Setting area = perimeter of rectangle, does it have any meaning?

[LearninDaMath]

Suppose I have a rectangle with length x and height y:

Can this ever be true? That xy = 2x+2y?

My guess would be that area of a shape can never equal the perimeter of that shape. And that would be confirmed by the fact that the equation is always false (except if both variables equal zero, but then there is no shape or perimeter which defeats the purpose of evaluating a shape's area or perimeter in the first place).

So I would say Area is apples and Perimeter is oranges. And trying to equate them doesn't work.

----------------------------------------------------------------

However, if the above conclusion is correct, how come I have an assignment that calls for the substitution of Area into Perimeter? How can mixing apples and oranges all of a sudden make sense? See following image:

Can this be? Why so? Why does the scenario in the second image seem to conflict with the scenario in the first image?

 

[checkitagain]

Can this ever be true? That xy = 2x+2y? ... ///...

So I would say Area is apples and Perimeter is oranges. And trying to equate them doesn't work.

----------------------------------------------------------------

Yes, it is "apples and oranges."

However, the problem (or instructor) should have stated
that it is wanted to be known where numerically
those quantities are equal.

The number of square units (area) can be equal to the
number of units (perimeter) for proper x and y values.

 

[phinds]

Can this ever be true? That xy = 2x+2y?

I take it you are not very adept at simple arithmetic.

x=4, y=4

 

[LearninDaMath]

Yes, it is "apples and oranges."

However, the problem (or instructor) should have stated
that it is wanted to be known where numerically
those quantities are equal.

The number of square units (area) can be equal to the
number of units (perimeter) for proper x and y values.

Okay, so 5*5 ≠ 2(5) + 2(5) because 25 ≠ 20

and 5*3 ≠ 2(5) + 2(3) because 15 ≠ 16

5*2 ≠ 2(5) + 2(2) because 10 ≠ 14

So you would be saying that at some x and y value(s), the equation xy = 2x + 2y.

However, the only way I can imagine finding those numbers would be a long list of trial and error (an infinite list of trial and error)


If I try to solve the equation for one of the variables, for instance y, I am just getting an equation that states 0 = 0. Everything is canceling everything and I'm left with a statement that just varifies that the equation is true.

So how could I actually go about finding the numerals at would make this equation true?

 

[LearninDaMath]

I take it you are not very adept at simple arithmetic.

x=4, y=4

You would be 100% correct. I didn't do much arithmetic as a kid or even as I got older, and I'm not happy about it now. But I'm putting the time in now to really try to understand the basics even as I'm going through the more "advanced" classes i.e. calculus 1.


Do you mind explaning how you get x=4 and y = 4? How come I seem to be getting 0 = 0?

This is how I proceeded:

xy = 2x + 2y

solved for x to get:

x = 2y/y-2

then subbed:

y(2y/y-2) = 2(2y/y-2) + 2y

then doing some algebra, I get:

2y = 4 + 2y - 4

which becomes 0=0

Is my algebra wrong or is the final answer correct, but just not what I'm supposed to do in order to find the specific numerals x = 4 and y = 4?

 

[phinds]

You would be 100% correct. I didn't do much arithmetic as a kid or even as I got older, and I'm not happy about it now. But I'm putting the time in now to really try to understand the basics even as I'm going through the more "advanced" classes i.e. calculus 1.


Do you mind explaning how you get x=4 and y = 4? How come I seem to be getting 0 = 0?

This is how I proceeded:

xy = 2x + 2y

solved for x to get:

x = 2y/y-2

then subbed:

y(2y/y-2) = 2(2y/y-2) + 2y

then doing some algebra, I get:

2y = 4 + 2y - 4

which becomes 0=0

Is my algebra wrong or is the final answer correct, but just not what I'm supposed to do in order to find the specific numerals x = 4 and y = 4?

Your algebra is quite correct and an obvious waste of time. A single equation in two unknowns cannot be solved and the process you used will ALWAYS end up with 0=0

I just tried a couple of numbers, noticed immediately that 2 and 5 gave opposite relationships so CLEARLY there was an answer between 2 and 5 so I tried 4 and voila !

 

[phinds]

However, the only way I can imagine finding those numbers would be a long list of trial and error (an infinite list of trial and error)

You need more practice. The infinite list of trial and error turned out to require 3 tries which I find to be a bit less than infinity.

 

[phinds]

And I should add for both of you, that my choice of 2 and 5 was AFTER approximately 1/10th of a second of thought in which it was apparent to me that 1 was way too low and 10 was way to high, thus the attempts at 2 and 5.

I actually expected the result to be a real number, not an integer and was surprised that 4 jumped right out.

 

[checkitagain]

This is how I proceeded:

xy = 2x + 2y

solved for x to get:

x = 2y/y-2. \ \ \ \ \ \ \ x = 2y/(y - 2)

then subbed:

y(2y/y-2) = 2(2y/y-2) + 2y. \ \ \ \ \ \ \ y[2y/(y - 2)] = 2[2y/(y - 2)] + 2y

As inserted into the quote box above, you must use grouping symbols
because of the Order of Operations.


