A couple of questions regarding AREAS (and getting the formulas,etc.)

  • Thread starter Thread starter mathzeroh
  • Start date Start date
  • Tags Tags
    Areas Couple
AI Thread Summary
The discussion revolves around solving a problem involving maximizing the area of a garden enclosed by 80 feet of fencing, with various configurations including a barn. Participants clarify the correct equations for calculating area, emphasizing the importance of using the right quadratic forms and finding the vertex for maximum area. There is confusion over the application of the quadratic formula and completing the square, with some members correcting each other's calculations. Ultimately, the correct dimensions and areas for the different garden designs are confirmed, leading to a resolution of the initial frustrations. The conversation highlights the collaborative effort to understand and solve the mathematical problem effectively.
mathzeroh
Messages
98
Reaction score
0
OK this problem has been giving me nonstop headaches and nightmares for over 4 days! :cry:

I just can't seem to understand something here! :frown:

Here's the actual problem:

Suppose that you have 80 feet of fence to enclose a garden. For each garden design below, find the dimensions of the garden which produce the maximum area. The "bold" lines indicate where the fence must go. Note: diagrams below not to scale.

1. Barn on one side. No fence necessary there.

max area = _______________


2. Garden with fence divider in the middle, barn on one side.

max area = _______________


3. No barn, fence on all four sides, and two center dividers.

max area = _______________


4. L - shaped barn used for one side and 10 ft. of the other side. No fence required for 3 ft. gate.

max area = _______________
 

Attachments

  • 1.JPG
    1.JPG
    3 KB · Views: 419
  • 2.JPG
    2.JPG
    4.8 KB · Views: 440
  • 3.JPG
    3.JPG
    5.1 KB · Views: 425
Physics news on Phys.org
And here is the image for the fourth question:
 

Attachments

  • 4.JPG
    4.JPG
    5.8 KB · Views: 413
For number one, the width let's say is w, that would make the length of the side that is not connected to the barn:

80-2w

So the area is:

A=w(80-2w)
A=(2w^2)-80w <----***TYPO! it should say: (-2W^2)+80W


And I'm sure that can go into the quadratic formula. that would give us the value for w and from there, we can find the missing side. BUT, if we find the vertex of the graph of this problem using this formula:

V=-b/2a

is this the same thing as getting the value of w using the quadratic formula?

And for nmber 2. this is what i got:

A=w(80-3w)
A=(-3w^2)+80w

Vertex:

V=-80/(2(-3))
V=40/3

And then i plugged that number (40/3) into the original equation to get the value for the Area. Which was: 533.333333333333...

so was that right? i highly doubt it! :cry: BUT WHY IS THAT WRONG? :confused:


and number 3:

well let's just say that it was a total distaster, so I'm not even going to put that one up. and just for the record, i gave up after that and number four wasn't even attempted. :frown:
 
Last edited:
mathzeroh said:
For number one, the width let's say is w, that would make the length of the side that is not connected to the barn:

80-2w

So the area is:

A=w(80-2w)
A=(2w^2)-80w

That's incorrect.You shouldn't have multiplied with "-1",because u changed the whole thing... :wink: Now your area function will have a minimum and not a maximum... :wink:

And for nmber 2. this is what i got:

A=w(80-3w)
A=(-3w^2)+80w

Vertex:

V=-80/(2(-3))
V=40/3

And then i plugged that number (40/3) into the original equation to get the value for the Area. Which was: 533.333333333333...

so was that right? i highly doubt it! :cry:

It's correct... :smile:

Daniel.
 
mathzeroh said:
OK this problem has been giving me nonstop headaches and nightmares for over 4 days! :cry:

I just can't seem to understand something here! :frown:

Here's the actual problem:

Suppose that you have 80 feet of fence to enclose a garden. For each garden design below, find the dimensions of the garden which produce the maximum area. The "bold" lines indicate where the fence must go. Note: diagrams below not to scale.

1. Barn on one side. No fence necessary there.

max area = _______________


2. Garden with fence divider in the middle, barn on one side.

max area = _______________


3. No barn, fence on all four sides, and two center dividers.

max area = _______________


4. L - shaped barn used for one side and 10 ft. of the other side. No fence required for 3 ft. gate.

max area = _______________
OK, let's try and stop your nightmares...

