Quadratic Regression Word Problems

In summary: AreaIn summary, the maximum area a rectangular parking lot can have is 636.08247422674x²+394123.71134017x-26319587.628863.
  • #1
Seviper
25
0

Homework Statement


A rectangular parking lot is to be placed with one side along a building where no fence is required. If 1000 yards of wire are available for fencing the other 3 sides of the lot, what
is the maximum area the lot may have, and what are the dimensions of the lot of a maximum area?

Homework Equations





The Attempt at a Solution


i used my graphing calculator to figure out the equation or at least what i think is the equation -x² +1000, i think you have t subtract something because you don't need one side but I am not sure what you subtract, maybe its -250 since its 1/4 of it, I am not sure
 
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  • #2
Seviper said:

Homework Statement


A rectangular parking lot is to be placed with one side along a building where no fence is required. If 1000 yards of wire are available for fencing the other 3 sides of the lot, what
is the maximum area the lot may have, and what are the dimensions of the lot of a maximum area?

Homework Equations





The Attempt at a Solution


i used my graphing calculator to figure out the equation or at least what i think is the equation -x² +1000, i think you have t subtract something because you don't need one side but I am not sure what you subtract, maybe its -250 since its 1/4 of it, I am not sure

##-x^2+1000## isn't an equation, it is an expression. And it has nothing to do with the problem anyway. And how do you use a graphing calculator in the first place without knowing what to graph?? You need to set up appropriate equations for the perimeter and area and use calculus.
 
  • #3
oh no my bad i mean -x² + 1000x, i used a ti 83 and i made a table . Is the answer -636.08247422674x²+394123.71134017x-26319587.628863
 
  • #4
Seviper said:
oh no my bad i mean -x² + 1000x, i used a ti 83 and i made a table . Is the answer -636.08247422674x²+394123.71134017x-26319587.628863

No. And -x² + 1000x is still just an expression, not an equation. And the question asks for the dimensions, not a polynomial in x. Draw a picture, label the unknowns, and write down some equations, complete with = signs to get started. Put the graphics calculator away.
 
  • #5
Seviper said:
oh no my bad i mean -x² + 1000x, i used a ti 83 and i made a table . Is the answer -636.08247422674x²+394123.71134017x-26319587.628863

What is x supposed to represent? How on Earth do you start with -x2 + 1000x and end up with -636.08247422674x²+394123.71134017x-26319587.628863? That makes no sense at all.

Suggestion: start by drawing a picture, clearly labelling the items of interest. Assign symbols such as x, y, z, etc., to the various unknown quantities, then try to see if there are some relationships between them which allows you to decrease their number. In particular: you have 1000 yds. of wire. How do you express that fact in terms of your unknowns? What is the enclosed area as expressed in terms of your unknowns? Try to be systematic, and to stop along the way to explain what you are doing.

RGV
 
  • #6
im so confused could someone just explain it to me very clearly, my whole life math has been my favourite subject but this is the only topic in math that i don't understand, i have a reputation to maintain, everyone in class is expecting me to figure this out and if i don't i lose major street cred
 
  • #7
LCKurtz said:
No. And -x² + 1000x is still just an expression, not an equation. And the question asks for the dimensions, not a polynomial in x. Draw a picture, label the unknowns, and write down some equations, complete with = signs to get started. Put the graphics calculator away.

Seviper said:
im so confused could someone just explain it to me very clearly, my whole life math has been my favourite subject but this is the only topic in math that i don't understand, i have a reputation to maintain, everyone in class is expecting me to figure this out and if i don't i lose major street cred

Why don't you try was has been suggested to you, either by me above or Ray Vickson another post? Show us the equations you start with and maybe we can see where you are stuck.

Also, since this is an academic type forum we encourage the use of correct English such as I instead of i for the personal pronoun. Sorry about the "street cred".
 
  • #8
ok, forget about the x and 1000 i was way of. i made a table with 4 columns: in the first column i labeled it length, second length, third width and last area.
Length Length Width Area
100 100 800 8000000
200 200 600 24000000
300 300 400 36000000
400 400 200 32000000
450 450 100 20250000

I got the maximum area to be approximately 310, 310, 380, 36518000
 
  • #9
Seviper said:
ok, forget about the x and 1000 i was way of. i made a table with 4 columns: in the first column i labeled it length, second length, third width and last area.
Length Length Width Area
100 100 800 8000000
200 200 600 24000000
300 300 400 36000000
400 400 200 32000000
450 450 100 20250000

I got the maximum area to be approximately 310, 310, 380, 36518000

So you apparently aren't interested in learning how to solve the problem. And, unfortunately, you can't even calculate areas correctly. I hope none of your peers see this because it will surely hurt your street cred. :frown:
 
  • #10
Seviper said:
ok, forget about the x and 1000 i was way of. i made a table with 4 columns: in the first column i labeled it length, second length, third width and last area.
Code:
Length             Length               Width                  Area
 100                100                  800                  8000000
 200                200                  600                 24000000
 300                300                  400                 36000000
 400                400                  200                 32000000
 450                450                  100                 20250000

I got the maximum area to be approximately 310, 310, 380, 36518000
OK, so you seem to have figured out how to calculate the width of the rectangle if you're given a number for the length. Now say the length is equal to x. Can you tell us what the width is in terms of x?

