A wire of length \( 20\,\text{m} \) is to the cut into two pieces. A piece of length \( l_{1} \) is bent to make a square of area \( A_{1} \) and the other piece of length \( l_{2} \) is made into a circle of area \( A_{2} \). If \( 2A_{1}+3A_{2} \) is minimum then \( (\pi l_{1}):l_{2} \) is equal to:
