Answer:
the second option : w should be 25 units
Step-by-step explanation:
the area of the rectangle is length×width = L×W
the perimeter of a rectangle = 2L + 2W
now, we know that the perimeter is 100 units.
and we have to find the best length of W, that will then define L (to keep the 100 units of perimeter) and maximizes the area of the rectangle.
in other words, what is the maximum area of a rectangle with perimeter of 100 (and what are the corresponding side lengths)?
now, w = 625 is impossible. that side alone would be bigger than the whole perimeter.
W = 0 would render the whole rectangle to a flat line with L = 50 because of
100 = 2L + 2W = 2L + 0 = 2L
L = 50
and A = L×W = 50×0 = 0
an area of 0 is for sure not the largest possible area.
w = 50 would cause L = 0
100 = 2L + 2W = 2L + 2×50 = 2L + 100
0 = 2L
L = 0
and with L = 0 the same thing happens as with W = 0 : a flat line with 0 area.
so, the only remaining useful answer is W = 25
100 = 2L + 2W = 2L + 2×25 = 2L + 50
50 = 2L
L = 25
A = L×W = 25×25 = 625 units²
and indeed, the maximum area for a given perimeter is achieved by arranging the sides to create a square.
