Let x be the shorter side, and y be the longer side
There would be 4 fences along the shorter side, and 2 fences along the longer side
4x + 2y = 800
Rewrite in terms of y:
y = 400 − 2x
The area of the rectangular field is
A = x*y
Replace Y with the equation above:
A = x(400 − 2x)
A = − 2x^2 + 400x
The area is a parabola that opens downward, the maximum area would occur at the parabola vertex.
At the vertex
x = −b/2a
= −400/[2(−2)]
= 100
y = 400 −2x
y = 400 -2(100)
y = 400-200
y = 200
The dimension of the rectangular field that maximize the enclosed area is 100 ft x 200 ft.
The dimension of the rectangular field that maximize the enclosed area is 100 ft by 200 ft.
The formula for calculating the vertex of a quadratic equation is given as:
x = -b/2a
From the given question
Let the shorter side be x
Let the longer side of the field be y
If there are 4 fences along the shorter side, and 2 fences along the longer side, hence;
4x + 2y = 800
Write the resultinq equation in slope-intecept form
2y = -4x + 800
y = -2x + 400
y = 400 − 2x
Area = x*y
Substitute the expression above into the area to have:
A = x(400 − 2x)
A = − 2x^2 + 400x
Since the parabola opens downward, the maximum area would occur at the parabola vertex as shown;
x = −b/2a
x = −400/2(-2)
x = 400/4
x = 100
y = 400 −2x
y = 400 -2(100)
y = 400-200
y = 200
Hence the dimension of the rectangular field that maximize the enclosed area is 100 ft by 200 ft.
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