anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
what do you understand about the derivative of a function?
anonymous
  • anonymous
Not much to be honest..I am really not doing so well with algebra...it's tough
amistre64
  • amistre64
thats fine, I can work with that :)

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amistre64
  • 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
  • anonymous
okayu
amistre64
  • 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
  • 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
  • anonymous
Yes, that makes sense..
anonymous
  • anonymous
My pc is running a tad slow..
amistre64
  • 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
  • anonymous
I was under the assumption that both sides are part of the slope; however the one on the right seeems straight
amistre64
  • 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
  • amistre64
flat desert
anonymous
  • anonymous
Are you referring to the x-axis
amistre64
  • 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
  • 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
  • anonymous
Makes sense yes..thank you for being patient
amistre64
  • amistre64
;)
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
  • amistre64
at any "given" point.... typos persist lol
anonymous
  • anonymous
I believe so...
amistre64
  • 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
  • anonymous
Okay, that maeks sense
amistre64
  • amistre64
good :) Now lets see how it applys to your problem with the fencing :)
anonymous
  • anonymous
okay
amistre64
  • 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
  • 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
  • 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
  • anonymous
Okay, got you so far
amistre64
  • 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
  • anonymous
Okay....im hanging in there
amistre64
  • 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
  • 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
  • anonymous
Wow.,,.rough
amistre64
  • 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
  • amistre64
squared feet lol
anonymous
  • 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
  • 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
  • amistre64
Youre welcome :) I hope it helps
amistre64
  • amistre64
The procedure was good, but I see a spot where I messed up the math :)
amistre64
  • amistre64
y = (320-3x)/2 ; I wrote /3 in the rest and messed up the numbers lol
anonymous
  • anonymous
lol...
anonymous
  • anonymous
So, should I replug in the numbers then?
amistre64
  • 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
  • amistre64
that worked out pretty lucky; 320/6 actually reduces to 160/3..... dont ask me how that worked in my favor lol
amistre64
  • 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
  • anonymous
So, shall I keep the original information to use?
amistre64
  • 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
  • anonymous
Wow, too funny.
anonymous
  • anonymous
Do you think you could help me out with something else since you are so good at explanation?
amistre64
  • amistre64
I can, but please post a new question for it :)
anonymous
  • anonymous
Oh, okay, I can do that...

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