$y^2 + x^2 = 4$
$A(x) = 2xy$ (make base of semicircle = $2x$)
plug it in: $A(x) = (\space 2\sqrt{(4-y^2)}\space)\cdot y$
Final derivative: $$\begin{align} A'(x) & = \frac{-2y^2}{\sqrt{4-y^2} + 2\sqrt{(4-y)}} \\ & \\ & \\ 0 & = \frac{-2y^2}{\sqrt{4-y^2} + 2\sqrt{(4-y)}} \\ \end{align}$$ $y = \frac{2\sqrt 2}{3} $
but the answer is 1.42 x 0.71
how do you get this? did i make a mistake?
For starters, $A(x,y)= 2xy$ where $2x$ is the base of the rectangle, not the semicircle. Also, if $x=\sqrt{4-y^2}$, you mean to say that $A(y) = 2y\sqrt{4-y^2}$.
Now, your derivative isn't correct. You should have gotten
$$A^{\prime}(y) = 2\sqrt{4-y^2} - \frac{2y^2}{\sqrt{4-y^2}} = \frac{8-4y^2}{\sqrt{4-y^2}}.$$
The maximizing value then is $y=\sqrt{2}$ and thus $x=\sqrt{2}$. So the dimensions of the rectangle are $2\sqrt{2}\times \sqrt{2}$.
