}{\sqrt{x^2-16}} + \dfrac{136x-4x^3}{8\sqrt{68x^2-x^4-256}} = \dfrac{2x}{\sqrt{x^2-16}} + \dfrac{34x-x^3}{2\sqrt{68x^2-x^4-256}}$$ $$=\dfrac{4x\sqrt{68x^2- x^4-256 }+(34x-x^3)\sqrt{x^2 - 16}}{2\sqrt{x^2 - 16}\sqrt{68x^2 -x^4 - 256}}$$
Clearly, the numerator equals zero when $x=0$. That's because $x = 0$ gives the minimum possible area! But you need the maximum area, and you need the solution such that $4\lt x \lt 8$. Put the numerator equal to zero, in a separate equation, and then solve for $x$.
$$4x\sqrt{68x^2 - x^4 - 256} + (34x - x^3)\sqrt{x^2 - 16} = 0$$ $$\iff 4\sqrt{68x^2 - x^4 - 256} + (34 - x^2)\sqrt{x^2 - 16} = 0$$
A: Hint
Use the fact that the derivative of $\sqrt{u(x)} $ is $\frac{u'(x)}{2 \sqrt{u(x)}}$. Apply this to the two parts of the expression. It will probaly look ugly but reducing to same denominator and squaring could help.  
I am sure you can take from here.
A: $A^{\prime}\left(x\right)=2x\left(x^2-16\right)^{-1/2}+\frac{1}{8}\left(136x-4x^3\right)\left(68x^2-x^4-256\right)^{-1/2}$
$A^{\prime}\left(x\right)=0$ when what?
You should double check $A^{\prime\prime}(x)<0$ for it to be a maximal.
