# A wire of length 36 cm is cut into two pieces, one of the pieces is turned in the form of a square and other in the form of an equilateral triangle.

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A wire of length 36 cm is cut into two pieces, one of the pieces is turned in the form of a square and 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.

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Let the length of one piece be x, then the length of the other  piece will be 36 - x.

Let from first piece we make the square, then

x = 4y

⇒ $y=\frac{x}{4},$ ...(i)

Where y is the side of the square

From the second piece of length (36 - x) we make an equilateral triangle, then side of the equilateral triangle = $(\frac{36-x}{3})$ Now combined area of the two =

A = $(\frac{x}{4})^2\,+$ $\frac{\sqrt{3}}{4}$ $(\frac{36-x}{3})^2$

Differentiating with respect to x, we have, For maximum/minimum, we have

$\frac{dA}{dx}=0$ Thus, length of one piece is $x=\frac{144}{4+3\sqrt 3}$ and the length of other piece is :

$36-\frac{144}{4+3\sqrt 3}$

$= \frac{144+108\sqrt3-144}{4+3\sqrt{3}}$

$= \frac{108\sqrt3}{4+3\sqrt{3}}$ cm.