A rectangle has a length at least four more than five times the width. If the perimeter is at least 44 units, find the least possible width.

Answer 1

Greetings! Lets solve this shall we ?

So, we must find the the last possible width using the formula of a Perimeter such that

2l + 2w ≥ P, for the least possible perimeter. Then, the length is l ≥ 5x + 4 and width = x with a perimeter of P = 44 units to obtain

2(5x + 4) + 2x ≥ 44.............Using the Distributive Property such that

10x + 8 + 2x ≥ 44............Add like terms such that

12x + 8 ≥ 44..........Then subtract 8 such that

12x ≥ 36..............Finally divide such that

x ≥ 3 which means the width should be at least 3 units!

I hope this helped!