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?

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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?

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Answer 1 · 2018-03-20T22:56:15

The length of wire required for the square will be $\frac{80 \sqrt{3}}{72 + 32 \sqrt{3}}$ $[80sqrt3]/[72+32sqrt3$ and the length of wire required for the triangle will be $10 - \frac{80 \sqrt{3}}{72 + 32 \sqrt{3}}$ $10 - [80sqrt3]/[72+32sqrt3$ . [ Answers in m]

Explanation:

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

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

$4 x + 3 y = 10$ $4x+3y=10$ , therefore, $y = \frac{10 - 4 x}{3}$ $y=[10-4x]/[3]$ .............. $[1]$ $[1]$ .
Area of equilateral triangle of side length $y$ $y$ = $\frac{\sqrt{3}}{4} y^{2}$ $sqrt3/4 y^2$ . [This derived from Pythagoras theorem].
The total area of both square and triangle = $x^{2} + [\frac{\sqrt{3}}{4}] {[\frac{10 - 4 x}{3}]}^{2}$ $x^2 + [sqrt3/4][[10-4x]/3]^2$ .............. $[2]$ $[2]$ Since $y = \frac{10 - 4 x}{3}$ $y= [10-4x]/3$ from ...... $[1]$ $[1]$

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

$d \frac{A}{d x}$ $dA/dx$ = $72 x - 80 \sqrt{3} + 32 \sqrt{3} x = 0$ $72x-80sqrt3+32sqrt3x=0$ [ for max/min] ,and after manipulation results in $x = \frac{80 \sqrt{3}}{72 + 32 \sqrt{3}}$ $x=[80sqrt3]/[72+32sqrt3$

$d^{2} \frac{a}{{d x}^{2}}$ $d^2a/dx^2$ = $72 + 32 \sqrt{3}$ $72+32sqrt3$ , which is positive [ whatever the value of x] and so by the second derivative test , the value of $x$ $x$ obtained from the first derivative will minimise the area function.

Hope this helps.

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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?

1 Answer

Explanation:

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