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

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