I see a lot of examples without radicals for $y$, but I'm not finding the solution to the answer for this one at all. Although I know the answer is supposed to be $4\sqrt 2$.
Consider a rectangle in the $xy$-plane with its lower-left vertex at the origin and its upper-right vertex on the graph of $y = \sqrt{6-x}$ as indicated in the figure above. What is the maximum area of such a rectangle?
My current approach:
$y = \sqrt{6-x}$
$A = x \sqrt{6-x}$
$A' = \sqrt{6-x} + \left(x\cdot\frac 1 2\cdot(6-x)^{-\frac 1 2}\right)$
$A' = \sqrt{6-x} + \dfrac x {2\sqrt{6-x}}$
$A' = \dfrac {2(6-x) + x} {2\sqrt{6-x}} = \dfrac {12-x+x} {2\sqrt{6-x}} = \dfrac 6 {\sqrt{6-x}}$
If I set $x = 0$, $x$ then becomes zero, which I know is not correct.
Update
$A' = \sqrt{6-x} - \dfrac x {2\sqrt{6-x}}$
$A' = \dfrac {2(6-x) - x} {2\sqrt{6-x}}$
$A' = \dfrac {12-2x-x} {2\sqrt{6-x}} = \dfrac {12-3x} {2\sqrt{6-x}} = \dfrac {3(4-x)} {2\sqrt{6-x}}$
$0 = \dfrac {3(4-x)} {2\sqrt{6-x}} $
$x = 4$
$A = 4\sqrt{6-4} = 4\sqrt 2$