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

Solution:

Suppose the wire, which is to be made into a square and a triangle, is cut into two pieces of length $x$ and $y$, respectively. Then,

$x+y=20$ .... (1)

Perimeter of square, 4 (Side) $=x$

$\Rightarrow$ Side $=\frac{y}{3}$

Area of triangle $=\frac{\sqrt{3}}{4} \times(\text { Side })^{2}=\frac{\sqrt{3}}{4} \times\left(\frac{y}{3}\right)^{2}=\frac{\sqrt{3} y^{2}}{36}$

Now,

$z=$ Area of square $+$ Area of triangle

$\Rightarrow z=\frac{x^{2}}{16}+\frac{\sqrt{3} y^{2}}{36}$

$\Rightarrow z=\frac{x^{2}}{16}+\frac{\sqrt{3}(20-x)^{2}}{36}$      [From eq. (1)]

$\Rightarrow \frac{d z}{d x}=\frac{2 x}{16}-\frac{2 \sqrt{3}(20-x)}{36}$

For maximum or minimum values of $\mathrm{z}$, we must have

$\frac{d z}{d x}=0$

$\Rightarrow \frac{2 x}{16}-\frac{\sqrt{3}(20-x)}{18}=0$

$\Rightarrow \frac{9 x}{4}=\sqrt{3}(20-x)$

$\Rightarrow \frac{9 x}{4}+x \sqrt{3}=20 \sqrt{3}$

$\Rightarrow x\left(\frac{9}{4}+\sqrt{3}\right)=20 \sqrt{3}$

$\Rightarrow x=\frac{20 \sqrt{3}}{\left(\frac{9}{4}+\sqrt{3}\right)}$

$\Rightarrow x=\frac{80 \sqrt{3}}{(9+4 \sqrt{3})}$

$\Rightarrow y=20-\frac{80 \sqrt{3}}{9+4 \sqrt{3}}$       [From eq. (1)]

$\Rightarrow y=\frac{180}{9+4 \sqrt{3}}$

$\frac{d^{2} z}{d x^{2}}=\frac{1}{8}+\frac{\sqrt{3}}{18}>0$

Thus, $z$ is minimum when $x=\frac{80 \sqrt{3}}{(9+4 \sqrt{3})}$ and $y=\frac{180}{9+4 \sqrt{3}}$.

Hence, the wire of length $20 \mathrm{~cm}$ should be cut into two pieces of lengths $\frac{80 \sqrt{3}}{(9+4 \sqrt{3})} \mathrm{m}$ and $\frac{180}{9+4 \sqrt{3}} \mathrm{~m}$.

Disclaimer: The solution given in the book is incorrrect. The solution here is created according to the question given in the book.