Edit:

LearninDaMath,

you arrived at x = 2y/(y - 2) [my correction].


It was never stated that the solutions must be integer.


Here is a quiz-like question to you from me:

If there is not this restriction, then are you limited in the
total number of solutions?

 

[LearninDaMath]

Phinds, thanks for all the feedback. By using the numbers y =4 and x = 4, I was able reason it out by letting the area = 16, and working backwards so that:

16 = 2(2y/y-2) + 2y

so y = ±4

Then subbing y into xy=2x+2y

so x = 4.

So it seems that if you are given the parimeter or area from the start, you can then use algebra to find the value of the variables. However, if you are not given the area or perimeter from the start, you have to be able to use a descent toolset of arithmetic to logic your way to the x and y values. Awesome! Thanks for your help.

 

[phinds]

It was never stated that the solutions must be integer.

Here is a quiz-like question to you from me:

If there is not this restriction, then are you limited in the
total number of solutions?

With the restriction, you get one solution. Without the restriction, you get one solution.

Doesn't seem like much of a restriction in this particular case.

 

[LearninDaMath]

As inserted into the quote box above, you must use grouping symbols
because of the Order of Operations.Edit:

LearninDaMath,

you arrived at x = 2y/(y - 2) [my correction].It was never stated that the solutions must be integer.Here is a quiz-like question to you from me:

If there is not this restriction, then are you limited in the
total number of solutions?

Well, prior to reading your post, I did some searching on google and found that there are only two possible sets of integers that work and they are: (4,4) and (6,3). And there are actually an infinite number of solutions, but they are in decimal form. So I wouldn't be able to answer you as though I hadn't looked it up already. However, I'm pretty sure my thoughts prior to finding that info on google were this:

-(4,4) are the only solutions.. although it seems as though there should be more solutions (possibly infinitely more solutions), so I can't understand why there could only be one solution. -

I'm pretty sure that's how I would have responded to your quiz had I not looked it up on google, and it would only be a gut feeling that there "should" be more solutions somehow, but I wouldn't be able to apply any sort of meaningful logic to confirm it one way or the other.

-----------------------

However, I'm still not sure about the second part.

If area is apples and perimeter is oranges, and they can only be equated on a numerical basis, how come you can solve for x in the area equation, and substitute it into the perimeter equation as the second image shows?

If they are apples and oranges, and are only equatable in certain cases, can the scenario in the second image only work in certain cases as well? Or in any case?

 

[phinds]

OOPS

My "there's only one solution" statement was based on a SQUARE when the problem clearly stated RECTANGLE.

 

[LearninDaMath]

Phinds,

I'm still confused how the fact that xy = 2x + 2y is true sometimes, in certain cases, such as (4,4) and (3,6), yet it is still possible to substitute Area into Perimeter always. Why is it not the case that Area would only be able to be substituted into Perimeter sometimes as well?

 

[Mentallic]

Phinds,

I'm still confused how the fact that xy = 2x + 2y is true sometimes, in certain cases, such as (4,4) and (3,6), yet it is still possible to substitute Area into Perimeter always. Why is it not the case that Area would only be able to be substituted into Perimeter sometimes as well?

I'm not sure exactly what you're trying to ask but I'll take a stab at it. If we want to find when it is true for say, x=y (which would make it a square if both sides are equal length) then we substitute x for all y we find in that equation, so we get

x^2=2x+2x

This is a quadratic that we would solve by moving everything to one side and equating it to 0,

x^2-4x=0

and then factorizing,

x(x-4)=0

which means that x=0 or x=4, so those are the only solutions for a square, although, I'd say a square with 0 length sides isn't really a square.

What if we want to know when the length is twice the width? So when 2x=y, then we substitute that into the equation as well,

2x^2=2x+4x

equating to zero,

2x^2-6x=0

then factorizing,

2x(x-3)=0

So this tells us that either x=0 or x=3, and since we already know that 2x=y, then when x=3, y=6, so that's another solution.

We could go on and do this process for an infinite number of values, so let's just make it that y=mx for some positive value m, whatever that may be. Following the same procedure we get

mx^2=2x+mx

mx^2-(2+m)x=0

x(mx-2-m)=0

So the solutions are always going to be x=0, y=0 or mx-2-m=0 thus x=\frac{2+m}{m} y=m\left(\frac{2+m}{m}\right)=2+m

And for any integer m, y is always going to be an integer, so if we want to choose m such that x is also an integer, we just can split x up as x=\frac{2}{m}+1 which means either m is going to be 1 or 2, because any other value of m will not make x an integer.

Just a little insight into why the integer solutions are the way they are.

 

[phinds]

Phinds,

I'm still confused how the fact that xy = 2x + 2y is true sometimes, in certain cases, such as (4,4) and (3,6), yet it is still possible to substitute Area into Perimeter always. Why is it not the case that Area would only be able to be substituted into Perimeter sometimes as well?

I don't understand where you get the idea that you CAN do that substitution. It is obviously false, so what led you to it? Perhaps you are misreading the statement of the original problem that you presented.

