Posted by **steven** on Sunday, July 12, 2009 at 10:53am.

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.

- math -
**Marth**, Sunday, July 12, 2009 at 11:07am
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

- math -
**Reiny**, Sunday, July 12, 2009 at 11:12am
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

- Marth ! -
**Reiny**, Sunday, July 12, 2009 at 11:13am
Hey, we think alike, lol

- math -
**Marth**, Sunday, July 12, 2009 at 11:26am
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.

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