The length of a rectangle is three times its width. If the perimeter is at most 112 centimeters, what is its greatest possible v

Question

Schach [20]

2 years ago

6

The length of a rectangle is three times its width. If the perimeter is at most 112 centimeters, what is its greatest possible v

alue for width?

Mathematics

2 answers:

Nana76 [90] · Answer 1 · 2022-07-13T10:39:42+03:00

Nana76 [90]2 years ago

7 0

Answer:

Step-by-step explanation:

Perimeter= 2L+2W. If L=3W(three times the width) then 2(3W)+2W=112. Or 6W+2W=112 which is equal to 8W=112. To solve for the width, divide both sides by 8 and find 112/8=14, therefore W=14.

lisabon 2012 [21] · Answer 2 · 2022-07-13T12:15:33+03:00

Answer:

14 cm

Step-by-step explanation:

Given the length is 3 times the width

If each width is denoted by W, then each length is L = 3W

Perimeter (add up all lengths and widths),

= L + L + W + W

= 3W + 3W + W + W

= 8W

Given that the max perimeter is 112cm

hence 8W = 112,

W = 14 cm at most