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3 years ago

the length of a rectangle is 3 cm more than twice its width. Find the largest possible width if the perimeter is at most 66cm

2 answers:

7 0

Let the length be L

Let the width be W.

L is 3cm more than twice its width.

L = 2W + 3

P = 2*(L + W)

P = 2*(2W + 3 + W)

P = 2*(3W + 3)

P = 6W + 6

Since Perimeter is at most 66, so P is less than or equal to 66.

P ≤ 66

6W + 6 ≤ 66

6W ≤ 66 - 6

6W ≤ 60

W ≤ 60/6

W ≤ 10

So the largest possible width is 10cm.

Largest possible width is 10cm.

I hope this helped.

Vanyuwa [196]3 years ago

6 0

If you let the width be x, then the length will be 2x + 3. The perimeter P is twice the width plus twice the length, i.e., P = 2*x + 2*(2x + 3) = 2x + 4x + 6 = 6x + 6. So if the perimeter is 66 cm then 66 = 6x + 6 ---> 6x = 60 ---> x = 10 cm., and so if the perimeter is at most 66 cm., the width can be at most 10 cm..

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$Collect\ together = \frac{1}{2} + \frac{9}{10}\\\\Collect\ together = \frac{1 \times 5}{2 \times 5} + \frac{9}{10}\\\\Collect\ together = \frac{5}{10} + \frac{9}{10}\\\\Collect\ together = \frac{9+5}{10}\\\\Collect\ together = \frac{14}{10} = \frac{7}{5} = 1.4\\\\In\ mixed\ fractions,\\\\Collect\ together = 1\frac{2}{5}$

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