A wire of length 22m is to be cut into two pieces. One of the...

Let the side of square be $\dfrac{22 - l}{4}$ and side of triangle be $\dfrac{l}{3}$ ,

So, total area is given by $\Delta = \dfrac{\sqrt{3}}{4}{\left(\dfrac{l}{3}\right)}^{2} + {\left(\dfrac{22 - l}{4}\right)}^{2}$

$⇒ \Delta = \dfrac{1}{12\sqrt{3}} \cdot {l}^{2} + \dfrac{1}{16}{\left(22 - l\right)}^{2}$

Differentiating with respect to $l$ to get maxima and minima points,

$⇒ \dfrac{d\Delta}{dl} = \dfrac{1}{6\sqrt{3}} \cdot l - \dfrac{1}{8}\left(22 - l\right) = 0$ $⇒ l = \dfrac{66\sqrt{3}}{4 + 3\sqrt{3}}$

Also$\dfrac{{d}^{2}\Delta}{d{l}^{2}} = \dfrac{1}{6\sqrt{3}} + \dfrac{1}{8}$ ($+ ve$. which is case of minima)

So, side of triangle will be $= \dfrac{l}{3} = \dfrac{22\sqrt{3}}{4 + 3\sqrt{3}} = \dfrac{66}{4\sqrt{3} + 9}$