SOLUTION: A farmer has 6000m of fencing and wishes to create a rectangular field subdivided into 4 congruent plots of land. Determine the dimensions of each plot of land if the area to be en

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: A farmer has 6000m of fencing and wishes to create a rectangular field subdivided into 4 congruent plots of land. Determine the dimensions of each plot of land if the area to be en      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 1201500: A farmer has 6000m of fencing and wishes to create a rectangular field subdivided into 4 congruent plots of land. Determine the dimensions of each plot of land if the area to be enclosed is a maximum.
Found 2 solutions by josgarithmetic, math_tutor2020:
Answer by josgarithmetic(39620) About Me  (Show Source):
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

It will depend on what layout the farmer goes with.

Are the plots laid in a single straight line (configuration A)?
Or are the plots in a 2x2 grid (configuration B)?

-----------------------

Configuration A:


Perimeter = sum of exterior sides
Perimeter = 8x+2y
8x+2y = 6000
2(4x+y) = 6000
4x+y = 6000/2
4x+y = 3000
y = -4x+3000

area of one plot = xy
area of four plots = 4xy
area of four plots = 4x(-4x+3000)
area of four plots = -16x^2+12000x

We wish to max out this area.
Use a graphing tool such as GeoGebra or Desmos to graph the parabola.
The vertex is located at (375, 2250000)
This is the highest point on the parabola.

This means a single plot that has horizontal width of x = 375 meters leads to a total max area of 2,250,000 square meters.
This applies only if configuration A is used.

The y value would be
y = -4x+3000
y = -4*375+3000
y = 1500

Each smaller plot is 375 meters by 1500 meters.
area of one plot = 375*1500 = 562,500 square meters
area of four plots = 4*562500 = 2,250,000 square meters

-----------------------

Configuration B:

Perimeter = sum of exterior sides
Perimeter = 4x+4y
4x+4y = 6000
4(x+y) = 6000
x+y = 6000/4
x+y = 1500
y = -x+1500

area of one plot = xy
area of four plots = 4xy
area of four plots = 4x(-x+1500)
area of four plots = -4x^2+6000x

Follow similar steps as the previous configuration.
This time the vertex is located at (750,2250000)
This means a horizontal width of x = 750 leads to a max area of 2250000 square meters.
This applies only if configuration B is used.

y = -x+1500
y = -750+1500
y = 750

Each smaller plot is 750 meters by 750 meters.
area of one plot = 750*750 = 562,500 square meters
area of four plots = 4*562500 = 2,250,000 square meters


--------------------------------------------------------
--------------------------------------------------------

Summary:

Configuration A has each smaller plot of 375 meters by 1500 meters.
Configuration B has each smaller plot of 750 meters by 750 meters.

Both configurations lead to a max total area of 2,250,000 square meters when considering all four plots.