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.
This seems to be an optimization problem. Does the river count as one side that doesn't need to be fenced?
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.
So, 2WidthCattle + 2LengthSheep + 2WidthCattle + 2 LengthSheep = 240 yards of fencing.
And we know that length * width = area.
Ok, Im following so far
We also know that squares work the best, in terms of area.
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.
We would do well to maximize both the squares so in the end, we can have 8lengths = 240 yards.
Can you solve the length of each side from there?
30
Exactly, so each side is 30, which gives you an area of.? (A=Length*Width)
900
But I thought I had to use optimization
And since you have two pens, the max would be 900*2 sq. yards
Squares are the optimization.
so we don't have to use the derivative or anything like that?
Is this right?! Are we forgetting about the fence in the middle that separates the animals
It's two separate pens.
we are double counting the 30 side where the 2 squares are joined
Or, that is what I am getting?
two separate pens
So we should be able to get a bigger area if we share one side
I just thought we had to use the derivative of some function to get the max area
I think I solved it...if we share sides we can get a bigger area. We will need to use the derivative to solve.
Ok, that is probably right because that is what we are doing
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
You with me so far? Sorry about the crappy drawing.
yea i am
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
To be 40*60 or 2400
What do you think? It looks to be right to me but double check yourself.
yea that sounds right thanks a lot!
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