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# Please help!! Maths for 10 points

Please help me on this question:

A piece of wire of length 8 cm is cut into two pieces and each piece is bent into a sqaure. if the total area enclosed is minimum, find the length of the sides of each square

Please give me a detailed solution as I don't know how to do it

### 3 Answers

- 1 decade agoFavorite Answer
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.

- ?Lv 71 decade ago
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