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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 let lengths be x and 36-x. There4, area(square) = (x/4)^2 ; area(tri) = 0.5x((36-x)/3)^2 x sin60'. A(total) = x^2 /16 + sqrt3 /4 x (36-x)^2 /9. Now dA/dx = x/8 - 2sqrt3 x (36-x) / 36 = 0 for stat pt. x~15.66. lengths are 16 and 20 cm (0dp)

i realise i took a more difficult route, which may have led me to the wrong ans, but ... meh

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