Pre-Calc enclosure problem, with a parallel halfway the enclosure

[Chaim]

A person has 300 feet of fencing to make a rectangle enclose. The person wants to use some fencing to split the enclose into 2 parts with a fence parallel to 2 of the sides. What dimensions should the enclose have to have max possible area?

I was stuck on this for a while.
I thought it was like a rectangle with a line in the middle
Then did 3l + 2w = 300
Then I got lost, I tried substitution but that just cancels the 300 which leaves me with 0
I think I did it all wrong.
Can anyone help?

 

[tkhunny]

Area(l,w) = l*w

Use your fence-length formula to make it a function of a single variable.

 

[Chaim]

Area(l,w) = l*w

Use your fence-length formula to make it a function of a single variable.

Do you mean my formula?
so f(x) = 3l + 2w?
Sorry, Im confused

 

[tkhunny]

No, Why did you do that?

You have 3l+2w = 300 and I gave you Area(l,w) = l*w

Use the first to create Area(w).

l = (300-2w)/3, so Area(w) =

 

[Chaim]

No, Why did you do that?

You have 3l+2w = 300 and I gave you Area(l,w) = l*w

Use the first to create Area(w).

l = (300-2w)/3, so Area(w) =

Then l = 100 - (2/3)w
So I plug that back into the equation right?
3(100-(2/3)w) + 2w = 300
300 - (6/3)w + 2w = 300
(6/3) = 2
300 - 2w + 2w = 300
Then everything = 0?
Cause -2w + 2w cancel
and if I subtract 300 from both side, they cancel too.

 

[tkhunny]

No. Anytime you substitute into the equation you just solved, you will get this result. It is meaningless. You have not yet used the Area formula. You're just dealing with the linear measurements.

3l+2w = 300

Area(l,w) = l*w

w = (300-3l)/2

Area(l) = l*(300-3l)/2 = 150*l - (3/2)l^2

You should recognize this last form as being related to a parabola, opening downward. Find its maximum value.

 

[Chaim]

No. Anytime you substitute into the equation you just solved, you will get this result. It is meaningless. You have not yet used the Area formula. You're just dealing with the linear measurements.

3l+2w = 300

Area(l,w) = l*w

w = (300-3l)/2

Area(l) = l*(300-3l)/2 = 150*l - (3/2)l^2

You should recognize this last form as being related to a parabola, opening downward. Find its maximum value.

Oh ok thanks!
Sorry, I'm not really good at math.
So what I was thinking was (-b/2a)?
So I plugged in (-150/2(-3/2)) = 50
Then I plugged everything back in to get 75
Thanks

 

[tkhunny]

Sorry, I'm not really good at math.

Please never say or think that again. You never will succed with this attitude. Just stop and think rather than just shooting in the dark.

So what I was thinking was (-b/2a)?

What were you thinking about that? Is it applicable? What are 'a' and 'b'?

So I plugged in (-150/2(-3/2)) = 50

What is 50?

Then I plugged everything back in to get 75

What is 75?

Did you solve the problem? It's just not clear. Lay out explicitly what you have done and what you have concluded.

Be more careful with notation. It WILL help you.

 

[Chaim]

Please never say or think that again. You never will succed with this attitude. Just stop and think rather than just shooting in the dark.



What were you thinking about that? Is it applicable? What are 'a' and 'b'?



What is 50?



What is 75?

Did you solve the problem? It's just not clear. Lay out explicitly what you have done and what you have concluded.

Be more careful with notation. It WILL help you.

Oh, since I was searching for the max possible, I used the vertex formula which is (-b/2a) since the equation you got was a parabola.
So basically I just used that equation and plugged it into the vertex formula which finds the maximum/minimum point, which then I use the number resulted form the vertex formula, then plugged it back into 3l * 2w, which then lets me plug 50 into the equation, then divide everything by 2 to get w.