Question

A wire of length $20$ m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is

• A. $\frac{5}{2+\sqrt{3}}$
• B. $\frac{5}{3+\sqrt{3}}$
• C. $\frac{10}{3+2\sqrt{3}}$
• D. $\frac{10}{2+3\sqrt{3}}$

Answer 1

The correct option is C $\frac{10}{3+2\sqrt{3}}$

Let the side of square be $a$ and that of hexagon be $b$.
Then, $4a+6b=20\cdots \left(1\right)$
Area of square $=a^2$
and area of hexagon $=\frac{3\sqrt{3}}{2}{b}^2$
$\therefore$ Total area, $A=a^2+\frac{3\sqrt{3}}{2}{b}^2$
$\Rightarrow A={\left(\frac{10-3b}{2}\right)}^2+\frac{3\sqrt{3}}{2}{b}^2$ $\left(\text{From}\right(1\left)\right)$
For minimum area, $\frac{dA}{db}=0$
$\Rightarrow \frac{dA}{db}=2\left(\frac{10-3b}{2}\right)(-\frac{3}{2})+2\left(\frac{3\sqrt{3}}{2}\right)b=0$
$\Rightarrow (10-3b)(-3)+6\sqrt{3}b=0$
$\Rightarrow b=\frac{10}{(3+2\sqrt{3})}$
Also, $\frac{d^{2}A}{d{b}^2}=\frac{9}{2}+3\sqrt{3}>0$
As $\frac{d^{2}A}{d{b}^2}>0$
Hence, area is minimum for $b=\frac{10}{3+2\sqrt{3}}$