Answer :
The 2 widths make 2x
The length is (7000 - 2x) (not just x)
Area = (7000 - 2x) * x
Area = 7000x - 2x^2
Area = - 2(x^2 - 1,400x)
Area = -2 (x^2 - 1,400x + (1,400/2)^2 ) + 7812.5
Area = -2 (x - 62.6)^2 + 7812.5
The maximum area = 7812.5
when
the width = 62.5 and
the length = 250 - 2*62.6
the length = 250 - 125
the length = 125
The length is (7000 - 2x) (not just x)
Area = (7000 - 2x) * x
Area = 7000x - 2x^2
Area = - 2(x^2 - 1,400x)
Area = -2 (x^2 - 1,400x + (1,400/2)^2 ) + 7812.5
Area = -2 (x - 62.6)^2 + 7812.5
The maximum area = 7812.5
when
the width = 62.5 and
the length = 250 - 2*62.6
the length = 250 - 125
the length = 125
Let
x--------> the length side of the rectangular plot (assume side along the river)
y------> the width side of the rectangular plot
we know that
the perimeter of the rectangular plot is equal to
[tex]P=x+2y[/tex]
[tex]P=7,000\ m[/tex]
so
[tex]7,000=x+2y[/tex]
Clear variable y
[tex]y=(7,000-x)/2[/tex] -------> equation [tex]1[/tex]
The area of the rectangular plot is
[tex]A=x*y[/tex] ------> equation [tex]2[/tex]
substitute equation [tex]1[/tex] in equation [tex]2[/tex]
[tex]A=x*[(7,000-x)/2][/tex]
[tex]A=x*[(7,000-x)/2]\\\\ A=3,500x-0.5x^{2}[/tex]
we know that
To find the larger area that can be enclosed------> Find the vertex of the quadratic equation
The quadratic equation is a vertical parabola open down
so
the vertex is a maximum
using a graphing tool
see the attached figure
the vertex is the point [tex](3,500,6,125,000)[/tex]
that means
For [tex]x=3,500\ m[/tex]
the largest area is [tex]6,125,000\ m^{2}[/tex]
Find the value of y
[tex]y=(7,000-3,500)/2=1,750\ m[/tex]
the dimensions of the rectangular plot are
[tex]Length=3,500\ m\\Width=1,750\ m[/tex]
therefore
the answer is
The largest area that can be enclosed is [tex]6,125,000\ m^{2}[/tex]
