1. $= \dfrac{4-x}{4} $m

2. Area of square $=(\dfrac{4-x}{4})^2$

Area of circle $= \pi. r^2$

but circumference $= \pi.d$

$x= \pi.d$

$d= \dfrac{x}{\pi}$

$ r= \dfrac{x}{2}.\pi$

$A = \dfrac{(\pi.x^2)}{4}\pi^2 = \dfrac{x^2}{4\pi}$

adding 2 areas= $\dfrac{x^2}{4\pi }+ [\dfrac{(4-x)}{4}]^2 $

$= (\dfrac{1}{16} + \dfrac{1}{4\pi}).x^2 - \dfrac{x}{2} + 1$

3. $\dfrac{dA}{dx} = 2.x.(\dfrac{1}{16} + \dfrac{1}{4\pi}) - \dfrac{1}{2} = 0$

, solving $x= 3.52$m for circle and $4-3.52= 0.48$m for square