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