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what do you understand about the derivative of a function?
Not much to be honest..I am really not doing so well with algebra...it's tough
thats fine, I can work with that :)
do you know what slope is? like if you look at a roof, how steep the roof is is determined by its slope right?
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?
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
Yes, that makes sense..
My pc is running a tad slow..
I was under the assumption that both sides are part of the slope; however the one on the right seeems straight
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
Are you referring to the x-axis
Oh, wow...yes, I see it...sorry, must be blind..I assumed it was just the x-axis. Okay, on the same page
Makes sense yes..thank you for being patient
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?
at any "given" point.... typos persist lol
I believe so...
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...
Okay, that maeks sense
good :) Now lets see how it applys to your problem with the fencing :)
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:
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.
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
Okay, got you so far
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?
Okay....im hanging in there
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 :)
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...
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
squared feet lol
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.
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 ;)
Youre welcome :) I hope it helps
The procedure was good, but I see a spot where I messed up the math :)
y = (320-3x)/2 ; I wrote /3 in the rest and messed up the numbers lol
So, should I replug in the numbers then?
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....
that worked out pretty lucky; 320/6 actually reduces to 160/3..... dont ask me how that worked in my favor lol
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
So, shall I keep the original information to use?
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 ;)
Wow, too funny.
Do you think you could help me out with something else since you are so good at explanation?
I can, but please post a new question for it :)
Oh, okay, I can do that...