The falseness of the substitution is immediately apparent if you consider a rectangle that gets thiner and thinner and longer and longer. As that process takes place, the area can be made to stay the same while the perimeter just gets bigger and bigger

 

[checkitagain]

Phinds, thanks for all the feedback. By using the numbers y =4 and x = 4,
I was able reason it out by letting the area = 16, and working backwards so that:

16 = 2(2y/y-2) + 2y \text{That is not correct. I already showed you that }
\text{in post #9. Grouping symbols must be around "y - 2."}

x = \dfrac{2y}{y - 2}


x = 2 + \dfrac{4}{y - 2}


The denominator must divide 4. There are only a finite number
of positive integer y-values that make x an integer.

But again, the problem never stated that the solutions
were to be integers, so please type the question as fully as
possible (if it wasn't already), to reduce ambiguities.

And all that "apples and oranges" here means is that square
units cannot be equal to linear units, hence the use of the
word "numerically."

 

[LearninDaMath]

x = \dfrac{2y}{y - 2}x = 2 + \dfrac{4}{y - 2}The denominator must divide 4. There are only a finite number
of positive integer y-values that make x an integer.

But again, the problem never stated that the solutions
were to be integers, so please type the question as fully as
possible (if it wasn't already), to reduce ambiguities.

And all that "apples and oranges" here means is that square
units cannot be equal to linear units, hence the use of the
word "numerically."

well if area is to equal 16 and perimeter is to equal 16, I would have:

Area = xy = 16
Perimeter = 2x +2y = 16

And since 16 = 16, we could say:

x = Area/y or x = 16/y

And subbing that into Perimeter, we can say:

16 = 2(Area/y) + 2y or 16 = 2(16/y) + 2y

And then y = 4.

4 is the y value of a rectangle (or square) in which the perimeter is equal to area.

And subbing y = 4 into either:

Area = xy = 16 or Perimeter = 2x + 2y = 16 yields:

x = 4, which is the x value of the rectangle (or square) where the perimeter equals area.

And if I perform the same exact steps for xy = 18 = 2x + 2y, will I not yield y = 6 and y = 3?

Which I can then substitute each value into xy = 18 or 2x + 2y = 18 and yield:

x = 6 and x = 3, confirming that (3,6) or (6,3) are x and y values that satisfy xy = 2x + 2y?

And wouldn't that prove that a rectangle (or square) whose area = perimeter are satisfied by the numerals 6 and 3?

----------------------------------------------------------------------------------------------------

Also, I don't understand the difference between what you are writing and what I'm writing. You are telling me to write 2y/(y-2)...when I am writing exactly that. The only difference I can see is that you are using brackets while I'm using parenthesis, but aren't we still describing the same exact equation and concept? And if not including the brackets is totally incorrect, if we were to assume I had written it correctly as per the Order of Operations, where then would exist the confusion of the concept itself?

-------------------------------------------------------------------------------

I don't understand where you get the idea that you CAN do that substitution. It is obviously false, so what led you to it? Perhaps you are misreading the statement of the original problem that you presented.

The falseness of the substitution is immediately apparent if you consider a rectangle that gets thiner and thinner and longer and longer. As that process takes place, the area can be made to stay the same while the perimeter just gets bigger and bigger

Here is the question:

Optimization: Find dimensions of a rectangle with area 1000meters^2, whose perimeter is as small as possible.

Step 1: Identify that perimeter = 2x + 2y

Step 2: Determine constraint: Area = 1000 = xy

Step 3: Write perimeter in terms of x by rearranging Area: y = Area/x or y = 1000/x

Thus, Perimeter = 2x + 2(1000x)

Step 4: Evaluate derivative of perimeter and find minimum critical value(s).

Step 5: plug critical value back into Area to find y value of rectangle.This is the problem. The way to solve it is to use exactly the fact that area = perimeter and to arrange area in such a way that it can be substitued into perimeter to set the perimeter equation in terms of x.

So why is this the correct way of solving this problem if it is false/wrong to substitute area into perimeter?

By the way, although my question is stemming from calculus, my confusion and question itself is not the calculus, its the precalculus fundamentals.

 

[checkitagain]

Also, I don't understand the difference between what you are writing and what I'm writing.
You are telling me to write 2y/(y-2)...when I am writing exactly that.

Nope. You never wrote that or its equivalent. You have no parentheses
(or any other grouping symbols) around your denominator.

Look at posts # 5 and #10. Here is part of post #5:

This is how I proceeded:

xy = 2x + 2y

solved for x to get:

x = 2y/y-2 <-------

then subbed:

y(2y/y-2) = 2(2y/y-2) + 2y <--------

2y/y - 2 = \dfrac{2y}{y} - 2 = 2 - 2 = 0.

And \ 2(2y/y - 2) =

2\bigg(\dfrac{2y}{y} - 2\bigg) =

2(2 - 2) = 2(0) = 0

Divisions are done before the subtractions, and with your
"y - 2" not being grouped, it is not one unit, and therefore
is not being divided into 2y.
Just the y is being divided into 2y. Please accept this
and move on to the next step of the solution.

