math
posted by steven on .
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.

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 
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(12002x)
= 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(x300)^2 + 180000 
Hey, we think alike, lol

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.

a farmer has a rectangle field with an area of 3/4 square like the field is 1/2 mule wide what is the length of the field