A wire 10 cm long is cut into two pieces, one of length x and the other of length 10 - x, as shown in the figure. Each piece is bent into the shape of a square.

Question

A wire 10 cm long is cut into two pieces, one of length x and the other of length
10 − x,
as shown in the figure. Each piece is bent into the shape of a square.

 

 

Answer 1

If your question is "what is the minimum area enclosed by these two squares?"

 

assume one section is 4x and the other is 4y=10-4x ( changed your line definition a bit). we form two squares of sides x and y.

 

A = x^2 + y^2    --- (1)

L = 4x + 4y             (note: L = 10cm)

solving x and substitute into (1),

Area = (L/4 -y)^2 +y^2 = L^2/16 - L/2y +2y^2

 

dA/dy = -L/2 +4y      (note:A=Area)

d^2A/dy^2 = 4         (second order directive is always positive)

when dA/dy = 0 ===> when y = L/8 we will have a minimum area, of L^2/32 = 100/32cm^2

 

We will not have a maximum area if we cut the wire by 2. For no matter how small we cut x square, we can increase the area of y square by still smaller x.