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

Question:

A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the we should be cut so that the sum of the areas of the square and triangle is minimum?

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.