The only difference I can see is that you are using brackets while I'm using parenthesis,
but aren't we still describing the same exact equation and concept?

No, again, you didn't type any grouping symbols at all around "y - 2"
in those posts. Please acknowledge that, and move on with the next step
of the solution.



-------------------------------------------------------------------------------

This is the problem. The way to solve it is to use exactly the fact that
area = perimeter

No. The area is not equal to the perimeter. The number of units
of area is being required to equal the number of units of perimeter.
You are not substituting area into perimeter. You are substituting an expression
for the number of units of area into an expression involving the
number of units of perimeter. And acknowlege this and move on to the
next step of the solution.

The number of units of area can/may equal the number of
units of perimeter, but no areas can equal any perimeters.

Example:

3 \ square \ miles \ \ne \ 3 \ miles.

But, the number of units are equal to each other.

 

[phinds]

So why is this the correct way of solving this problem if it is false/wrong to substitute area into perimeter?

OK, I now believe I was correct in thinking that perhaps you misunderstood the wording of the original problem. You have made the statement that area = perimeter, but nowhere in the statement of the problem does it say or imply that, it is simply a misinterpretation on your part.

Do you dispute that the statement (area=perimeter) is in fact obviously false?

 

[LearninDaMath]

OK, I now believe I was correct in thinking that perhaps you misunderstood the wording of the original problem. You have made the statement that area = perimeter, but nowhere in the statement of the problem does it say or imply that, it is simply a misinterpretation on your part.

Do you dispute that the statement (area=perimeter) is in fact obviously false?

I agree that the statement, area = perimeter, is false if you mean in terms of the units they represent. Of course 5 meteres don't equal 5 meters^2

However I disagree that the statement area = perimeter should be false, if you mean in terms of the equations that describe them: Area = xy and Perimeter = 2x + 2y. I think they can be set equal to each other. If one is to be used as a substitute into another, how can they not be set equal to each other?

 

[phinds]

I agree that the statement, area = perimeter, is false if you mean in terms of the units they represent. Of course 5 meteres don't equal 5 meters^2

However I disagree that the statement area = perimeter should be false, if you mean in terms of the equations that describe them: Area = xy and Perimeter = 2x + 2y. I think they can be set equal to each other. If one is to be used as a substitute into another, how can they not be set equal to each other?

As has already been clearly demonstrated, they CAN be set equal to each other and solved algebraically (or, as I did, by hit/miss).

What I THOUGHT you were saying is that the statement is simply logically true (thus saying that they are ALWAYS the same), not that it is a valid algebraic equation with some solutions.

 

[aznboy]

To clear this up for you:

In most shapes/cases the area does NOT equal the perimeter

xy = 2x + 2y, substitutes for most x's and y's do not satisfy the eqn

BUT in some shapes where for e.g. the side length is 4, the area will satisfy the eqn, hence perimeter = area. It is in this sense, that a perimeter can be calculated with known side length and area (which relates to your 2nd question).

In the above eqn it can account for any rectangle or square ==> ie any side length.

For your 2nd question,

Perimeter = 2x + 2(area)/x

If you think of a computer screen with known LENGTH and known (surface) AREA. You can obviously find a value for the BREADTH.

Hope that helps. Btw this is my first time on this forum =]

 

[HallsofIvy]

To clear this up for you:

In most shapes/cases the area does NOT equal the perimeter

xy = 2x + 2y, substitutes for most x's and y's do not satisfy the eqn

BUT in some shapes where for e.g. the side length is 4, the area will satisfy the eqn, hence perimeter = area. It is in this sense, that a perimeter can be calculated with known side length and area (which relates to your 2nd question).

That is a "sense" which most people here would not accept, starting with your statement "for e.g, the side length is 4". A length is never "4". It is "4 feet" or "4 meters" or "4 kilometers". In the first case, if both sides of a rectangle had length 4 feet, then the area would br "16 square kilometers" and the perimeter would be "16 kilometers" and those are NOT the same.

In the above eqn it can account for any rectangle or square ==> ie any side length.

For your 2nd question,

Perimeter = 2x + 2(area)/x

If you think of a computer screen with known LENGTH and known (surface) AREA. You can obviously find a value for the BREADTH.

Hope that helps. Btw this is my first time on this forum =]

No, I don't believe that helps. Lengths or areas are NOT "numbers" they are "measurements" and are meaningless without their units.

 

[LearninDaMath]

To clear this up for you:

In most shapes/cases the area does NOT equal the perimeter

xy = 2x + 2y, substitutes for most x's and y's do not satisfy the eqn

BUT in some shapes where for e.g. the side length is 4, the area will satisfy the eqn, hence perimeter = area. It is in this sense, that a perimeter can be calculated with known side length and area (which relates to your 2nd question).

In the above eqn it can account for any rectangle or square ==> ie any side length.

For your 2nd question,

Perimeter = 2x + 2(area)/x

If you think of a computer screen with known LENGTH and known (surface) AREA. You can obviously find a value for the BREADTH.

