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What is the shortest length of fence that the rancher can use?

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2 Answers

I used the same target equation and the same constraint about the area and I obtained a minor length and the same area. 
I do not think this is really a question for elementary algebra if you want it solved exactly. Trust me I am doing this "de bona fide" because is the way of getting the exact answer to this problem, one can make adjustments but will never get the exact minimum.

Objective Cell (Min)
Cell Name Original Value Final Value
$E$3 3x+2y = 5 6928.22

Variable Cells
Cell Name Original Value Final Value Integer
$E$4 x 1 1154.70
$E$5 y 1 1732.06
The area is 2000000, the dimensions are 3 sides of 1154.70 and two sides of 1732.06 and the total length of fence will be 6928.22 feet.

Cell Name Cell Value Formula Status Slack
$E$6 xy 2000000 $E$6=2000000 Binding 0
Let x = the length of one side of the rectangular field and let y = the length of the other side.  The Area of the enclosed rectangular field is:
Area = x*y = 2,000,000 ft2
Therefore y = 2,000,000/x
The length of the fence enclosing the area is 2x + 2y.  Add another x for the fence dividing the area in half.  The Length of the fence, then, is:
L = 2x + 2y + x = 3x + 2y
Since y = 2,000,000/x
L = 3x + 2(2,000,000/x) = 2x + (4,000,000/x)
To find the minimum length, take the derivative of L, set it to zero, and solve for x:
dL/dx = 2 - (4,000,000/x2)
0 = 2 - (4,000,000/x2)
2,000,000 =  x2
103√2 ≅ 1414.2 ft = x      [-1414.2 is also a solution, but we can't have a negative length of fence]
y = 2,000,000/x = 2*106/103√2 = 103√2 ≅ 1414.2 ft        [the enclosed area is a square]
L = 3x + 2y = 5(103√2) = 5000√2 ≅ 7071.1 ft
Area = (103√2)*(103√2) = 2*106 = 2,000,000 ft2


Philip, correct me if I'm wrong. For a given area, the perimeter of a rectangle is minimum when the rectangle is a square. Therefore the simplest solution would be √2,000,000. I think this is an Algebra I or II question.
I don't think kids get into derivatives at this stage. Just my opinion and my way of thinking.