SOLUTION: I am stuck on this problem, any help would be greatly appreciated. Given a rectangle with a fixed perimeter of 372 feet, what dimensions would yield the maximum area? What is t

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I am stuck on this problem, any help would be greatly appreciated. Given a rectangle with a fixed perimeter of 372 feet, what dimensions would yield the maximum area? What is t      Log On


   

Question 463628: I am stuck on this problem, any help would be greatly appreciated.
Given a rectangle with a fixed perimeter of 372 feet, what dimensions would yield the maximum area? What is the maximum area?
Found 2 solutions by rfer, lwsshak3:
Answer by rfer(16322) About Me  (Show Source):
You can put this solution on YOUR website!
a square is max
372/4=93 ft per side
A=93^2
A=8649
sq ft

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
I am stuck on this problem, any help would be greatly appreciated.
Given a rectangle with a fixed perimeter of 372 feet, what dimensions would yield the maximum area? What is the maximum area
...
Standard form for parabola: y=A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex.
let x=length of rectangle
let y= width of rectangle
2x+2y=372
x+y=186
y=186-x
...
Area=x*y=x(186-x)=186x-x^2
Area=-x^2+186x
completing the square
-(x^2-186x+(186/2)^2)+(186/2)^2
-(x-93)^2+(93)^2=-(x-93)^2+8649
This equation is a parabola with vertex at (93,8649)
Dimensions:
x=93 ft
y=186-x=93 ft
maximum area=8649 sq ft