Question

A Rancher wants to enclose 2 rectangular area near a river, one for sheep and one for cattle. There is 240 yd. of fencing available. What is the largest total area that can be enclosed.

Answer 1

This seems to be an optimization problem. Does the river count as one side that doesn't need to be fenced?

Answer 2

Yes it is an optimization problem. No I don't believe the river counts as a side that doesn't need to be fenced. It just says near a river so..I don't think so.

Answer 3

So, 2WidthCattle + 2LengthSheep + 2WidthCattle + 2 LengthSheep = 240 yards of fencing.

Answer 4

And we know that length * width = area.

Answer 5

Ok, Im following so far

Answer 6

We also know that squares work the best, in terms of area.

Answer 7

So let's simplify our perimeter equation. If all four sides are the same for both pens (to form a square = maximum area), we can simplify the perimeter to 4lenghts + 4lenghts = 240 yards of fencing.

Answer 8

We would do well to maximize both the squares so in the end, we can have 8lengths = 240 yards.

Answer 9

Can you solve the length of each side from there?

Answer 10

30

Answer 11

Exactly, so each side is 30, which gives you an area of.? (A=Length*Width)

Answer 12

900

Answer 13

But I thought I had to use optimization

Answer 14

And since you have two pens, the max would be 900*2 sq. yards

Answer 15

Squares are the optimization.

Answer 16

so we don't have to use the derivative or anything like that?

Answer 17

Is this right?! Are we forgetting about the fence in the middle that separates
the animals

Answer 18

It's two separate pens.

Answer 19

we are double counting the 30 side where the 2 squares are joined

Answer 20

Or, that is what I am getting?

Answer 21

two separate pens

Answer 22

So we should be able to get a bigger area if we share one side

Answer 23

I just thought we had to use the derivative of some function to get the max area

Answer 24

I think I solved it...if we share sides we can get a bigger area. We will need to use the derivative to solve.

Answer 25

Ok, that is probably right because that is what we are doing

Answer 26

First set up your sides as W = width. You will have 3 of these. You will also have two lengths. So you will have 2L+3W = 240. So L=120-3/2 W
120 - 3/2w
_____________
| | |
| w | w | w
______________
120 - 3/2w

Answer 27

You with me so far? Sorry about the crappy drawing.

Answer 28

yea i am

Answer 29

Ok, that was the hard part. Now I think all you have to do is find the Area function:

A = L*W or A = (120 - 3/2W) * W = \[120W - 3/2W^2 = A\]

Then take the derivative:
\[A \prime = 120 - 3W\]

Now find where the derivative = 0 which will be it's maximum or minimum.
\[0 = 120 - 3W \]

From here you get the answer of W = 40. Now you can just punch it in and find the total area to b

Answer 30

To be 40*60 or 2400

Answer 31

What do you think? It looks to be right to me but double check yourself.

Answer 32

yea that sounds right
thanks a lot!