A wire of length 22m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then,…

Let the side of square be $\frac{22-l}{4}$422−l​ and side of triangle be $\frac{l}{3}$3l​ ,

So, total area is given by $\Delta =\frac{\sqrt{3}}{4}{(\frac{l}{3})}^{2}+{(\frac{22-l}{4})}^{2}$Δ=43​​(3l​)2+(422−l​)2

$$\Rightarrow \Delta =\frac{1}{12\sqrt{3}}\cdot {l}^{2}+\frac{1}{16}{(22-l)}^{2}$$⇒Δ=123​1​⋅l2+161​(22−l)2

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

$\Rightarrow \frac{d\Delta }{dl}=\frac{1}{6\sqrt{3}}\cdot l-\frac{1}{8}(22-l)=0$⇒dldΔ​=63​1​⋅l−81​(22−l)=0 $\Rightarrow l=\frac{66\sqrt{3}}{4+3\sqrt{3}}$⇒l=4+33​663​​

Also$\frac{{d}^{2}\Delta }{d{l}^{2}}=\frac{1}{6\sqrt{3}}+\frac{1}{8}$dl2d2Δ​=63​1​+81​ ($+\mathrm{ve}$+ve. which is case of minima)

So, side of triangle will be $=\frac{l}{3}=\frac{22\sqrt{3}}{4+3\sqrt{3}}=\frac{66}{4\sqrt{3}+9}$=3l​=4+33​223​​=43​+966​