The width of a rectangle is 5 feet more than its length. The perimeter is, at most, 50 feet. What is the largest possible width of the rectangle?

Question

Can you answer it on paper

Answer 1

Hope this helps...

Perimeter (of a RECT) = 2L + 2W

The problem says that the width is 5 more than the length, so W = L + 5

So you need to replace W in the Perimeter equation with the function L + 5

This will isolate the term for L

P = 2L + 2W

P = 2L + 2(L + 5)

P = 2L + 2L + 10

P given as 50ft

50 = 4L + 10

Isolate the L term...

4L = 50 - 10 = 40

So...L = 40/4 = 10 ft. - that the most the length can be!