SOLUTION: Consider the set of all rectangles whose perimeter is 50 cm. For any one rectangle, once its width W is measured, it is possible to do some computations that yield the area of the

Question 1053197: Consider the set of all rectangles whose perimeter is 50 cm. For any one rectangle, once its width W is measured, it is possible to do some computations that yield the area of the rectangle. Verify this assertion by producing an explicit expression for area A as a function of width W. Find the domain and range of your function.

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
p = 2 * l + 2 * w
a = l * w

l = length
w = width
a = area
p = perimeter

if the perimeter is always 50, then p = 2 * l + 2 * w becomes:

50 = 2 * l + 2 * w

if you divide both sides of this equation by 2, you get

25 = l + w

if you know the width, then you can solve for l to get:

l = 25 - w

since a = l * w, you can replace l with 25 - w to get:

a = (25 - w) * w

simplify to get:

a = 25w - w^2

an example of how this works is as follows:

let's say that the perimeter is 50 and the width is 5.

you can solve for area by using the formula of a = 25w - w^2

since w = 5, this means that a = 25*5 - 25.

this results in a being equal to 100.

let's see if this makes sense.

p = 50 and w = 5

p = 2 * l + 2 * w

you get 50 = 2 * l + 10

solve for l to get l = 40/2 = 20

you get l = 20 and w = 5

p = 2*l + 2*w = 40 + 10 = 50 ***** perimeter checks out ok.

you get a = l*w = 20*5 = 100 ***** area check out ok.

looks like it works.

as long as the perimeter is always 50 and as long as you know the width, you can solve for the area using the formula of a = 25w - w^2.

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