SOLUTION: A rectangular field is to be enclosed by a fence and divided into two parts by another fence. Find the maximum area that can be enclosed and seperated in this way with 800 m of fen

Question 67213: A rectangular field is to be enclosed by a fence and divided into two parts by another fence. Find the maximum area that can be enclosed and seperated in this way with 800 m of fencing
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A rectangular field is to be enclosed by a fence and divided into two parts by another fence. Find the maximum area that can be enclosed and separated in this way with 800 m of fencing.
:
Because of the two areas, separated by a fence, the total fence would be:
2L + 3W = 800
:
2L = 800 - 3W
Divide both sides by 2 and you have
L = 400 - 1.5W
:
Area = L * W
:
Substitute (400-1.5W) for L:
Area = W(400-1.5W)
Area = 400W - 1.5W^2
:
In the form of a quadratic equation:
y = -1.5x^2 + 400x
:
Find the vertex: x = -b/(2a)
In our equation; a = -1.5 and b = 400
:
x = -400/(2*-1.5)
x = -400/-3
x = 133.33 is the width at which max area occurs
:
To find the max area, substitute 133.33 for the W in the area equation:
Area = 400W - 1.5W^2
Area = 400(133.33) - 1.5(133.33^2)
Area = 53332 - 1.5(17776.89)
Area = 53332 - 26665.33
Area = 26666.67 sq meters is the max area
:
:
Check solution by finding area:
L = 400 - 1.5W
L = 400 - 1.5(133.33)
L = 400 - 200
L = 200 meters:
Area = 200 * 133.33 = 26666
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