Hope that helps. Btw this is my first time on this forum =]

So is it valid to say that a rectangle of any given area has exactly one and only one set of dimensions (x,y) such that perimeter = area?

For instance, for a rectangle with area 18 meters^2, there is some given (x,y) that satisfies [Area = 18 = xy] and [Perimeter = 18 = 2x + 2y]. That xy being (3,6) or (6,3). So must the area of 18.0001, 18.00002, 19, 19.5, 20 and so on, all have their own respective (x,y) values that satisfy [Area = xy] and [Perimeter = 2x + 2y]?

It just feels that if a rectangle of area 18 has a perimeter and area of (6,3), then a rectangle of every possible area should have some values x,y that satisfies the same criteria. Would this assumption be valid?

 

[LearninDaMath]

I'm not sure exactly what you're trying to ask but I'll take a stab at it. If we want to find when it is true for say, x=y (which would make it a square if both sides are equal length) then we substitute x for all y we find in that equation, so we get

x^2=2x+2x

This is a quadratic that we would solve by moving everything to one side and equating it to 0,

x^2-4x=0

and then factorizing,

x(x-4)=0

which means that x=0 or x=4, so those are the only solutions for a square, although, I'd say a square with 0 length sides isn't really a square.

So are you saying that there is only one specific rectangle in existence with the property that its area and perimeter share the same value for the x and y variables? In other words, are you saying that there is only one size square in the known universe in which the area equals perimeter? (note: by one size, mean in numeric terms, not units). And since that one size square which satisfies area=perimeter (or xy = 2x + 2y) is when the dimensions are (4,4), it, again, means that out of all the numerals (between -∞ and ∞) that could represent the dimensions of height and length of a square, the only valid numeral is 4?

What if we want to know when the length is twice the width? So when 2x=y, then we substitute that into the equation as well,

2x^2=2x+4x

equating to zero,

2x^2-6x=0

then factorizing,

2x(x-3)=0

So this tells us that either x=0 or x=3, and since we already know that 2x=y, then when x=3, y=6, so that's another solution.

And here, are you saying that out of all the infinite rectangles with the specific property where its base is twice its length (or visa versa), there is only one specific set of (x,y) dimensions that satisfy Area = Perimeter?

We could go on and do this process for an infinite number of values, so let's just make it that y=mx for some positive value m, whatever that may be. Following the same procedure we get

mx^2=2x+mx

mx^2-(2+m)x=0

x(mx-2-m)=0

So the solutions are always going to be x=0, y=0 or mx-2-m=0 thus x=\frac{2+m}{m} y=m\left(\frac{2+m}{m}\right)=2+m

And for any integer m, y is always going to be an integer, so if we want to choose m such that x is also an integer, we just can split x up as x=\frac{2}{m}+1 which means either m is going to be 1 or 2, because any other value of m will not make x an integer.

Just a little insight into why the integer solutions are the way they are.

For this last part of your post, I'm going to have to take some more time to comprehend what you are saying. However, could you still comment on whether I am understanding the first 2 parts of your post?

 

[LearninDaMath]

Also, is it valid to say that every value of area can be arranged into specific rectangular dimensions, such that the area's numeric value = perimeter's numeric value?

So for instance, suppose I pull some random value out of the sky, say 856.92. And let that value represents the area of some rectangle. Does that mean that there is some specific height and length of this rectangle such that the value of its area and perimeter are equal, numerically?

 

[Mentallic]

So are you saying that there is only one specific rectangle in existence with the property that its area and perimeter share the same value for the x and y variables? In other words, are you saying that there is only one size square in the known universe in which the area equals perimeter? (note: by one size, mean in numeric terms, not units). And since that one size square which satisfies area=perimeter (or xy = 2x + 2y) is when the dimensions are (4,4), it, again, means that out of all the numerals (between -∞ and ∞) that could represent the dimensions of height and length of a square, the only valid numeral is 4?

Yes, that's what I'm saying. Except I wouldn't take the numerical values between -∞ and ∞ because it doesn't make any physical sense to have a square without positive length sides.

You can also see why this is true because y_1=x^2 grows faster than y_2=4x, so since at say, x=1, y₁=1 and y₂=4, if y₁ grows faster then it's going to catch up to y₂ at some point, and then overtake it, only to grow ever faster. This point where they're equal is at x=4.

And here, are you saying that out of all the infinite rectangles with the specific property where its base is twice its length (or visa versa), there is only one specific set of (x,y) dimensions that satisfy Area = Perimeter?

Yes.

For this last part of your post, I'm going to have to take some more time to comprehend what you are saying. However, could you still comment on whether I am understanding the first 2 parts of your post?

All I'm doing is following the same procedure that I did to solve for a square and rectangle with twice the length of its width, but this time by taking any arbitrary rectangle with length being m times its width (m can be any positive number). If you can follow the earlier reasoning, then you should be able to follow this.

edit: but if you're not understanding why I did what I did, you can still get the same answer out of your re-arranged equation.

