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Brut [27]
3 years ago
7

Help me please

Mathematics
1 answer:
Andreyy893 years ago
7 0

Answers:

  • Part (a) upper bound for area = 4527.7056 square cm
  • Part (b) lower bound for perimeter = 278 cm

======================================================

Explanation:

Part (a)

The horizontal portion is shown to be 87.3 cm long. This is the result <em>after</em> rounding has occurred. Specifically, rounding to one decimal place (tenths). The question is: What could the number have been <em>before</em> rounding?

You have to think in reverse of the usual rounding process.

The values 87.25, 87.26, 8.27, ... 8.33, 87.34 all round to 87.3 when we round to one decimal digit. Note the hundredths digit is increasing by 1 each time. We can see that 87.34 is the largest possible value for the horizontal piece. After that, 87.35 will round to 87.4 which is just out of reach.  

Through similar logic/reasoning, the vertical side can be at most 51.84 cm.

The area of the rectangle could be 87.34*51.84 = 4527.7056 which is the upper bound for the area. This is the largest possible area, given the values in the diagram, since we made each value as large as possible.

In short, 4527.7056 square cm is the max area.

-----------------------------------

Part (b)

We use the idea of part (a), but we'll use the smallest possible values this time.

So we go with 87.25 for the horizontal portion and 51.75 for the vertical portion.

Compute the perimeter with these smallest dimensions.

P = 2*(length + width)

P = 2*(87.25 + 51.75)

P = 278

The smallest perimeter possible is 278 cm which is the lower bound of the perimeter. This is guaranteed to be the smallest perimeter possible because we made the sides as small as possible, while staying in the restrictions as discussed in the previous section.

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