A piece of wire 8 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.

Question

(a) How much wire should be used for the square in order to maximize the total area?

(b) How much wire should be used for the square in order to minimize the total area?

Answer 1

(a) Let x m is cutted for square, then (8 - x) m for triangle.

Then total area A(x) = (x/4)² + ((8-x)/3)²√3/4;

(b) A(x) = x²/16 + (8 - x)²√3/36;

A'(x) = x/8 - (8-x)√3/18 = 0; 9x + 4x√3 = 32√3; x = 32√3/(9 + 4√3).

And we have mini9mum because A''(x) = 1/8 + √3/18 > 0.

So, x = 32√3(9 - 4√3)/33 = 32(9√3 - 12)/33 ≈ 3.48 m

Answer 2

squares have larger area for a given perimeter than triangles.

for max area, take the wire ans have each side of the square = 2 m and each side of the triangle zero.

maximum area = 2^2 = 4 m^2

Area = A = s^2 + bh/2 where h=b(sqr3/2)

A = s^2 + b^2(sqr3)/4 s = (8-3b)/4 = 2-3b/4

= (2-3b/4)^2 + b^2(sqr3/4)

A = (2-3b/4)^2 + b^2(sqr3/4)

= 4 - 3b + 9b^2/16 + b^2(sqr3/4) = 4 - 3b + (sqr3/4 + 9/16)b^2

A'(b) = -3 + (sqr3/2 + 9/8)b = 0

b = 3/(sqr3/2 + 9/8) = about 1.05 = side of the equilateral triangle

3 sides = 3.15. leaving 8-3.15 = 4.85 for perimeter of a square with sides = 4.85/4 = 1.21

Area = 1.05^2 + (1.21)^2(sqr3/4) = 1.1+ .634 = 1.73 = minimum area

when you cut the 8 meter wire into 2 parts, one = 4.85 and the other part 3.15 m

This is subject to revision. But no one else tried this, so it's up for grabs. My stab at it, subject to endless possible arithmetic errors en route. but the basic method is correct. calculate area, find the derivative, set=0 solve