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 $s=t=\sqrt{2}\,r$.