Question

The length of a rectangle is three times its width. If the perimeter is at most 112 centimeters, what is the greatest possible value for the width? Solve the problem.

Answer 1

A.) 42 cm

B.)22.4 cm

C.) 14 cm

D.) 28 cm

Answer 2

14 cm

Answer 3

l = 3b
2(l+b) = 112
2(3b+b) = 112
2(4b)=112
4b=56
b=14

Answer 4

l=3b l=length b=breadth
perimeter at most is 112
perimeter=2(l+b)
2(l+b)=112
l+b=56
l=3b
3b+b=56
4b=14
b=14
that's the max value of width

Answer 5

4b = 56 @ash