A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the…

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

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

$\Rightarrow \Delta =\frac{1}{12\sqrt{3}}·l^2+\frac{1}{16}{\left(22-l\right)}^2$

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

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

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

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