Please help!! Maths for 10 points

The answer is 1cm

The total area enclosed is minimum when the 2 wires are of equal length, i.e. 4cm and 4cm.

Length of each side of square = 4cm divided by 4 = 1cm
Total area = 1cm x 1cm + 1cm x 1cm = 2cm2

If the wire is cut into 5cm and 3cm,
- Length of each side of square is 5cm divided by 4 = 1.25cm and 3cm divided by 4 = 0.75cm
- Total area is 1.25cm x 1.25cm + 0.75cm x 0.75cm = 2.125cm2

If the wire is cut into 6cm and 2 cm,
Length of each side of square is 6cm divided by 4 = 1.5cm and 2cm divided by 4 = 0.5cn
Total area is 2.25cm2 + 0.25cm2 = 2.5cm

If the wire is cut into 7cm and 1 cm
Length of each side of square is 7cm divided by 4 = 1.75cm and 1 cm divided by 4 = 0.25cm
Total area is 3.0625cm2 + 0.0625cm2 = 3.125cm2

Therefore, total area enclosed is minimum when the wire is cut into two equal pieces (4cm and 4cm). The length of the side of each square is therefore 1cm.

Let X be the lenght of one of the piece of wire.
Then 8-X be the lenght of the other piece of wire.

Side of the squares: X/4 and (8-X)/4

Area of the squares
= X/4 * X/4 + (8-X)/4 * (8-X)/4
= (X^2 + 64 - 16X + X^2) / 16
= (2X^2 -16X + 64) / 16

To find the minimum, take derivative with respect to X, and set it to zero
d Area / d X = (4X - 16) / 16 = 0
X = 4

Therefore, the wire is cut into two pieces with the same lenght 4cm.
Side of the two squares = 4cm / 4 = 1cm

教育資訊站-數學網
含數學史、教學資源、活動及相關網址等資訊。
www.edp.ust.hk/math -