Question

A rectangular field is to be enclosed by a fence. Two fences parallel to one side of the field divide the field into three rectangular fields. If 2400m of fence are available find the dimensions giving the max area.

Answer 1

let the width of the whole rectangle be x m (there will be 4 of these)
let the length be y m
then 4x + 2y = 2400
2x + y = 1200
y = 1200 - 2x

Area = xy
= x(1200-2x)
= -2x^2 + 1200x

Now, I don't know if you are studying Calculus.
If you do, then
d(Area)/dx = -4x + 1200
= 0 for a max area
x = 300

then y = 600
and the max area is (300(600) = 180000

If you don't know Calculus, complete the square on the above quadratic
you should end up with
Area = -2(x-300)^2 + 180000

Answer 2

Interesting...I did not think to complete the square to find the maximum area though. Based on steven's other questions I was not sure whether he was in Calculus.

Answer 3

Because there are 2 dividers, the perimeter = 2l + 4w, where l is the length of the field, and w is the width of the field (and the dividers)

2l + 4w = 2400

let x = l
2x + 4w = 2400
4w = 2400 - 2x
w = 600 - 0.5x

Now you can find the area.
A = l*w
A = x(600 - 0.5x)

Now you need to optimize area. A calculator is the easiest way, although if you are in Calculus you would likely have to show more work.

If you can use a graphing calculator, enter the equation in and find the maximum value on the graph.

I will demonstrate the calculus way below:
A = 600x - 0.5x^2
dA/dt = 600 - x

Find where dA/dt = 0

600 - x = 0
x = 600

dA/dt changes signs from + to - at x = 600, so there is a relative maximum at x = 600

Because x = length, the length is 600
w = 600 - 0.5x
w = 600 - 300
w = 300

The dimensions are 300m * 600m

Answer 4

Hey, we think alike, lol