Still confused...so sorry...am not sure how to differentiate to find points where d(area)/dx has maxima. Yet another weakness..please help
Suppose that 320 feet of fencing are available to enclose a rectangular field and that one side of the field must be given double fencing. What are the dimensions of the field of maximum area?

- anonymous

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- amistre64

what do you understand about the derivative of a function?

- anonymous

Not much to be honest..I am really not doing so well with algebra...it's tough

- amistre64

thats fine, I can work with that :)

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- amistre64

do you know what slope is? like if you look at a roof, how steep the roof is is determined by its slope right?

- anonymous

okayu

- amistre64

the derivative of a funtion tells us the slope at any given point. So we can see how steep it is. does that make sense?

- anonymous

i just did slopes..i didn't do so great, but at least got a C that week ..I do understand the idea of it i supposee

- anonymous

Yes, that makes sense..

- anonymous

My pc is running a tad slow..

- amistre64

can you tell me which one of the green lines has a "0" slope? in other wirds, it is perfectly flat?

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- anonymous

I was under the assumption that both sides are part of the slope; however the one on the right seeems straight

- amistre64

is the one on the right flat? we are looking for the line that is straight across fromleft to right with no slope; like a flay desert

- amistre64

flat desert

- anonymous

Are you referring to the x-axis

- amistre64

here is another example; the places where the green lines are flat across the screen from left to right; are where there is "0" slope. Does that make sense?

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- anonymous

Oh, wow...yes, I see it...sorry, must be blind..I assumed it was just the x-axis. Okay, on the same page

- anonymous

Makes sense yes..thank you for being patient

- amistre64

;)

- amistre64

so the derivative of an equation tells us the "slope" at any gien point; can you see that when the slope = 0 that we have a high spot or a low spot in the graph?

- amistre64

at any "given" point.... typos persist lol

- anonymous

I believe so...

- amistre64

when the derivative of an equation is equal to 0; we have 3 possible conditions.
The point is a MAX
The point is a MIN
or the point is giving us a false reading and is an inflection point between concavities...

- anonymous

Okay, that maeks sense

- amistre64

good :)
Now lets see how it applys to your problem with the fencing :)

- anonymous

okay

- amistre64

Suppose that 320 feet of fencing are available to enclose a rectangular field and that one side of the field must be given double fencing. What are the dimensions of the field of maximum area?
We have a total of 320 feet of fencing; and we know that one side is double fenced. We also know that it is a rectangle: Like this:

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- amistre64

x + 2x + y + y = 320 feet around the outside right?
and the Area of the rectangle is: A = xy
lets use these to our benefit.

- amistre64

3x + 2y = 320
y = (320-3x)/3 is what we get for a "value" of y right?
Lets use this "y" value in our Area equation

- anonymous

Okay, got you so far

- amistre64

A = x(320-3x)/3
A = (320x)/3 -(3x^2)/3
Now we can use rules of derivatives to find the derivative of A with respect to x. Okay?

- anonymous

Okay....im hanging in there

- amistre64

The Power rule for derivatives states:
Dx(Cx^n) = Cn x^(n-1)
A' = 320/3 -6x/3 = 320/3 - 2x
When 320/3 - 2x = 0, we have either a MAX or a MIN condition :)

- amistre64

320/3 - 2x = 0 ; multiply by 3
320 - 6x = 0 ; subtract 320
-6x = -320 ; divide by -6
x = 320/6 = 53' 2/6 = 53' 1/3 feet; lets see if this works in our problem :)
And if you have any questions on it let me know...

- anonymous

Wow.,,.rough

- amistre64

3x + 2y = 320 ; x = 320/6 for simplicities sake :)
3(320/6) + 2y = 320
320/2 + 2y = 320
2y = 320-320/2
y = (320/2)/2
y = 320/4 = 80.
double check:
3(320/6) + 2(80) = 320
320/2 + 160 = 320
320/2 + 320/2 = 640/2 = 320 :)
So whats our Area then...
A = x(y)
A = 320(80)/6
A = 4266' 2/3 feet

- amistre64

squared feet lol

- anonymous

Wow...You know your stuff. I can use this to help me with the others. I really appreciate you taking the time to explain this to me.

- amistre64

The key is to find out what x and y are in terms of one variable:
We found that y = (320-3x)/2 right?
Then we used that "value" in the Area formula of a rectangle to determine its equation in terms of just the x variable.
A = 320x/3 - x^2
We then take the derivative to find a slope of "0"
The value of "x" that we find in the derivative is the MAX condition for x;
Use that value to determine the your real "y" value :)
then compute the area for xy ;)

- amistre64

Youre welcome :) I hope it helps

- amistre64

The procedure was good, but I see a spot where I messed up the math :)

- amistre64

y = (320-3x)/2 ; I wrote /3 in the rest and messed up the numbers lol

- anonymous

lol...

- anonymous

So, should I replug in the numbers then?

- amistre64

y = (320-3x)/2
A = x(320-3x)/2
A = 160x -(3/2)x^2
A' = 160 - 3x = 0
x = 160/3
...................
3x +2y = 320
3(160/3) + 2y = 320
160 + 2y = 320
2y = 160
y = 80
...........................
x = 160/3 ; y = 80
Area = 160(80)/3
Area = 4266' 2/3
I got different number for x bu tthe same values lol....

- amistre64

that worked out pretty lucky; 320/6 actually reduces to 160/3..... dont ask me how that worked in my favor lol

- amistre64

I see what I did; I transposed the x and y; got the resulting values which would have been the same regardless; and produced identical results...
Im a genius lol

- anonymous

So, shall I keep the original information to use?

- amistre64

the original is good; I actually used the transposed numbers the second time. So use work the first time as the answer if you wanna get a taste for what I did; if you wanna see how my stupidity plays in my favor for the second attempt, by all means, parse thru that one as well ;)

- anonymous

Wow, too funny.

- anonymous

Do you think you could help me out with something else since you are so good at explanation?

- amistre64

I can, but please post a new question for it :)

- anonymous

Oh, okay, I can do that...

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