A piece of wire 24m long is to be used to form a square and or a rectangle whose length is three Times its width determine their maximum and minimum combined area

When it forms a square and a rectangle, Max area = 27 + 36 = 63 m^2

When it forms a square or a rectangle, min area = 27 m^2

Step-by-step explanation:

Since the whole length of the wire. =24 m

To form a square from it, all the four sides of the square should be the same, i.e

Each side of the square = 24/4

Each side of the square, l = 6 m

The area of the square = l^2

Area of the square = 6^2 = 36 m^2

For the rectangle, if the length = 3 * width

There are two equal lengths and two equal widths in a rectangle

2l = 3(2w)

Since the length is described to be 3 times the width, only 18 and 6 are three times each other that can be added to make 24.

I.e. 2l = 18, 2w = 6

l = 9 m, w = 3 m

The area of the rectangle = l*w

Area of the rectangle = 9*3 = 27 m^2

When it forms a square and a rectangle, Max area = 27 + 36 = 63 m^2

When it forms a square or a rectangle, min area = 27 m^2

entering that problem on the tube might get you an answer