SOLUTION: Consider a rectangle with perimeter 28 (units). Let the width of the rectangle be w (units) and let the Area of the region enclosed by the rectangle be A ( square units). Expre

Question 1162897: Consider a rectangle with perimeter 28 (units). Let the width of the
rectangle be w (units) and let the Area of the region enclosed by the
rectangle be A ( square units).
Express A as a function of w and state the domain and range of the
function.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

P = 2L+2W is the perimeter of any rectangle with length L and width W

Solve for L to get
P = 2L+2W
2L+2W = P
2L = -2W+P
L = (-2W+P)/2

Now plug in the given perimeter P = 28
L = (-2W+P)/2
L = (-2W+28)/2
L = (-2W/2)+(28/2)
L = -W + 14
L = 14 - W
Whatever the width (W) is, subtract it from 14 to get the length L. To ensure L is positive, we must restrict W < 14. This is the largest W can be. The smallest is W = 0. More specifically W is in the interval 0 < W < 14 which is the domain of the function we'll construct below.

A = Area of rectangle
A = Length*Width
A = L*W
A = (14-W)*W
A = -W^2 + 14W

If you were to graph that function, or convert to vertex form, you'd find that (7,49) is the vertex point. It is the highest point for this particular function. We see that A = 49 is the largest possible area. So the range is 0 < A < 49.

Side note: the rectangle area maxes out when all four sides are the same length (in this case 7). So we simply divide the perimeter 28 by 4 to get the side length to form a square of maximum area. This idea is useful if you are given a fixed amount of fencing to work with and you want to build the largest rectangular enclosure possible.

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Answers:
A = -W^2 + 14W
Domain = 0 < W < 14
Range = 0 < A < 49

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