maximum area

L

Lizzie

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Joined
Sep 8, 2005
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317
<u>The Problem:</u>
"A rancher has 300 feet of fencing to enclose a pasture bordered on one side by a river. The river side of the pasture needs no fence. Find the dimensions of the pasture that will produce a pasture with a maximum area. Use the second derivative to show that your area is a maximum."
<u>Where I am at and Where I got stuck:</u>
Ok, so my real problem is getting started on this. We did a problem similar to this in class, but I am not sure how to change what we did to work for this problem. I honestly do NOT want someone to do this for me. If anyone could just explain how to get started, I am pretty sure that I could figure out the rest of it. Any help is greatly appreciated. Thank you!
 
Hello, Lizzie!

A rancher has 300 feet of fencing to enclose a pasture bordered on one side by a river.
The river side of the pasture needs no fence.
Find the dimensions of the pasture that will produce a pasture with a maximum area.
Use the second derivative to show that your area is a maximum.
Click to expand...
Did you make a sketch?
Code: 
    ~ ~ ~ ~`~ ~ ~ ~ ~ ~ ~ ~
      |                 |
    y |                 | y
      |                 |
      * - - - - - - - - *
               x
He has 300 feet of fencing: \(\displaystyle \,x\,+\,2y\:=\:300\)

The area of the pasture is: \(\displaystyle \,A\:=\:xy\)


Is that enough to get you started?
 
How do you know that it is x+2y and not 2x+y? Or does it not matter??
 
All right, I now have two equations. . . and that helps, but my real problem is figuring out what to do with them to begin with. I am going to guess here, but maybe I will be correct. Would I somehow combine the two and go from there with derivatives and such?
 
Lizzie said:
All right, I now have two equations. . . and that helps, but my real problem is figuring out what to do with them to begin with. I am going to guess here, but maybe I will be correct. Would I somehow combine the two and go from there with derivatives and such?
Click to expand...
In short, yes.

If you solve for x in the perimiter and plug into the area, you will be able find the y value when the area is at maximum (same for y-x). The fact that it will be a quadratic should tell you how to find the max (even without calculus).

But once you have your A= f(x) or A = f(y) you need to differentiate A wih respect to whatever variable it is a function of. Then there is a method to tell whether a function is at an extrema (tangent line slope = 0), and then with the second derivative test, a way to tell if it is a max or a min, but not both.

I hope that helps,
Daon
 
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