1. ## qUEstion

A rectangle is inscribed in semicircle of radius 4m. Find the max possible area for such a rectangle

2. Let $AB$ be the diameter, $O$ the center of the circle, $MNPQ$ the rectangle, with $M,Q\in[AB]$.
Let $PQ=x,OQ=y$. Then the area is $A=2xy$.
We have $x^2+y^2=16$.
If the area is maximum, then the squared area is also maximum.
$A^2=4x^2y^2=4x^2(16-x^2)$
Let $x^2=t$ and $f(t)=4t(16-t)=-4t^2+64t$
The maximum value of f is $\displaystyle f_{max}=-\frac{\Delta}{4a}$, where $\Delta=b^2-4ac, \ a=-4, \ b=64, \ c=0$.
So, $f_{max}=256$, for $t_{max}=-\frac{b}{2a}=8$.
Then $x=y=2\sqrt{2}$

3. Originally Posted by red_dog
Let $AB$ be the diameter, $O$ the center of the circle, $MNPQ$ the rectangle, with $\color{red}{M,Q\in[AB]}$.
How can you claim that the statement in red is necessary?

4. Originally Posted by Isomorphism
How can you claim that the statement in red is necessary?
Because opposite vertices of an inscribed rectangle must lie on a diameter.

RonL

5. Here is what I was doing until I realized I used the wrong radius.

Sorry if it seems redundant since RedDogs fine answer, but I worked it out earlier and do not want my attempt to go in vain.

Let x and y be the dimensions of the rectangle. Then area is A=xy

But $(\frac{x}{2})^{2}+y^{2}=16$

This means $y=\sqrt{16-\frac{x^{2}}{4}}=\frac{1}{2}\sqrt{64-x^{2}}$

Sub into A and we have:

$\frac{x}{2}\sqrt{64-x^{2}}$

$\frac{dA}{dx}=\frac{32-x^{2}}{\sqrt{64-x^{2}}}$

$32-x^{2}=0, \;\ x=4\sqrt{2}, \;\ y=2\sqrt{2}$

Max area is then $(4\sqrt{2})(2\sqrt{2})=16$

Now, try the same problem with an inscribed trapezoid