SOLUTION: A large open area is to have a section surrounded by a rectangular fence. This rectangle is then divided into six smaller rectangles, using one dividing fence parallel to its lengt

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Question 1012462: A large open area is to have a section surrounded by a rectangular fence. This rectangle is then divided into six smaller rectangles, using one dividing fence parallel to its length and two fences parallel to its breadth. If the total length of fencing available is 1200m, find the maximum possible area.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A large open area is to have a section surrounded by a rectangular fence.
This rectangle is then divided into six smaller rectangles, using one dividing fence parallel to its length and two fences parallel to its breadth.
If the total length of fencing available is 1200m, find the maximum possible area.
:
This means there will be 3 sections = to the length and 4 sections equal to the width, therefore
3L + 4W = 1200
simplify, divide by 4
3%2F4L + W = 300
we can use .75 here
.75L + W = 300
W = -.75L + 300
:
Area = L*W
Replace W with -.75L+300
A = L(-.75L+300)
A = -.75L^2 + 300L
Max area occurs at the axis of symmetry, x = -b/(2a), using a & b from this equation
L = %28-300%29%2F%282%2A-.75%29
L = +200m is the length for max area
and
W = -.75(200) + 300
W = 150m is the width
:
Find the actual max area: 200 * 150 = 30,000 sq/m