A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = $$x$$ units and a circle of radius = $$r$$ units. If the sum of the areas of the square and the circle so formed is minimum, then:

Solution :
$$\eqalign{ & 4x + 2\pi r = 2\quad \Rightarrow 2x + \pi r = 1 \cr & S = {x^2} + \pi {r^2} \cr & S = {\left( {\frac{{1 - \pi r}}{2}} \right)^2} + \pi {r^2} \cr & \frac{{dS}}{{dr}} = 2\left( {\frac{{1 - \pi r}}{2}} \right)\left( {\frac{{ - \pi }}{2}} \right) + 2\pi r \cr & \Rightarrow \frac{{ - \pi }}{2} + \frac{{{\pi ^2}r}}{2} + 2\pi r = 0\quad \Rightarrow r = \frac{1}{{\pi + 4}} \cr & \Rightarrow x = \frac{2}{{\pi + 4}}\quad \Rightarrow x = 2r \cr} $$