A wire of length 36cm is cut into two pieces,one of the pieces is turned in d form of a square and d other  in d form of equilatral tringle.find the length of each piece so that d sum of areas of d two be minimum.? reply fast. 

We have a 36-cm-long wire.

Let a piece of length x cm be cut out from the wire to form a square.

Then, the length of the other piece, with which an equilateral triangle is made, is (36 − x) cm.

Now, we have

Side of square = 

Side of equilateral triangle = 

Combined area (A) of square and equilateral triangle is given by

Now, gives 

It is clear that when 

Therefore, by second derivative test, A is the minimum when 

Thus, the combined area is the minimum when the length of the piece used for making the square is  and the length of the other piece iscm