Question

A wire of length $36cm$ is cut into two pieces, one of the pieces is turned in form of a square and the other in the form of a equilateral triangle. Find the length of each piece such that the sum of the areas of the two are minimum.

Answer 1

Let the perimeter of the square be $xcm$.

Then the perimeter of the triangle is $(36-x)cm$.

$\therefore$ side of the square $=\frac{x}{4}cm$.

And, side of the triangle $=\frac{1}{3}(36-x)cm$.

$\therefore A=\frac{x^2}{16}+\frac{\sqrt{3}}{4}{(12-\frac{x}{3})}^2=\frac{x^2}{16}+\frac{\sqrt{3}}{4}(144+\frac{x^2}{9}-8x)$

$\Rightarrow A=(\frac{\sqrt{3}}{36}+\frac{1}{16})x^2-2\sqrt{3}x+36\sqrt{3}$

$\Rightarrow \frac{dA}{dx}=\frac{(4\sqrt{3}+9)}{144}\times 2x-2\sqrt{3}$ and $\frac{d^{2}A}{d{x}^2}=\frac{4\sqrt{3}+9}{72}>0$

So. for minimum value of length such that the area of the compound shape is minimum

$\therefore \frac{dA}{dx}=0\iff x=\frac{144\sqrt{3}}{(4\sqrt{3}+9)}cm$.