To use Alan's expression mmm
I have edited this because I had forgot the half in the area of a triangle 
I hope I ahve fixed that error and not created any more mistakes :/
A piece of wire 5 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.
How much wire should be used for the square in order to maximize the total area?
How much wire should be used for the square in order to minimize the total area?
Let the perimeter of the equilateral triangle be 6x then the perimeter of the square will be 5-6x metres
We know that
\(0<6x<5\\ 0
The side length of the triangle is 2x so the AREA will be \(2x\cdot \sqrt3x=2\sqrt3x^2\)
If all of the wire is used for the triangle then x=5/6 and the area will be \(\color{red}{0.5*}\color{black}{2\sqrt3\cdot \frac{25}{36}}=\frac{25\sqrt3}{36}\approx 1.2m^2\)
The side length of the square is \(\frac{5-6x}{4}\) so its AREA will be \(\left[\frac{5-6x}{4}\right]^2=\frac{25+36x^2-60x}{16}\)
If all the wire is used for the square then x=0 the area will be area will be \(\frac{25}{16}=1.5625\)
So the total Area
\(A=\sqrt3x^2+\frac{25+36x^2-60x}{16}\\ A=\sqrt3x^2+2.25x^2-3.75x+\frac{25}{16}\\ \text{This is a concave up parabola so there will only be a minimum, }\\\text{the max will be at an end of domain point.} \\ \frac{dA}{dx}=2\sqrt3x+4.5x-3.75 \\\frac{dA}{dx}=(2\sqrt3+4.5)x-3.75\\ min\;\; when\;\; \frac{dA}{dx}=0 \\ 0=(2\sqrt3+4.5)x-3.75\\ x=\frac{3.75}{(2\sqrt3+4.5)}\\ x\approx 0.47086\\ \text{which is in the required domain}\)
So in order to minimize total area the this much of the wire must be used for the square.
\(square\;\; perimeter \approx 5-6*0.47086\\ square\;\; perimeter \approx 2.17 metres\)
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So the minimum total area will be when the square gets approx 2.17m of the total 5m
And the maximum total area will be be gained as the the square perimerter appoaches 5
+118713 