If xy=2x+2y then solving for y, we get xy-2y=2x y(x-2)=2x y=\frac{2x}{x-2}

At this point you can just ask yourself, ok, since I started with the equation Area=Perimeter, I assumed this, thus this equation is algebraically trying to tell me that if I chose a value for x (the length) of 4, then the value of y (the width) such that area=perimeter must be y=\frac{2x}{x-2}=\frac{8}{2}=4

So a rectangle with x=4, y=4 satisfies area=perimeter.
What about for x=5? Then y=10/3. What about for any positive x? Then just plug in x to find your y value.

But what about if we plugged in x=2? Well, this would force us to divide by zero, and that can't be done. But physically, what does this mean exactly? It means that area=perimeter can be achieved for any positive value of x (and its corresponding y) EXCEPT for x=2. At length of 2 there is no possible width that can go with it to make the area = perimeter.

Now, the only reason I suggested we introduce the value m was because it made it easy to answer some types of questions, such as if I want my length to be twice the size of my width, then what are the values that make it area=perimeter? Well, from the equation y=\frac{2x}{x-2} we would need to plug y=2x in, then solve for x, which is time consuming. Introducing the value of m forced us to create an equation to find the answer quickly and easily.

 

[Mentallic]

Also, is it valid to say that every value of area can be arranged into specific rectangular dimensions, such that the area's numeric value = perimeter's numeric value?

So for instance, suppose I pull some random value out of the sky, say 856.92. And let that value represents the area of some rectangle. Does that mean that there is some specific height and length of this rectangle such that the value of its area and perimeter are equal, numerically?

Well that question is more interesting. It turns out that if the area is less than 16, there is no possible way for the area to be equal to the perimeter.

For example, if we take A=10, then we have that A=xy, thus 10=xy. We also need to satisfy the equation xy=2x+2y, so if we rearrange the area equation to make x or y the subject, then y=10/x, and plugging this into the area=perimeter equation, we get 10=2x+\frac{20}{x}

Now we multiply through by x and have a quadratic to solve.

10x=2x^2+20

x^2-5x+10=0

x=\frac{5\pm\sqrt{5^2-4\cdot 10}}{2}

x=\frac{5\pm\sqrt{-15}}{2}

But notice that we can't take the square root of a negative value, so there are no values of length and width that can give us an area of 10 and also be area=perimeter.

If we follow the same procedure for any area, denoted A, then we get A=xy, y=A/x, then A=2x+\frac{2A}{x}

Ax=2x^2+2A

2x^2-Ax+2A=0

x=\frac{A\pm\sqrt{A^2-16A}}{4}

Thus, for there to exist a real value of x, we need the part in the square root to not be negative. This means A^2-16A\geq 0

A(A-16)\geq 0

Thus, A\geq 16 (ignoring A\leq 0 because we want a positive value of Area).

So to answer your question, yes there is a specific length and width that gives an area of 856.92 that will also give the same perimeter.

 

[LearninDaMath]

Okay, so in a series of statements, I'm going to attempt to confirm that I understand everything you are saying.


Statement 1: For any given area of 16 and greater, there will always be one (and only one) specific set of dimensions (x,y) such that area = perimeter (xy = 2x + 2y).

Statement 2: Given the shape of a rectangle, if you let x = any positive value, there will always be some corresponding positive y value, such that the dimensions (x,y) satisfy area = perimeter.

Statement 3: The ratio between height and length of a rectangle can be used to rewrite xy = 2x + 2y into an equation in terms of x, (such as x = y so that x(x) = 2x + 2(x)). So that the x dimension of the described rectangle can be found, which can then be used to find the specific y value that let's the equation area = perimeter be true.

Statement 4: And in the example of ratio x = y, [x(x) = 2x + 2x] ==> [x^2 = 4x] where x = 4 and y = 4. You can not use the fact that the sides are of a specific proportion in order to assume that any value of x will satisfy Area = Perimeter. In other words, just because the sides are x = y, it doesn't mean that 5 = x and 5 = y will satisfy Area = Perimeter. As a second example, just because (6,3) satisfies [y = 2x] and thus [x(2x) = 2x + 2(2x)] = [Area = Perimeter], it doesn't mean that (8,4), (10,5), (12,6) are going to satisfy Area = Perimeter.

Statement 5: I can be given a random Area (greater than 16) and by way of algebra, find the two dimensions that satisfy Area = Perimeter.

Statement 6: I can start with a value representing x, and from there, use algebra to determine the a y value such that area = perimeter, even if I haven't yet determined the specific value of area. And once I have found the y value corresponding to the x value, a simple multiplication will tell me the area (and of course perimeter).

Statement 7: Given [Area = xy] and [Perimeter = 2x + 2y] Area and Perimeter do not have to be equal to each other in order to substitute one into the other. Confirmed by a sample case where if I have a rectangle with area 50 and give it dimensions x = 5 and y = 10, the corresponding perimeter will be 30, and since 50 ≠ 30, then Area ≠ Perimeter. However, I can still arrange Area = 10(5) into Area/10 = 5 and substitute into [Perimeter = 2x + 2y] to get [Perimeter = 2x + 2(Area/10)] or [30 = 2(10) + 2(50/10)] ==> [30 = 30]


The accuracy of the above statements should determine what I'm understanding and if there is anything I'm still confused about. Hopefully this confirms I've understood everything discussed thus far.

