SOLUTION: The perimeter of a rectangle is 30 m. If one side is x, express the area of the rectangle in terms of x. Show that there is no value of x such that the area is 60m^2. Thanks

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The perimeter of a rectangle is 30 m. If one side is x, express the area of the rectangle in terms of x. Show that there is no value of x such that the area is 60m^2. Thanks      Log On


   

Question 785549: The perimeter of a rectangle is 30 m. If one side is x, express the area of the rectangle in terms of x. Show that there is no value of x such that the area is 60m^2.

Thanks
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
x = length of one side in meters
y = length of the adjacent side in meters
(one is the length of the rectangle, and the other one is the width)
The perimeter (in meters) is
2x%2B2y=30 --> x%2By=15 --> y=15-x
The area is
area=x%2Ay
Substituting 15-x for y, we get
area=x%2815-x%29 <--> area=15x-x%5E2 (area expressed in terms of x)
THat is a quadratic function.
It graphs as a parabola, and the vertex is a maximum.
(It is realy a portion of a parabola, because we must only define it for
0%3Cx%3C15 to have positive numbers for width odf the rectangle.
area=15x-x%5E2-->area=-x%5E2%2B15x-56.25%2B56.25-->area=-%28x%5E2-15x-56.25%29%2B56.25-->area=-%28x-7.5%29%5E2%2B56.25
The maximum area is found when x=7.5, and it is 56.25.
At that point y=15-7.5=7.5 and the rectangle is a square.
For any other value of x, tyhe area is less than that:
%28x-7.5%29%5E2%3E0, -%28x-7.5%29%5E2%3C0, and area=-%28x-7.5%29%5E2%2B56.25%3C56.25