# A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is minimum?

Mar 20, 2018

The length of wire required for the square will be [80sqrt3]/[72+32sqrt3 and the length of wire required for the triangle will be 10 - [80sqrt3]/[72+32sqrt3. [ Answers in m]

#### Explanation:

Let the side of the square have a length $x$ m and the side of the triangle have a length of $y$ m.

The combined lengths of the square and rectangle will equal $4 x + 3 y$ and this is equal to $10$ [given].

$4 x + 3 y = 10$, therefore, $y = \frac{10 - 4 x}{3}$..............$\left[1\right]$.
Area of equilateral triangle of side length $y$ =$\frac{\sqrt{3}}{4} {y}^{2}$. [This derived from Pythagoras theorem].
The total area of both square and triangle =${x}^{2} + \left[\frac{\sqrt{3}}{4}\right] {\left[\frac{10 - 4 x}{3}\right]}^{2}$..............$\left[2\right]$ Since $y = \frac{10 - 4 x}{3}$ from ......$\left[1\right]$

So differentiating ..$\left[2\right]$ with respect to A [area] to find a turning point and hence a max or min, using the chain rule .

$\frac{\mathrm{dA}}{\mathrm{dx}}$ = $72 x - 80 \sqrt{3} + 32 \sqrt{3} x = 0$ [ for max/min] ,and after manipulation results in x=[80sqrt3]/[72+32sqrt3

${d}^{2} \frac{a}{\mathrm{dx}} ^ 2$ = $72 + 32 \sqrt{3}$, which is positive [ whatever the value of x] and so by the second derivative test , the value of $x$ obtained from the first derivative will minimise the area function.

Hope this helps.