SOLUTION: This is a quadratic word problem.
A garden is to be laid out in such a way that 3 rectangular sections are created by fences. If the total amount of fencing available is 200 ft
Algebra ->
Problems-with-consecutive-odd-even-integers
-> SOLUTION: This is a quadratic word problem.
A garden is to be laid out in such a way that 3 rectangular sections are created by fences. If the total amount of fencing available is 200 ft
Log On
Question 247981: This is a quadratic word problem.
A garden is to be laid out in such a way that 3 rectangular sections are created by fences. If the total amount of fencing available is 200 ft., find the dimensions that will maximize the total area. Found 2 solutions by Tiffany.Marie, Alan3354:Answer by Tiffany.Marie(1) (Show Source):
You can put this solution on YOUR website! My answer is that you should ask nicely. You should say "please help me with this question."
And Maybe you'll get what you need. I'm just saying. Anywho, If you want help with math dont ask me I fail at math. :D Sorry
Tiff
You can put this solution on YOUR website! A garden is to be laid out in such a way that 3 rectangular sections are created by fences. If the total amount of fencing available is 200 ft., find the dimensions that will maximize the total area.
---------------
I don't know why Tiff is here if she can't offer assistance.
Anyway,
The 200 feet of fence will be in 6 pieces. Call W the width of 2 pieces, and L the length of the 4 pieces.
2L + 4W = 200
The total area is L*W
2L + 4W = 200 --> L + 2W = 100 --> L = 100 - 2W
Sub for L in the 2nd eqn
Area = L*W
Area = W*(100 - 2W) = 100W - 2W^2
Find W to maximize the area
It's a parabola, and the max is at the vertex.
The vertex is as W = -b/2a = -100/-4
W = 25 feet
---------
L = 50 feet