 

[checkitagain]

x^2-5x+10=0

a \ = \ 1
b \ = \ -5
c \ = \ 10

x=\frac{5\pm\sqrt{(-5)^2-4\cdot 10}}{2} \ \ \ &lt;--- It \ \ should \ \ be \ \ this \ \ instead.

\text{b = -5, not 5, so when you substitute the values, you cannot}
\text{sub in 5, regardless that (5)^2 = (-5)^2. In general, when a step }
\text{is shown for the substitution of values into an algebraic expression, }
\text{it must be correct, else that would not be shown.}
\text{And if it is the latter case, then show a step which follows from it. }
\text{What was shown is a type of unintentional fudging.}


Actually, to be consistent, either type:


x = \frac{-(-5) \pm\sqrt{(-5)^2 - 4 \cdot 1 \cdot 10}}{2(1)} \ \ \ or


x = \frac{5 \pm \sqrt{25 - 40}}{2}

...

 

[checkitagain]

Statement 2: Given the shape of a rectangle, if you let x =
any positive value, there will always be some corresponding positive
y value, such that the dimensions (x,y) satisfy area = perimeter.
x = 2 for the given situation is a counterexample. Remember that?

I am addressing only your statement 2 at present. My not addressing
your other numbered statements at this time doesn't necessarily
support them or not.

 

[LearninDaMath]

Looking forward to Mentallic's input. He seems to be the only person who is able to provide some needed (and appreciated) clarity on the confusions we all seem to be having on this specific topic.

Mentallic, when you get a chance, can you provide your thoughts on post #30.

 

[checkitagain]

Looking forward to Mentallic's input.

He seems to be the only person who is able to provide some needed
(and appreciated) clarity on the confusions we all seem to be
having on this specific topic.

Mentallic, when you get a chance, can you provide your thoughts on post #30.

"We" don't have confusions. You are to welcome and appreciate the time, effort,
and expertise given by the hints given by me, phinds, aznboy, and HallsofIvy as well.

If you fail to correct your errors after they are being repeatedly emphasized,
then that is on you.

This post of yours essentially treats the others' posts as if they did not exist.

 

[Mentallic]

Looking forward to Mentallic's input. He seems to be the only person who is able to provide some needed (and appreciated) clarity on the confusions we all seem to be having on this specific topic.

Mentallic, when you get a chance, can you provide your thoughts on post #30.

You'll have to wait another 12 hours, I have a pretty heavy workload right at the moment.

 

[LearninDaMath]

"We" don't have confusions. You are to welcome and appreciate the time, effort,
and expertise given by the hints given by me, phinds, aznboy, and HallsofIvy as well.

If you fail to correct your errors after they are being repeatedly emphasized,
then that is on you.

This post of yours essentially treats the others' posts as if they did not exist.

That is untrue. No need to argue with you, as anyone who reads through this thread can conclude. Have a nice day.

 

[LearninDaMath]

You'll have to wait another 12 hours, I have a pretty heavy workload right at the moment.

Thanks Mentallic. Looking forward to your feedback. Good luck on your work/studies.

 

[Mark44]

Looking forward to Mentallic's input. He seems to be the only person who is able to provide some needed (and appreciated) clarity on the confusions we all seem to be having on this specific topic.

"We" don't have confusions. You are to welcome and appreciate the time, effort,
and expertise given by the hints given by me, phinds, aznboy, and HallsofIvy as well.

If you fail to correct your errors after they are being repeatedly emphasized,
then that is on you.

This post of yours essentially treats the others' posts as if they did not exist.

That is untrue. No need to argue with you, as anyone who reads through this thread can conclude. Have a nice day.

I have read many of the posts in this thread, and am inclined to agree with checkitagain. It seems rude to me to single out one person (Mentallic) for thanks, and cavalierly ignore the efforts of the others who have participated.

And you projected your confusion onto the people who posted responses when you said "the confusions we all seem to be having on this specific topic." I can see why checkitagain took umbrage.

 

[LearninDaMath]

I have read many of the posts in this thread, and am inclined to agree with checkitagain. It seems rude to me to single out one person (Mentallic) for thanks, and cavalierly ignore the efforts of the others who have participated.

And you projected your confusion onto the people who posted responses when you said "the confusions we all seem to be having on this specific topic." I can see why checkitagain took umbrage.

