# Homework Help: A couple of questions regarding AREAS (and getting the formulas,etc.)

1. Feb 6, 2005

### mathzeroh

OK this problem has been giving me nonstop headaches and nightmares for over 4 days!!

I just cant seem to understand something here! :yuck:

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. Feb 6, 2005

### mathzeroh

And here is the image for the fourth question:

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3. Feb 6, 2005

### mathzeroh

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! BUT WHY IS THAT WRONG?

and number 3:

well lets 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.

Last edited: Feb 6, 2005
4. Feb 6, 2005

### dextercioby

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

It's correct...

Daniel.

5. Feb 6, 2005

### christinono

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.

6. Feb 6, 2005

### mathzeroh

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.

thank you!

REALLY?!?!?! :surprised :surprised thanks man!!!

7. Feb 6, 2005

### dextercioby

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"... :grumpy:

Daniel.

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

Disregard this EDIT... :yuck:

Last edited: Feb 6, 2005
8. Feb 6, 2005

### christinono

Thanx for giving the answer to my question to him... I was trying to get him to think...

9. Feb 6, 2005

### dextercioby

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... :tongue2:

Last edited: Feb 6, 2005
10. Feb 6, 2005

### christinono

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...

11. Feb 6, 2005

### mathzeroh

wait, $$\frac{x}{2})$$?? ok now im 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! :uhh:

christinono, i was going to answer to ur post, and i had it edited on the other post, but i guess it was wrong.

12. Feb 6, 2005

### mathzeroh

that's what i was thinking too. :uhh: :uhh:

13. Feb 6, 2005

### christinono

I still think we are right... :grumpy:
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. Feb 6, 2005

### dextercioby

THEY ARE BOTH CORRECT AND SHOULD YIELD THE SAME MAXIMUM AREA.

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

Daniel.

15. Feb 6, 2005

### christinono

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. Feb 6, 2005

### mathzeroh

christinono completing the square wont 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, im sorry for all the trouble u guys had to go through.

17. Feb 6, 2005

### dextercioby

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... :tongue2:

Daniel.

18. Feb 6, 2005

### dextercioby

$$y(x)=-2x^{2}+80x$$

Complete the square...

Daniel.

19. Feb 6, 2005

### mathzeroh

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. :uhh:

20. Feb 6, 2005

### christinono

What in the world is b, and why does it equal -2????? :grumpy: