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

[dextercioby]

Yes,it's correct.

Daniel.

 

[mathzeroh]

Yes,it's correct.

Daniel.

who? me?

 

[christinono]

Wasn't the length of the fence supposed to be 80 feet? I think you used 40...

 

[dextercioby]

who? me?

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

Daniel.

 

[christinono]

Sorry, I made a mistake... You were right.

 

[mathzeroh]

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

Daniel.

yes! that is unbelievable.

christinono ur just trying to scare me.


im going to work on number four now.

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...

 

[christinono]

did it work? (just kidding...)

 

[christinono]

I did question 2 and got the same answer. Is that unbelievable too?

 

[dextercioby]

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.

 

[christinono]

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...

 

[mathzeroh]

Hahahaha, yes it did, and yes, that is amazing still!


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!

 

[mathzeroh]

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.

 

[christinono]

Yes, it is ABSOLUTELY INCREDIBLY AMAZING!

 

[christinono]

k now I don't know if we've got the right answer... I though we did.

 

[christinono]

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!

 

[dextercioby]

What's the area function for the rectangle in the right (similar to problem #1).Isn't it
$$S(x)=x(73-2x)$$

Daniel.

 

[christinono]

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...

 

[mathzeroh]

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)

 

[christinono]

hmm..i got this:

x=93-2y

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

A=y(93-2y)

Same thing I did! (just different variables)

 

[mathzeroh]

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.

 

[mathzeroh]

Same thing I did! (just different variables)

exactly.

 

[dextercioby]

Okay guys the total LENGTH OF THE WIRE/FENCE NEEDS TO BE 80,REMEMBER?

Take the rectangle from the right (similar with the one in the first problem).The total legth available is 80-(10-3)=73 ft...

The area function is
$$S(x)=x(73-x)$$

which gives a maximum area of 666.125ft squared.

If u add the rectangle which has fence only on one out of the 4 sides,the total area becomes 921.625.

Daniel.

 

[dextercioby]

NOTE:There's no fence along the 10 feet side,simply because the bolded line/contiur does not include that portion...

Daniel.

 

[mathzeroh]

true. But this is what i did:

there were three 3 sides to be dealt with:

y-3
y-10
x

these. (and maybe christinono used the same thing except switched the variables).

so this is what i did, i put the ys together:

y-3+y-10

put the like terms together:

2y-13

and that's how i went forward with it...is that right?

 

[christinono]

Okay guys the total LENGTH OF THE WIRE/FENCE NEEDS TO BE 80,REMEMBER?

Take the rectangle from the right (similar with the one in the first problem).The total legth available is 80-(10-3)=73 ft...

The area function is
$$S(x)=x(73-x)$$

which gives a maximum area of 666.125ft squared.

If u add the rectangle which has fence only on one out of the 4 sides,the total area becomes 921.625.

Daniel.

Daniel, I don't agree. We are looking for the area of the WHOLE garden, not just the area enclosed by the fence. When I said that:
80 = (x-10) + (x-3) + y, I was saying the total length of the fence was 80 ft. What was represented by "x" was the total bottom side (same as the top), not just the length of the wire on the bottom.

 

[dextercioby]

No.You should add:
10-3+y+2x=80

And the total area is y(x+7).

Daniel.

 

[christinono]

No.You should add:
10-3+y+2x=80

And the total area is y(x+7).

Daniel.

I don't understand what you mean or what is wrong with how we solved the problem.

 

[christinono]

Daniel, could you define what "x" is in your equation?

 

[dextercioby]

Daniel, I don't agree. We are looking for the area of the WHOLE garden, not just the area enclosed by the fence.

The fence does not inclose an area,because it's not an enclosed curve...

80 = (x-10) + (x-3) + y, I was saying the total length of the fence was 80 ft. What was represented by "x" was the total bottom side (same as the top), not just the length of the wire on the bottom.

Yes,we're not talking about the same area...

This problem is confusing...

Daniel.

 

[dextercioby]

Daniel, could you define what "x" is in your equation?

"x" is the length of the part of the fence which is left "outside" to form the exterior rectangle...

Daniel.