Given the following equation where A = Area of a rectangle and w = width of the rectangle, what value of 'w' would maximize the area? A = LW
P = 2L+2W
P = 100
w should be 625 units
w should be 25 units
w should be 0 units
w should be 50 units

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.

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb