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

In summary, the conversation involves a person struggling with a math problem involving finding the maximum area of a garden with a limited amount of fence. They discuss various equations for finding the area and attempt to solve the problem together.
  • #1
mathzeroh
98
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 = _______________
 

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  • #2
And here is the image for the fourth question:
 

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  • #3
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:
 
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  • #4
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.
 
  • #5
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:
 
  • #6
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:
 
  • #7
Find the "x" & "y" for the vertex of the following parabola
[tex] y(x)=x(80-2x) [/tex]

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...
 
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  • #8
dextercioby said:
Find the "x" & "y" for the vertex of the following parabola
[tex] y(x)=x(80-2x) [/tex]

Daniel.
Thanx for giving the answer to my question to him... I was trying to get him to think... :devil:
 
  • #9
For the forst problem your equation.It should be
[tex] A(x)=x(80-\frac{x}{2}) [/tex]

Maximize this one...

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

Daniel.

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

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
[tex] A(x)=x(80-\frac{x}{2}) [/tex]

Maximize this one...

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

Daniel.

wait, [tex] \frac{x}{2}) [/tex]?? 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, [tex] \frac{x}{2}) [/tex]?? 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 [tex]2\pi x[/tex] 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:
[tex] y(x)=-2x^{2}+80x [/tex]

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:
[tex] y(x)=-2x^{2}+80x [/tex]

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:
 
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  • #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.
 

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