For the first question, I think you have the wrong equation. What you need to do is come up with an equation for the area (which will be quadratic), complete the square and find the vertex (the x value, that is).
You know the sum of the 3 sides will be 80 feet. 2 of the sides will have the same length (according the the attachment). So the sides will have the following lengths: x, x, 80-2x. Now try and find an equation for the area. Let me know what you got. :smile:
 
dextercioby said:
That's incorrect.You shouldn't have multiplied with "-1",because u changed the whole thing... :wink: Now your area function will have a minimum and not a maximum... :wink:

could you please elaborate a bit further on that? i mean if its no trouble for you. i just don't know where i multiplied with -1. :confused:

thank you! :smile:

dextercioby said:
It's correct... :smile:

Daniel.

REALLY?! :bugeye: thanks man! :cool:
 
Find the "x" & "y" for the vertex of the following parabola
y(x)=x(80-2x)

There's been a missunderstanding.I chose another variable to be the "x"...

Daniel.

EDIT:This is not the function u'll need to solve the problem #1.

Disregard this EDIT...
 
Last edited:
dextercioby said:
Find the "x" & "y" for the vertex of the following parabola
y(x)=x(80-2x)

Daniel.
Thanx for giving the answer to my question to him... I was trying to get him to think... :devil:
 
For the forst problem your equation.It should be
A(x)=x(80-\frac{x}{2})

Maximize this one...

Disregard my previous post (the EDIT,of course).

Daniel.

EDIT:See the next post... :-p
 
Last edited:
  • #10
dextercioby said:
For the forst problem your equation (function for area is wrong).It should be
A(x)=x(80-\frac{x}{2})

Maximize this one...

Disregard my previous post (except the EDIT,of course).

Daniel.
Correct me if I'm wrong, but wasn't the first equation you posted right,
A = 80-2x(x) ? According to my diagram, it is... :confused:
 
  • #11
dextercioby said:
For the forst problem your equation (function for area is wrong).It should be
A(x)=x(80-\frac{x}{2})

Maximize this one...

Disregard my previous post (except the EDIT,of course).

Daniel.

wait, \frac{x}{2})?? ok now I am totally lost. i thought Area=Length*Width

So its not this?:

A=LW
A=x(80-2x)
A=80x-(2x^2)

I even edited my post up there because of that typo. this is what i'd gotten while working on it and now it looks like its all wrong! :rolleyes:


christinono, i was going to answer to ur post, and i had it edited on the other post, but i guess it was wrong. :frown:
 
  • #12
christinono said:
Correct me if I'm wrong, but wasn't the first equation you posted right,
A = 80-2x(x) ? According to my diagram, it is... :confused:

that's what i was thinking too. :rolleyes: :rolleyes:
 
  • #13
mathzeroh said:
wait, \frac{x}{2})?? ok now I am totally lost. i thought Area=Length*Width

So its not this?:

A=LW
A=x(80-2x)
A=80x-(2x^2)

I even edited my post up there because of that typo. this is what i'd gotten while working on it and now it looks like its all wrong! :rolleyes:


christinono, i was going to answer to ur post, and i had it edited on the other post, but i guess it was wrong. :frown:
I still think we are right...
I remember doing the same type of questions in Math 20IB a couple years ago...
Anyways, do you know how to find the vertex by completing the square (or you could just use your calc)?
 
  • #14
THEY ARE BOTH CORRECT AND SHOULD YIELD THE SAME MAXIMUM AREA.

I took another "x" than the OP...

Daniel.
 
  • #15
dextercioby said:
THEY ARE BOTH CORRECT AND SHOULD YIELD THE SAME MAXIMUM AREA.

I took another "x" than the OP...

Daniel.
I get it, but still don't understand WHY you'd want to do it that way. It's much simpler to define to of the sides as "x". I guess if you wanted, you could define those sides as 2\pi x and solve it that way...
 
  • #16
christinono completing the square won't do anything because of this:

b=-2, right?

so you have to add this to both sides:

(-2/2)^2

which is 1, so it would be this:

A+1=(-2x^2)+80x+1

so now u have to subtract the one from that side where the A is and you get the original equation...unless i missed something. this is so time consuming, I am sorry for all the trouble u guys had to go through. :frown:
 
  • #17
Yes,of course,there's an infinity of ways of solving the problem correctly...Let's hope the OP finds at least one if them... :-p

Daniel.
 
