The smaller side of the rectangle with the largest area, that can be inscribed inside a semi-circle of radius 2 units is of length

Let x be the larger side, y be the smaller side and A be the area of the rectangle inscribed in a semi circle of radius 2.

$\frac{x^2}{4}+y^2=4\Rightarrow y^2=4-\frac{x^2}{4}.$ 

Now $A=xy=x\sqrt{4-x^2/4}=\sqrt{4x^2-x^4/4}.$

Let $S=4x^2-\frac{x^4}{4}.$ Then $\frac{dS}{dx}=8x-x^3$ and $\frac{d^{2}S}{d{x}^2}=8-3x^2.$

$\frac{dS}{dx}=0\Rightarrow 8x-x^3=0\Rightarrow x(8-x^2)=0\Rightarrow x=0,\pm 2\sqrt{2}\Rightarrow x=2\sqrt{2}.$

${\left(\frac{{\displaystyle d^{2}S}}{{\displaystyle d{x}^2}}\right)}_{x=2\sqrt{2}}=8-24=-16<0.\therefore S$ has maximum when $x=2\sqrt{2}$

$\therefore A$ has maximum when $x=2\sqrt{2}.$ If A is maximum then $x=2\sqrt{2},y=\sqrt{2}$.