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a rectangular field is to be divided up into 3 equal sections, enclosed by 160m of fencing.
the field is beside a barn, and the fencing will attach to the wall of the barn.
the wall of the barn requires no fencing. determine the dimensions of the entire
rectangular field that will result in the maximum area
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Wording is not perfect, so it requires to be re-worded.
160 m of fencing include one side along the barn (parallel to the barn wall),
and 4 fencing perpendicular to the barn wall.
So, let x be the total length of the fencing along the barn and y is the length of every of 4
fencing perpendicular to the barn wall.
We have then
x + 4y = 160 meters (total fencing).
The total area is
x*y = (160-4y)*y = 160y - 4y^2.
We want the maximum area.
You have the quadratic function f(y) = -4y^2 + 160y.
It gets the maximum at
y = , where " a " is the coefficient at y^2 and " b " is the coefficient at y.
In your case, the dimensions are
y = = = 20 meters; x = 160 - 4y = 160-4*20 = 80 m.
ANSWER. 80 meters along the barn wall and 20 meters perpendicular to it provide the maximum area.
Solved.
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On finding the maximum/minimum of a quadratic function see the lessons
- HOW TO complete the square to find the minimum/maximum of a quadratic function
- Briefly on finding the minimum/maximum of a quadratic function
- HOW TO complete the square to find the vertex of a parabola
- Briefly on finding the vertex of a parabola
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