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O ARITHMETIC AND ALGEBRA REVIEWWord problem on optimizing an area or perimeter

O ARITHMETIC AND ALGEBRA REVIEWWord problem on optimizing an area or perimeter-example-1

1 Answer

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a) field 1:

Step-by-step explanation:

a) Area of the rectangular field = 100ft²

For field 1:

width = 4 ft

Area of a rectangle = length × width

length = area/width

length for field 1 = 100/4

length = 25 ft

perimeter of a rectangle = 2(length + width)

perimeter = 2(25 + 4) = 2(29)

perimeter = 58 ft

For field 2:

width = 5 ft

length = 100/5

length = 20 ft

perimeter = 2(20 + 5) = 2(25)

perimeter = 50 ft

For field 3:

width = 10ft

length = 100/10

length = 10

perimeter = 2(10 + 10) = 2(20)

perimeter = 40 ft

b) To fence a rectangular field, this means the length of the fence will be the perimeter of the field

As a result, length of fencing for field 1 = 58ft

length of fencing for field 2 = 50 ft

length of fencing for field 3 = 40 ft

The least amount of fencing out of the three fields is 40 ft

Hence, field 3 will require the least amount of fencing

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