Question

A piece of wire $28$ m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How much wire should be used for the square in order to maximize the total area?

Answer 1

Let the $2$ pieces of rope be x, $(28-x)$m

Area of square $=(28-x{)}^2$

Area of triangle $=\frac{\sqrt{3}}{4}{x}^2$

Total Area $=(28-x{)}^2+\frac{\sqrt{3}}{4}{x}^2$

Differentiating wrt x:-

$\frac{dy}{dx}=2(28-x)(-1)+\frac{\sqrt{3}}{2}x$

$\frac{d^{2}y}{d{x}^2}=-2(-1)+\frac{\sqrt{3}}{2}=-3+\frac{\sqrt{3}}{2}=-$ve

Since $\frac{d^{2}y}{d{x}^2}$ is negative, therefore maxima will be obtained.

To find critical points, $\frac{dy}{dx}=0$

$\Rightarrow 2(28-x)=\frac{\sqrt{3}}{2}x$

$\Rightarrow 4\times 28=(\sqrt{3}+2)x$

$\Rightarrow x=\frac{112}{\sqrt{3}+2}$ [value of x where maxima will be obtained

rope used in square

$=28-x=28-\frac{112}{\sqrt{3}+2}=\frac{28\sqrt{3}-56}{\sqrt{3}+2}$

$=\frac{28(\sqrt{3}-2)(\sqrt{3}-2)}{(\sqrt{3}+2)(\sqrt{3}-2)}=\frac{28[3+4-4\sqrt{3}]}{-1}=28(4\sqrt{3}-7)m$.