**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.