  • #18
mathzeroh said:
christinono completing the square won't do anything because of this:

b=-2, right?

so you have to add this to both sides:

(-2/2)^2

which is 1, so it would be this:

A+1=(-2x^2)+80x+1

so now u have to subtract the one from that side where the A is and you get the original equation...unless i missed something. this is so time consuming, I am sorry for all the trouble u guys had to go through. :frown:

This is your function:
y(x)=-2x^{2}+80x

Complete the square...

Daniel.
 
  • #19
dextercioby is the area 800? it has to be.

because y=40
and x=20

and yeah, I'm hoping that's the right answer. :rolleyes:
 
  • #20
mathzeroh said:
christinono completing the square won't do anything because of this:

b=-2, right?

so you have to add this to both sides:

(-2/2)^2

which is 1, so it would be this:

A+1=(-2x^2)+80x+1

so now u have to subtract the one from that side where the A is and you get the original equation...unless i missed something. this is so time consuming, I am sorry for all the trouble u guys had to go through. :frown:
What in the world is b, and why does it equal -2? :confused: :confused:
 
  • #21
mathzeroh said:
dextercioby is the area 800? it has to be.

because y=40
and x=20

and yeah, I'm hoping that's the right answer. :rolleyes:
Yes, that's right :!) :smile:
 
  • #22
mathzeroh said:
dextercioby is the area 800? it has to be.

because y=40
and x=20

and yeah, I'm hoping that's the right answer. :rolleyes:

Yes,the area is 800 and the sides are 20,20 & 40...

Daniel.
 
  • #23
dextercioby said:
This is your function:
y(x)=-2x^{2}+80x

Complete the square...

Daniel.

ur right, but I'm hoping i got the right answer not using the completing the square method... :rolleyes:

christinono i was on crack earlier, my equation wasn't in the standard form which as everybody but me knows, is crucial! :-p :biggrin:
 
  • #24
THANK YOU BOTH!

i'm going to work on the other ones and i'll get back to u guys. thanks again! wow! that took way less time than 4 days! THANK YOU!
 
  • #25
mathzeroh said:
ur right, but I'm hoping i got the right answer not using the completing the square method... :rolleyes:

christinono i was on crack earlier, my equation wasn't in the standard form which as everybody but me knows, is crucial! :-p :biggrin:
What's that supposed to mean...I'm an IB student...I know what I'm talking about! :mad: :mad:
 
  • #26
christinono said:
What's that supposed to mean...I'm an IB student...I know what I'm talking about! :mad: :mad:

Take it easy...Slow down... :-p You'll get yourself a "warning"... :wink: :devil:

Daniel.
 
  • #27
sorry... :blushing:
 
  • #28
christinono said:
sorry... :blushing:


i didn't mean anything offensive. :frown: if i offended u I am sorry.

what i meant was that this is the standard form of a quadtratic equation:

(ax^2)+bx+c

and that's what i meant with "b", the coefficient in front of x.

but what i was doing was mixing up my coefficients, and i was talking about myself. I am sorry if i offended u. :frown:









So here's my answers for number 3:

width=10
length=20
Area=200


Correct? :smile:
 
Last edited:
  • #29
mathzeroh said:
i didn't mean anything offensive. :frown: if i offended u I am sorry.

what i meant was that this is the standard form of a quadtratic equation:

(ax^2)+bx+c

and that's what i meant with "b", the coefficient in front of x.

but what i was doing was mixing up my coefficients, and i was talking about myself. I am sorry if i offended u. :frown:









So here's my answers for number 3:

width=10
length=20
Area=200


Correct? :smile:
No hard feelings, I was just kidding :-p
As for your answer for number3, I got:
width=10
length=40
Area=400
 
  • #30
Aaahhh crap, what did i do wrong this time??

ok here's my work:

x=width
y=length

A=x(40-2x)
A=40x-(2x^2)

then i used the vertex formula to get x:

-40/(2(-2))
=40/4
=10

then:

y=40-2(10)
=40-20
=20


A=20*10
=200

was that wrong?
 
  • #32
dextercioby said:
Yes,it's correct.

Daniel.

who? me? :rolleyes:
 
  • #33
Wasn't the length of the fence supposed to be 80 feet? I think you used 40...
 
  • #34
mathzeroh said:
who? me? :rolleyes:


Yes,u are correct...Does it sound unbelievebly well,or what...? :-p

Daniel.
 
  • #35
Sorry, I made a mistake... You were right.
 
  • #36
dextercioby said:
Yes,u are correct...Does it sound unbelievebly well,or what...? :-p

Daniel.

