Let the length of one piece be x cm, then the length of other piece will be (28 - x) cm.

Let from the first piece we make a circle of radius r and from the second piece we make a square of side y.

Then,

**Let A be the combined area of the circle and square then,**

A = πr^{2 }+ y^{2}

⇒ A = \(\pi(\frac{x}{2\pi})^2+\)\((\frac{28-x}{4})^2\) **...(ii)**

**Differentiating (ii) and with respect to 'x', we get**

**For maximum and minimum A' = 0**

**Since, **

A' = + ve for \(x= \frac{28\pi}{4+\pi}\)

∴ A is min for \(x= \frac{28\pi}{4+\pi}\)

**Thus, the required length of two pieces are**

\(x= \frac{28\pi}{4+\pi}\) cm and 28 - \(x\)

= 28 - \(\frac{28\pi}{4+\pi}\)

= \(\frac{192}{4+\pi}\)cm.