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
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
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