Next, as LCKurtz pointed out, you didn't calculate the area correctly. Think again about how you find the area of a rectangle. Do you see why that column of your table is wrong?
 
  • #11
oh i accidentally multiplied all of them together, ill do it again, thanks Vela

Mod note: Removed insult
 
Last edited by a moderator:
  • #12
So can you answer my first question? If x is the length of one side, what expression gives the area of the rectangle?
 
  • #13
lenth times width and what's an infraction
 
  • #14
i changed the area and it is 80000, 120000, 120000, 80000, but i took out the 450
 
  • #15
so would the equation be -2x² + 1000x and the maximum area is 125000 and the dimensions are 250,250,500, is this right
 
  • #16
Seviper said:
so would the equation be -2x² + 1000x and the maximum area is 125000 and the dimensions are 250,250,500, is this right

It is impossible to answer your questions (actually, not questions, since they are not followed by a question mark "?"---but we can guess what you mean). Why impossible? Because, for the simple reason that you never tell us what x is supposed to be. Also: just writing down a formula without explanation is no good: you need to explain how you get it. (That way, if the formula is wrong we can help you figure out why it is wrong.)

RGV
 
  • #17
Seviper said:
lenth times width and what's an infraction
Read the forum rules, the ones you agreed to when you signed up.

Seviper said:
i changed the area and it is 80000, 120000, 120000, 80000, but i took out the 450
I have no idea what this means.

Seviper said:
so would the equation be -2x² + 1000x and the maximum area is 125000 and the dimensions are 250,250,500, is this right
That is the correct expression for the area of the rectangle. When you say equation, there has to be an equal sign somewhere.

Your area and dimensions are incorrect because you didn't include any units. Is it 250 miles on one side? 250 inches? 250 meters? 250 nanometers? Numerically, though, they're fine, if you have the right units.
 
  • #18
i don't know what x is, could someone please explain it to me and tell me the answer so then i could work backwards please, I am not cheating i will work backwadr
 
  • #19
How can you not know what x is? What did you have in mind when you wrote down ##-2x^2+1000x##?
 
  • #20
i just put it into my graphing calculator (ti83) and it gave me an equation if you want i could tell you exactly what i did
 
  • #21
Yes, tell us exactly what you did.
 
  • #22
ok, i clicked on stat the press enter to edit, then in L1 i put in 100, 200, 300, 400 and in L2 i put in 80000, 120000, 120000, 80000, then i clicked stat plot(2nd y=) then pressed enter and pressed enter on the on icon, the pressed stat and moved to calc and clicked 5(quadreg) then pressed vars then moved over to y-vars and clicked function and clicked enter on y1 then pressed enter and i got the equation
 
  • #23
Oh, okay, so you entered your numbers from the table, and your calculator told you what function would fit it.

Can you answer the question I asked you back in post 12? Let's break it down a little further. Say x is the length of one side of the rectangle.
  1. What is the width of the rectangle in terms of x? Think about how you came up with 800 in the first line of your table given that the length of one side was 100.
  2. Given your expression for the width and the fact that x represents the length of the rectangle, what is the area of the rectangle?
Please answer the questions even if you see where this goes. I want to make sure you understand what's going on.
 
  • #24
1. x * y = area
length * width = area

2. area is length times width
 
  • #25
You're not answering the questions.

In the first line of your table, you have 100 in the first column and 800 in the second column. How did you come up with those numbers? In other words, how are they related?
 
  • #26
oh I am sorry since one side of the rectangle is against the bulding and you have 1000 yds for the wires i said l + l + w = 1000
 
  • #27
OK, good, so when solve that equation for w, you get ##w=1000-2l##, right? Now you know the length of the rectangle is ##l## and the width is ##1000-2l##. What expression do you get for the area of the rectangle? Compare that to what the calculator gave you.
 
  • #28
my calculator is -2x² + 1000 but what you just said is w= 1000 - 2l that's the difference, yours has an equal
 
  • #29
No, not quite. Answer question #2.
 
  • #30
the area is 12500 because that is the maximum for the equation which i found using my calculator, btw this is applied math not precal my teacher said this class is about using your calculator to get the answer not using algebra to get it like in precal
 
  • #31
Then why do you care what x is?
 
  • #32
i don't I am just asking if 250 250, 500 is the right asnwer
 
  • #33
Then why did you ask what x is in post 18?

Reread post 17.
 
  • #34
cuz i wanted to know more but whatever is that the right answer or not because i have a history and math test to study for and i have 4 hrs
 
  • #35
Figure it out.

I'll point out that if you bothered to understand the simple math behind what's going on, you wouldn't have to worry about whether you have the right answer or not. You'd be able to see whether it's correct or not on your own.
 

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