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ARITHMETIC AND ALCEBRA REVIEWWord problem on optimizing an area or perimeter

ARITHMETIC AND ALCEBRA REVIEWWord problem on optimizing an area or perimeter-example-1
User Klarissa
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1 Answer

15 votes
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region 1 will require the most paint

Step-by-step explanation:

length of tape = 24 ft

perimeter of the region to be painted = 24 ft

The region is rectangular, so we will apply the area of rectangle and perimeter to find the width.

perimeter = 2(length + width)

Area = length × width

We have 3 different regions with different lengths

For region 1:

length = 6 ft

from the formula for perimter, we can find the width:

perimeter = 2(length + width)

perimeter/2 = length + width


\text{width = }\frac{perimeter\text{ }}{2}-\text{ length}

width for region 1 = 24/2 - 6 = 12 - 6

width = 6

Area = 6 × 6 = 36 ft²

For region 2:

length = 7 ft

width = 24/2 - 7 = 12 - 7

width = 5 ft

Area = 7 × 5 = 35 ft²

For region 3:

length = 8 ft

width = 24/2 - 8 = 12 - 8

width = 4 ft

Area = 8 × 4 = 32 ft²

b) The region that will require the most paint is the one that has the highest area. The higher the area, the larger the amount of paint that will be needed.

From calculations above, the highest area is 36 ft²

Hence, the region that will require the most paint is region 1

User KOUSIK MANDAL
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