|
Question 919285: Hey guys im having a hard time with this word problem. Ive spent about 2 hours trying to figure it out with no success. Please show the steps, thank you!
A farmer with 12,000 feet of fencing wants to enclose a rectangular field and then divide it into two (non-equal) plots by running a fence parallel to one of its sides. What is the largest area that can be enclosed?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website!
total fencing is equal to 12,000 feet.
the perimeter of the fenced in area will be equal to 2L + 2W.
if you string a fence parallel to the length of the enclosure, then the total fencing will be 3L + 2W.
You have 2L for the outside length and 2W for the outside width and L for the string of fence that will be strung parallel to the length and between the two outside lengths.
since you have 12,000 of total fencing available, your formula is therefore:
12,000 = 3L + 2W for the fencing required.
the area of the enclosure will be L * W.
you want to maximize the total area and then you can subdivide as you wish to get 2 separate areas within the total enclosure that will have their individual areas add up to the entire area of the enclosure.
these individual areas can be equal, but don't have to be.
from 3L + 2W = 12,000, you can solve for W to get:
2W = 12,000 - 3L which then becomes W = 6,000 - 3/2 * L
in the formula for area, you get A = L * W becomes A = L * (6,000 - 3/2 * L) which becomes A = 6,000 * L - 3/2 * L^2
if you replace L with x, then you can recognize this as a quadratic equation of y = 6,000 * x - 3/2 * x^2 where:
y = the total area of the enclosure
x = the width of the enclosure.
this quadratic equation can be written as:
y = -3/2 * x^2 + 6,000 * x.
since this equation is in standard form of y = ax^2 + bx + c, then you get:
a = -3/2 and b = 6,000
the formula for the axis of symmetry is x = -b/2a which becomes x = -6000 / (2*(-3/2)) which becomes x = -6000 / (-6/2) which becomes x = -6000 / -3 which becomes x = 2,000.
when x = 2,000, y = -3/2 * (2,000)^2 + 6,000 * 2,000 which becomes y = -6,000,000 + 12,000,000 which becomes y = 6,000,000 square feet.
the vertex of the quadratic equation is (2,000,6,000,000).
the vertex of the equation is also the maximum point of the equation.
y represents the area and x represents the length of the enclosure.
area equals length times width.
you now know the area and you know the length because y represents the area and x represents the length.
to find the width, divide 6,000,000 by 2,000 and you get a width of 3,000 feet.
the dimensions of the enclosure are 2,000 feet in length by 3,000 feet in width.
3 times the length gives you 6,000 feet of fencing for the length and 2 times the width give you 6,000 feet of fencing for the width for a total of 12,000 feet of fencing.
the maximum area of the enclosure is 6,000,000 square feet.
you can put the inside fencing parallel to the length at any point in the width.
this doesn't affect the total area of the enclosure, since the total area of the enclosure is equal to the sum of the 2 sub-divided enclosures within it.
i am reasonably sure this is the answer you are looking for.
i checked the answer in excel by listing all possible length and width in 100 feet increments and excel confirmed that these measurements provide the maximum area of the enclosure.
your enclosure will look something like this:
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
x x
x x
x x
x x
x x
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
x x
x x
x x
x x
x x
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
the horizontal lines are each 2,000 feet in length.
the vertical lines that extend from the bottom horizontal to the top horizontal are each 3,000 feet in length.
that middle line parallel to the two outside horizontal lines can be positioned anywhere along the vertical lines without affecting the total area of the enclosure.
|
|
|
| |
