calculusgeometry
A rectangle of largest area is inscribed in a semicircle of radius $r$. What is the area of the rectangle?
I just need the hint to solve it. How can I get length and breadth of rectangle in terms of radius $r$? If I can get length and breadth in terms of radius $r$, then I can solve $\frac{d (\text{Area})}{dr} = 0$.
Then $|AC| = r$.
Using pythagoras theorem, I will get
$$r = \left(b^2 + \frac{l^{2}}{4} \right)^{\frac{1}{2}}$$
Thanks in advance.
N.B. $\operatorname{cs}(ACD)$ is the circle segment defined by the rays, $AC$ and $AD$.
In the piture below, we have that $$\operatorname{cs}(ACD)+\operatorname{cs}(BCD)$$ contains the part were looking for. However we have overcounted. Can you see by how much we overcounted (i.e. what we now need to substract)?
I think you have come till this equation: $$P=2x+\sqrt{4-x^2}$$
Further, $$\frac{dP}{dx}= 2-\frac{x}{\sqrt{4-x^2}}$$ Set $\frac{dP}{dx}=0$ for finding the maximum of P.
So, $$2=\frac{x}{\sqrt{4-x^2}}$$ $x=\sqrt{\frac{16}{5}}$ is the value of x for maximum value of $P$ (which gives the maximum perimeter).
Therefore, the perimeter of the rectangle when $x=\sqrt{\frac{16}{5}}$ is $P=2\sqrt{5}$
(by substituting the value of x in the first equation).
Finally, $$x=\frac{4}{\sqrt{5}}$$
and
$perimeter(max)=2\sqrt{5}$
Best Answer
It might be easier to deal with this using trigonometry.
Using your figure,
Notice that the area of the rectangle is four times the area of $\triangle{ABC}$.
Thus it is enough to maximize the area of the triangle.
Now if $\angle{BCA}$ is $\theta$, then the area of the triangle is $\frac{r^2}{2} \sin \theta \cos \theta = \frac{r^2}{4} \sin 2\theta$ (as $BC = r \cos \theta$ and $AB = r \sin \theta$).
Thus you need $\theta = \frac{\pi}{4}$ and the area of the rectangle is $r^2$