You have 800 feet of fencing and you want to make two fenced in enclosures by splitting one enclosure in half.?
November 14, 2009 - 11:21 am
I have been trying for 3 hours and my brain is about to explode. Can someone please show me how to solve this problem.
You have 800 feet of fencing and you want to make two fenced in enclosures by splitting one enclosure in half. What are the largest dimensions of this enclosure that you could build?
800/5 is 160
Make it 160 feet wide and 160 feet long and then put a fence down the middle of it
The other guy is probably right.
November 14th, 2009 at 5:02 pm
sketch it out….
the pen will be w by L, and you’ll have a middle length of w as well
so the total length of fence will be 2L + 3w = 800
or, 2L = 800 - 3w ==> L = 400 - 1.5w
Area = w * L = w (400 - 1.5w)
this is the function to maximize
this will be a parabola that opens downward with zeros at w = 0 and 400 - 1.5w = 0
400 = 1.5w
w = 400 / 1.5 = 800 / 3 = 266.67
halfway between 0 and 800 / 3 = 400 / 3 = 133.333
if w = 133.33, then L = 400 - 1.5w = 400 = 1.5(133.33) = 200
the pen will be 133.33 by 200 for an area of 26,666.67 sq. ft
there will be 2 lengths of 200 and 3 lengths of 133.33
****
or, with calculus goodness:
A = w(400 - 3/2 w) = 400w - 3/2 w^2
A ‘ = 400 - 3w
A’ = 0 when 400 - 3w = 0 ==> 400= 3w ==> w = 400 / 3
if w = 400 / 3, then L = 400 - 3/2 (400 / 3) = 400 - 200 = 200
largest area will be 200 by 400 / 3 (looks familiar), and the shorter length will be repeated in the middle to split the enclosure into 2 equal pieces
edit: to the answer below: if you make it 160 by 160, and split it down the middle (800 / 5), then A = 160^2 = 25600
this is LESS than a pen of 200 by 133.33 = 26,667
the maximum area will be when the amount of fence you use in each direction is equal
(2 * 200 = 3(133.33))
edit: thank you Bob for sketching it out!
References :
November 14th, 2009 at 5:47 pm
800/5 is 160
Make it 160 feet wide and 160 feet long and then put a fence down the middle of it
The other guy is probably right.
References :
November 14th, 2009 at 6:05 pm
imagine this:
_ _ _ _ h_ _ _ _ _
|. . . . . . |. . . . . |
|. . . . . . |. . . . . | w
|. . . . . . |. . . . . |
|. . . . . . |. . . . . |
- - - - - - - - - - - - -
the amount of fence used is 3w + 2h = 800 and the area of the enclosure is h*w.
h = (800-3w)/2
so area = w*(400-3/2*w)
Then you do some calculus and find the max of that graph.
References :