SOLUTION: a rancher has 1000m of fencing to create 6 rectangular corrals of equal area. What is the maxmum total area and what are the dimensions of the enclosed area? (there is a diagram th

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Question 190118: a rancher has 1000m of fencing to create 6 rectangular corrals of equal area. What is the maxmum total area and what are the dimensions of the enclosed area? (there is a diagram that has a rectangle with length=x and width=y and there are intesecting lines inbetween)
Answer by ankor@dixie-net.com(22740) (Show Source):
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a rancher has 1000m of fencing to create 6 rectangular corrals of equal area.
What is the maxmum total area and what are the dimensions of the enclosed area?
(there is a diagram that has a rectangle with length=x and width=y and there
are intersecting lines in between)
:
I picture this as having two layers of 3 rectangles inside a rectangle x by y
This would give you 3 lengths and 4 widths, therefore:
3x + 4y = 1000
4y = 1000 - 3x
y = - x
y = 250 - .75x
:
Area = x * y
substitute (250-.75x) for y:
A = x(250-.75x)
A = 250x - .75x^2
Written as a quadratic equation:
A = -.75x^2 + 250x
:
Maximum area occurs at the axis of symmetry. Use x = to find this.
In this equation a=-.75, b = 250
x =
x =
x = +166 meters is the length
:
Find the width:
y = 250 - .75(166.67)
y = 125 meters is the width
:
This would give a max area of 166.67 * 125 = 20,833 sq/meters