Question

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?

Answer 1

what do you understand about the derivative of a function?

Answer 2

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

Answer 3

thats fine, I can work with that :)

Answer 4

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

Answer 5

okayu

Answer 6

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?

Answer 7

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

Answer 8

Yes, that makes sense..

Answer 9

My pc is running a tad slow..

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

flat desert

Answer 14

Are you referring to the x-axis

Answer 15

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?

Answer 16

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

Answer 17

Makes sense yes..thank you for being patient

Answer 18

;)

Answer 19

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?

Answer 20

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

Answer 21

I believe so...

Answer 22

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...

Answer 23

Okay, that maeks sense

Answer 24

good :)

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

Answer 25

okay

Answer 26

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:

Answer 27

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.

Answer 28

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

Answer 29

Okay, got you so far

Answer 30

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?

Answer 31

Okay....im hanging in there

Answer 32

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 :)

Answer 33

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...

Answer 34

Wow.,,.rough

Answer 35

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

Answer 36

squared feet lol

Answer 37

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.

Answer 38

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 ;)

Answer 39

Youre welcome :) I hope it helps

Answer 40

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

Answer 41

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

Answer 42

lol...

Answer 43

So, should I replug in the numbers then?

Answer 44

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....

Answer 45

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

Answer 46

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

Answer 47

So, shall I keep the original information to use?

Answer 48

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 ;)

Answer 49

Wow, too funny.

Answer 50

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

Answer 51

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

Answer 52

Oh, okay, I can do that...