A wire of length 20m is to be cut into two pieces. One of the...

Let the wire is cut into 2 pieces of length $y \text{ and } 20 - y$

Therefore side of square will be $= \dfrac{y}{4}$ and hexagon will be $= \dfrac{20 - y}{6}$

Area of square$= {\left(\dfrac{y}{4}\right)}^{2}$

Area of hexagon$= 6 \times \dfrac{\sqrt{3}}{4}{\left(\dfrac{20 - y}{6}\right)}^{2}$

Total area$= \dfrac{{y}^{2}}{16} + \dfrac{\sqrt{3}{\left(20 - y\right)}^{2}}{24}$

On differentiating,

${A}^{'}\left(y\right) = \dfrac{2y}{16} + \dfrac{2\sqrt{3}\left(20 - y\right)\left(- 1\right)}{24}$

${A}^{'}\left(y\right) = 0aty = \dfrac{40\sqrt{3}}{3 + 2\sqrt{3}}$

Length of side of regular hexagon$= \dfrac{1}{6}\left(20 - y\right) = \dfrac{1}{6}\left(20 - \dfrac{40\sqrt{3}}{3 + 2\sqrt{3}}\right) = \dfrac{10}{3 + 2\sqrt{3}}$