Maximizing area of a rectangle inscribed in a semicircle 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.
 A: 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 $r^2$
A: The rectangle has upper corners on the semi-circle. Use this knowledge (and the knowledge about how to write an equation of the circle containing the semi-circle) to express the area $A=xy$ of the rectangle as a function of a single variable, then maximize it.
It will make life easier if you assume the center of the circle is at the origin.
A: If you don't like doing things by halves, draw a rectangle inscribed in the semicircle. Then reflect the rectangle, and its semicircle, in the diameter of the semicircle. We get a rectangle inscribed in a circle of radius $r$. The old rectangle has maximum area if and only if the doubled rectangle has maximum area.
Let the sides of the doubled rectangle be $s$ and $t$. By the Pythagorean Theorem, $s^2+t^2=4r^2$.  
The area of the doubled rectangle is $st$. We maximize this, or equivalently we maximize its square $s^2t^2$.  But $s^2=4r^2-t^2$, so we maximize $t^2(4r^2-t^2)$, which is $4r^2t^2-t^4$.  The usual calculus procedures tell us that the maximum is reached when $8r^2 t-4t^3=0$.  
The solution $t=0$ is obviously not good for a maximum. But it is easy to verify that the solution $t=r\sqrt{2}$ gives a maximum.  When $t=r\sqrt{2}$, we find that the maximum doubled rectangle is a square. Not a surprise!
So in the original semicircle, the inscribed rectangle of maximum area has base $r\sqrt{2}$ along the diameter, and height $r\sqrt{2}/2$. 
Remark: To avoid calculus, which presumably you don't want to do since this is a calculus course, go back to the equation  $s^2+t^2=4r^2$. We have
$$s^2+t^2=(s-t)^2+2st,$$
and therefore 
$$2st=4r^2-(s-t)^2.$$
To maximize $st$, it is enough to maximize $2st$. To maximize $2st$, we minimize $(s-t)^2$. The minimum of $(s-t)^2$ is clearly $0$, reached when $s=t$, and therefore when $s=t=\sqrt{2}\,r$. 
