The Rancher's Corral
A rancher has 400 feet of fencing material and needs to built a corral
that's divided into three equal rectangles. Here's a diagram...
What are the length and width (x and y) of each rectangle that will
maximize the total fenced area, and what is the area enclosed?
Ragknot
hjg
that's divided into three equal rectangles. Here's a diagram...
What are the length and width (x and y) of each rectangle that will
maximize the total fenced area, and what is the area enclosed?
Ragknot
hjg
Labels: SharedPuzzle
posted by Rajesh Lal at
10:31 PM
15 Comments:
max area is 5000 sq ft
x=100/3 ft
y=50 ft
y=66.66666666...
x=22.22222222...
maximum area = square of 66.666 by 66.666
Total sq foot is 10,000 100x100
3 Compounds 10,000 /3 = 3333.33
Therfore the sides of the compound must add up to 833.33 feet.
Therefore there are any number of combinations that would make x and Y and 1/3rd of the total area.
Cheers Limey
Sorry Typo for the perimeter of the compound is not 833 but should be should be 266.66 feet
x =33.3
y=50
Area = length x width
A=3x*y
6x+4y=400
3x+2y=200
2y=200-3x
y=100-1.5x
A=3x*y
A(x)=3x(100-1.5x)
A(x)=300x-4.5x^2
Maximum is where slope = 0
0=300-9x
9x=300
x=300/9 = 33.333...
then y=100-1.5x
y=100-1.5*100/3
y=50
x = 30
y = 54
there are 6 x's total 180
there are 4 y's total 220
180 + 220 = 400
hope this helps
sorry y =55
x=20 and y=70 so 20*6=120 feet add 70*4=280 feet total 400 feet
Wow. This has been totally over thought in my opinion.
Maximum area is achieved by equal sides... for example 10x10=100 rather than 9x11 which is only 99. So you get the most area with making all walls 40 feet. Each section is 160sq ft or 40x40.
I really don't see how you beat this and a good portion of the other calculations don't allow for the 10 walls required.
Technically these are 3 square corrals rather than rectangles but if it has to be rectangular than 9.999999999x10.000000001 seems the next best thing.
400' of fencing/6 sections of fence
=66.666666' of fencing per fence section
=22.222222'*66.666666'
=1481.481451851852 sqft per rectangular tract.
The overall dimensions are 50 by 100 = 5000 sqft. This takes up 300 feet of fencing.
The other 100 fencing is used to make two 50 segments dividing the 5000 sqft into 3 equal parts.
X=33.333... ft
Y=50 ft
A=5000
First we know that the total length of all the fences is 400ft.
6x+4y=400
The area can be found using the formula:
A=3xy
The next step is to solve for y so we are left with one variable
6x+4y=400
3x+2y=200
y=100-1.5x
A=3xy
A(x)=3x(100-1.5x)
A(x)=300x-4.5x²
This function will give us the area of the shape depending on how long is x. Now if we graph this formula, we will result in a parabola that faces down. The very top (absolute maximum/vertex) of the parabola gives you the value of the largest area possible. To find the value at the top you can use a calculator or just estimate using the graph. To find the exact answer there are two ways; the algebraic way or the calculus way.
**** Algebraic Way ****
To find the coordinates of the tip (vertex) we need to get the formula A(x) in the form f(x)=a(x-p)²+q where (p,q) is the coordinates of the vertex.
A(x)=300x-4.5x²
A(x)/-4.5=x²-66.6667x
A(x)/-4.5=x²-66.6667x+(33.3333²-33.3333²)
A(x)/-4.5=(x²-66.6667x+33.3333²)-33.3333²
A(x)/-4.5=(x-33.3333)²-1,111.1111
A(x)=(x-33.3333)²+5000
The coordinates of the vertex is (100/3, 5000). This means when x is 100/3ft long, the area is 5000ft².
To find y, we substitute it back into the formula
6x+4y=400
6(100/3)+4y=400
200+4y=400
4y=200
y=50ft
In conclusion
x=100/3
y=50
area=5000
**** Calculus Way ****
At the absolute maximum, the slope is equal to 0. to find the slope of the function A(x) we have to take the derivative.
A(x)=300-4.5x²
A'(x)=300-9x
Solve for when A'(x)=0
0=300-9x
x=100/3
This means when x=100/3 the function A(x) will be at its maximum, therefore the area is at its maximum.
To find y, we substitute it back into the formula
6x+4y=400
6(100/3)+4y=400
200+4y=400
4y=200
y=50
The area can then be found for substituting both variables in.
A=3xy
A=3(100/3)(50)
a=5000
In conclusion
x=100/3
y=50
area=5000
To: Sam
Excelleant except for when you typed...
...take the derivative
A(x)=300-4.5x²
It should have been...
A(x)=300x-4.5x²
For the large rectangle it is 100 x 50 making it 5000 sq ft. Then break it down into smaller rectangles 100/3 equals 33.33 * 50 equals 1666.66 sq ft for each rectangle.
Post a Comment
Links to this post:
Create a Link
<< Home