:bugeye: yes! that is unbelievable.

christinono ur just trying to scare me. :-p


im going to work on number four now. :wink:

so, so far these are the answers:

1. Area = 800 sq. ft.

2. Area = 533.3333333... sq. ft.

3. Area = 200 sq. ft.

and now four...
 
  • #37
did it work? :wink: (just kidding...)
 
  • #38
I did question 2 and got the same answer. Is that unbelievable too? :smile:
 
  • #39
For the last point (#4),i get the area
S=666.125 ft^{2}

if u don't include the rectangle of sides 7 and 36.5.

If u do,then the total area is 921.625 ft^{2}

Daniel.
 
  • #40
dextercioby said:
For the last point (#4),i get the area
S=666.125 ft^{2}

if u don't include the rectangle of sides 7 and 36.5.

If u do,then the total area is 921.625 ft^{2}

Daniel.
Really?
I got the sides to be: 13.25ft, 20.25ft, 46.5ft
and the area to be 1091.125 sq. ft.

But maybe I'm wrong...
 
  • #41
Hahahaha, yes it did, and yes, that is amazing still! :-p


OK, here's what i got for the last one:

width=y
length=x


Width = 23.25
Length = 46.5

Area = 1081.125

IS THAT AMAZING, TOO? please tell me it is or I am going to jump off the Golden Gate Bridge right now! :biggrin:
 
  • #42
uh oh, looks like we've got three different answers here...hmm...what happened? i blew it somewhere maybe, but i have a feeling i was right. :confused:
 
  • #43
Yes, it is ABSOLUTELY INCREDIBLY AMAZING! :biggrin: :approve: :smile:
 
  • #44
k now I don't know if we've got the right answer... I though we did. :confused:
 
  • #45
christinono said:
Really?
I got the sides to be: 13.25ft, 20.25ft, 46.5ft
and the area to be 1091.125 sq. ft.

But maybe I'm wrong...
Sorry, typed my answer for the area wrong. I got 1081.125 sq. ft. too!
 
  • #46
What's the area function for the rectangle in the right (similar to problem #1).Isn't it
S(x)=x(73-2x)

Daniel.
 
  • #47
dextercioby said:
What's the area function for the rectangle in the right (similar to problem #1).Isn't it
S(x)=x(73-2x)

Daniel.
The length is: 80 = (x-10) + (x-3) + y
y is: y = 80-(2x-13)
The area is: A = (x)(93-2x)
The x value of the vertex is 23.25

That's how i did it...
 
  • #48
dextercioby said:
What's the area function for the rectangle in the right (similar to problem #1).Isn't it
S(x)=x(73-2x)

Daniel.

hmm..i got this:

x=93-2y

so the area function that i got out of that was this:

A=y(93-2y)

:rolleyes: :rolleyes:
 
  • #49
mathzeroh said:
hmm..i got this:

x=93-2y

so the area function that i got out of that was this:

A=y(93-2y)

:rolleyes: :rolleyes:
Same thing I did! :smile: (just different variables)
 
  • #50
christinono said:
The length is: 80 = (x-10) + (x-3) + y
y is: y = 80-(2x-13)
The area is: A = (x)(93-2x)
The x value of the vertex is 23.25

That's how i did it...

pretty similar to how i did it except my variables in the equations that i worked with are switched from what u used. but hey, a rose by any other name is just as sweet eh? i mean a variable is a variable. :-p
 

Similar threads

Construction Short Dividing fence on my property: what are my options?
Replies
16
Views
3K
MHB Garden Materials Help: Calculate Circumference, Perimeter, Areas & Costs
Replies
1
Views
2K
What Is the Maximum Area a Farmer Can Fence Using Different Lengths of Material?
Replies
2
Views
2K
MHB How Can a Farmer Minimize the Cost of Fencing a Rectangular Field?
Replies
2
Views
3K
What Are the Dimensions to Minimize Fence Length for a 384 ft² Study Area?
Replies
2
Views
2K
Maximizing area word problem WITHOUT LP formula
Replies
1
Views
2K
How to Verify Maximum Area of a Rectangular Pen with Limited Fencing?
Replies
1
Views
2K
Really with word problem - maximizing area
Replies
1
Views
2K
MHB Flowerchild1993's question at Yahoo Answers regarding minimizing cost of fence
Replies
1
Views
2K
Area of segment cut from a circle
Replies
2
Views
1K

Hot Threads

Recent Insights

Back
Top