Find the dimensions of the largest rectangle that can be inscribed in a semicircle of radius r. I know that If I were to make a loose coordinate plane graph than the radius and (X,Y) of the rectangle would have to mixed into make an equation out of the whole thing. What exactly that equation is beyond me though.
Can someone help me make an equation out of this? I asked my professor and he said that the key to find the local Min-Max.
I'm not given any specific numbers either. Any ideas?
 A: We may assume $r=1$ without loss of generality.
It is simple to prove that the largest inscribed rectangle has a side on the diameter of the semicircle.
In particular, we may assume that a vertex of the largest inscribed rectangle lies at $(x,\sqrt{1-x^2})$ with $x\in[0,1]$ and its area is given by $2x\sqrt{1-x^2}$. On the other hand
$$ \text{argmax}\;2x\sqrt{1-x^2} = \text{argmax}\, x^2(1-x^2) $$
and by the AM-GM inequality $x^2(1-x^2)\leq \frac{1}{4}$, with equality attained only at $x=\frac{1}{\sqrt{2}}$. 
It follows that the largest inscribed rectangle is twice a square and its area is $1$ ($r^2$ in the original context).
A: Let the equation of semicircle be given by $x^2+y^2=r^2$ (Consider the upper half part)
Now, let the length of side of rectangle be $2a$ and width $b$ vertexes of rectangle will be $$(a,0)~ ;~ (a,b) ~;~ (-a,0) ~; ~(-a,b)$$

Area of rectangle : $A=(2a)(b)=2ab ~~\cdots (1)$
Also, $a^2+b^2=r^2$ (The point $(a,b)$ lies on the circle) , Put in (1)
$A=2ab=2a\Big(\sqrt{r^2-a^2}\Big)=2\sqrt{r^2a^2-a^4}=2\sqrt{\dfrac{r^4}{4}-\Bigg(a^2-\dfrac{r^2}{2}\Bigg)^2} \le 2\sqrt{\dfrac{r^4}{4}}$
$$\boxed{A \le {r^2} }$$
