A wire of length 36 cm is cut into two pieces. One of the pieces is turned in the form of a square and the other in the form of an equilateral triangle. Find the length of each piece so that the sum of the areas of the two be minimum.
Let the length of one piece of wire $= x$ cm
Then, length of remaining wire $= 36 - x$ cm
Let the wire of length x be turned in the form of a square.
$\therefore$ Length of side of square $= \dfrac{x}{4}$ cm
$\therefore$ Area of square $= \left(\dfrac{x}{4}\right)^2 = \dfrac{x^2}{16} \; cm^2$
$\left(36-x\right)$ cm wire is shaped as an equilateral triangle.
$\therefore$ Length of each side of triangle $=\dfrac{36-x}{3}$ cm
$\therefore$ Area of equilateral triangle $= \dfrac{\sqrt{3}}{4} \times \left(\dfrac{36-x}{3}\right)^2 = \dfrac{\sqrt{3}}{36} \left(36-x\right)^2 \; cm^2$
Sum of areas $= A = \dfrac{x^2}{16} + \dfrac{\sqrt{3}}{36} \left(36-x\right)^2$
For minimum area, $\dfrac{dA}{dx}=0$
Now, $\dfrac{dA}{dx} = \dfrac{2x}{16} + \dfrac{\sqrt{3}}{36} \times 2 \times \left(36-x\right) \times \left(-1\right)$
i.e. $\dfrac{dA}{dx} = \dfrac{x}{8} - \dfrac{\sqrt{3}}{18} \left(36-x\right)$ $\;\; \cdots$ (1)
$\therefore$ $\dfrac{dA}{dx} = 0$ $\implies$ $\dfrac{x}{8}- \dfrac{\sqrt{3}}{18} \left(36-x\right) = 0$
i.e. $9x = 4\sqrt{3}\left(36-x\right)$ $\implies$ $x = \dfrac{144 \sqrt{3}}{9+4\sqrt{3}}$
From equation (1), $\dfrac{d^2A}{dx^2} = \dfrac{1}{8} + \dfrac{\sqrt{3}}{18} > 0$
$\implies$ $x = \dfrac{144\sqrt{3}}{9+4\sqrt{3}}$ gives minimum area.
Length of second piece of wire $= 36 - \dfrac{144\sqrt{3}}{9+4\sqrt{3}} = \dfrac{324}{9+4\sqrt{3}}$
$\therefore$ Length of pieces of wire are $\dfrac{144\sqrt{3}}{9+4\sqrt{3}}$ cm and $\dfrac{324}{9+4\sqrt{3}}$ cm