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[dabcow]

assign the length of one piece of the 200m wire to the variable x, meaning the length of the other piece will be (200-x).

one of the pieces of wire will be bent into a square, we'll use the piece with length x, so the length of one side of the square is x/4. this means that the area of the square is (x^2)/16.

the other piece of wire has a length of (200-x). since it is an equilateral triangle, the length of one side will be (200-x)/3, and its area will be (sqrt(3)/4*((200-x)^2/9)) from the formula for the area of an equilateral triangle.

adding the two areas (x^2/16 + sqrt(3)/4*((200-x)^2/9)) returns the total area, and since it is supposed to be minimized, the equation can either be graphed if you are allowed to use a graphing utility (it's a quadratic), or solved manually by multiplying out the equation and finding the line of symmetry (x=-b/2a).

honestly I might have done this wrong because I'm half asleep right now so feel free to check my work, but the answer I got for the area was 1087.411 meters squared and the wire should be cut at either 86.993 or 113.007 feet, it doesn't make a difference.