SOLUTION: Cindy wants to construct five rectangular dog training areas side by side, using a total of 600 ft of fencing. What should the overall width and length be to maximize the area of t

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Question 1054244: Cindy wants to construct five rectangular dog training areas side by side, using a total of 600 ft of fencing. What should the overall width and length be to maximize the area of the five combined areas? What are the dimensions of each individual arena? What's the area of each arena?
I don't understand how to do the set up
Thanks!

Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Cindy wants to construct five rectangular dog training areas side by side, using a total of 600 ft of fencing. What should the overall width and length be to maximize the area of the five combined areas? What are the dimensions of each individual arena? What's the area of each arena?
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Sketch a rectangle that is wider than high.
Sketch in 4 vertical segments to create 5 equal-areas in the rectangle.
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You now have 6 vertical pieces plus a base and a top.
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Let each verticle be "x".
Each horizontal piece is (600-6x)/2 = 300-3x
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Area = x(300-3x) = -3x^2 + 300x
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Max occurs when x = -b/(2a) = -300/(2(-3)) = 50
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Dimensions of each small area::
height = x = 50 ft
width = (300-3x)/5 = (300-3*50)/5 = (300-150)/5 = 30 ft
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Each small area = 50 by 30 = 1500 sq ft.
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Cheers,
Stan H.


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +x+ = the sum of the 5
sides of the 5 rectangles.
Let +y+ = the short side of
the 5 combined rectangles
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+2x+%2B+2y+=+600+ is the "outside "
perimeter. I need to add in +4y+
for the internal sides of the 5 areas
+2x+%2B+2y+%2B+4y+=+600+
+2x+%2B+6y+=+600+
(1) +x+%2B+3y+=+300+
The formula for area is:
(2) +A+=+x%2Ay+
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(1) +x+=+300+-+3y+
Plug this into (2)
(2) +A+=+%28+300+-+3y+%29%2Ay+
(2) +A+=+-3y%5E2+%2B+300y+
This is a parabola with a positive peak
The y-value of the peak is:
+y%5Bmax%5D+=+-b%2F%282a%29+
+y%5Bmax%5D+=+-300%2F%28+2%2A%28-3%29+%29+
+y%5Bmax%5D+=+50+ ft
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(1) +x+%2B+3y+=+300+
(1) +x+%2B+3%2A50+=+300+
(1) +x+=+300+-+150+
(1) +x%5Bmax%5D+=+150+ ft
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The maximum area is:
+A%5Bmax%5D+=+x%5Bmax%5D%2Ay%5Bmax%5D+
+A%5Bmax%5D+=+150%2A50+
+A%5Bmax%5D+=+7500+ ft2
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The overall dimesions are:
50 x 150
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The dimensions of the individual areas are:
+x%2F5+ x +y+
+150%2F5+ x +50+
30 x 50
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Hope this is understandable & I got it right!