SOLUTION: I'm getting stuck at the end of this one...any help would be greatly appreciated!a pen with one side against a wall WITH 6M of fencing. X is the width and the length is (6-2x). The

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: I'm getting stuck at the end of this one...any help would be greatly appreciated!a pen with one side against a wall WITH 6M of fencing. X is the width and the length is (6-2x). The      Log On


   

Question 704198: I'm getting stuck at the end of this one...any help would be greatly appreciated!a pen with one side against a wall WITH 6M of fencing. X is the width and the length is (6-2x). The area is a=x(6-2x).
A). What dimensions of the pen will produce the maximum area?
B) what is the maximum area?
I have done the following
A=(6x-2x)
Max area occurs at axis of symmetry
X=-6/-4
3/2 is the max area
L=6-2(3/2)
Length=5.5?
I have a hard time working with fractions...any help would be greatly appreciated!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+A+=+x%2A%28+6+-+2x+%29+
+A+=+-2x%5E2+%2B+6x+
This is a parabola with a maximum
The x-coordinate of the vertex is at
+x%5Bv%5D+=+-b%2F%282a%29+
+x%5Bv%5D+=+-6%2F%282%2A%28-2%29%29+
+x%5Bv%5D+=+3%2F2+
width = 3/2
length = +6+-+2x+=+6+-+2%2A%283%2F2%29+%29+
+6+-+2%2A%283%2F2%29+=+3+
-----------------
So, the 2 sides are 3/2 + 3/2 = 3
and the length = 3
Total length = +3+%2B+3+=+6+
OK