Mark, a couple of semesters ago, I went to see my guidence counselor, which is encouraged for first time university students pursing an engineering degree. The engineering guidance office is located somewhere off campus. I knew what street number I needed to get to, and knew what general direction to head towards. So on my way, I rolled my window down at a red light and asked a guy in the car next to me if I should go left or right at the next intersection to get to such and such street. He paused for a moment as though he was going through a mental map of the city, so it was clear to me he must've known what street I was talking about. After a moment, he confidently told me I must turn right at the next light. Long story short, after pulling into a gas station to get a second opinion, I was finally on my way in the correct direction. I don't fault the guy who confidently looked me in the eyes and told me to turn right when I was actually supposed to go left (I simply wished he was confident enough to admit he didn't know what the particular street I was referring to). I just thank the gas station clerk who not only told me I was going in the wrong direction, but actually provided sound reasoning in how to get there by telling me what other streets I should pass and a landmark I should make another turn at.Mark, if you read the entire thread, each post consecutively, you will see how Mentallic was the only poster in this thread who provided clarity to my question. I am not faulting anyone to any degree for any wrong doing or lack of accuracy. I am simply giving credit where credit is due.

I can see why checkitagain may be slightly disgruntled. HallsofIvy contribution to this thread was in correcting aznboy's potentially incorrect information. HallsofIvy did not direct any information to me, directly, in this thread. For checkitagain to include HallsofIvy in his list of names means he is not following the context of the very thread in which he is participating (or he is purposely misrepresenting the context).

And if HallsofIvy corrects aznboy, so what. I'm sure aznboy is just as thankfull to HallsofIvy as I am to Mentallic in his thread. If Phinds told me something was obviously false, when Mentallic helped me discover that it was obviously NOT false, then kudos to Mentallic, why take that away from Mentallic. I'm sure phinds appreciates Mentallic's input just as much as I do. And I'm thankfull to Phinds for providing clarity on a few things too. And if checkitagain wants to be disgruntled that I am awaiting Mentallic's input instead of his on this particular topic, then maybe he shouldn't be diverting the topic by telling Mentallic he has a scratch in his paint when mentallic is generously providing directions to me on how I can get to my destination.

I appreciate anyone who adds their opinions to a thread. But just because information is posted doesn't mean that information is correct. Textbooks go through editing and still get published with honest mistakes, so nobody is infallible. And what is the big deal if someone is confused about something? If person A tells person B the answer is no when the answer is actually yes, and person B relies on person A's information, then both person A and person B are confused until person C introduces the correct information. Simple as that.

 

[Mentallic]

LearninDaMath, you seem to be giving credit where you believe it's deserved, but as a matter of fact, I'm only giving you special treatment because this doesn't look like it's a homework problem, but rather a curiosity that you have. In any other case, myself and other homework helpers would be giving you hints to nudge you in the right direction as opposed to explaining it in full detail.
Similarly, your example about getting help to find your destination isn't really a valid representation of what's happening. It is against the forum rules to just give students the answers to their homework problems (and yes, I've broken those per the reasoning above), so what you should be hoping to get at most is for that gas station manager to tell you you're heading in the wrong direction, and then stop right there. You'd have to figure the rest out for yourself, and if you are having further troubles, then you can ask for more help on those specific sub-problems.

Now, if someone has been leading you towards a wrong answer, then I can understand if you want to avoid their posts, but if they were honestly trying, then they at least deserve acknowledge in that regard. I'm not exactly sure what you were referring to when you said that someone was giving you the wrong answers though. I don't see any wrong answers, just confusion, as I had as well with my first post in this thread.


Anyway, now onto your questions:

Statement 1: For any given area of 16 and greater, there will always be one (and only one) specific set of dimensions (x,y) such that area = perimeter (xy = 2x + 2y).

The short answer is yes, but to understand why this is so, we'll need to introduce the concept of functions.
Firstly, the equation we're dealing with (xy=2x+2y) needs to be a function. If we make a variable x or y the subject (it's customary to make y the subject) then we have,
y=\frac{2x}{x-2}
And this is known to be a function. What that means is that for any x value, there is only 1 y value that will pop out.
But how do we know this? Well, this is called a rational function, but a little bit of algebra can simplify it in a way that'll be easier to grasp.

y=\frac{2x}{x-2}
=\frac{2x-4+4}{x-2}
=\frac{2(x-2)+4}{x-2}
=\frac{2(x-2)}{x-2}+\frac{4}{x-2}
=2+\frac{4}{x-2}

And it's trivial that \frac{4}{x-2} is a function, thus the equation we're dealing with is a function. This means for any x (length) we input, we will get an appropriate y value (again, except for x=2 or y=2).

Secondly, if we want to know that for any given A (with appropriate restrictions on A) there is a unique solution to the Area=Perimeter equation, we need to look back at the quadratic we got after plugging A=xy into the equation. So, from the previous post I made,

2x^2-Ax+2A=0

Now we want to know whether this is a function (given A\geq 16 and x&gt;0. Well, when we make x the subject, we get

x=\frac{A\pm\sqrt{A^2-16A}}{4}

So how do we know whether this is a function or not? That is, for any given A, is there just one value of x that pops out? If it's hard to tell, try breaking it up like this:

x=\frac{A}{4}\pm\frac{\sqrt{A^2-16A}}{4}

So we have x equalling to some number \pm another number, which means x equals 2 numbers! Unless the part after the \pm is equal to 0, but that only happens when A^2-16A=0 which is at A=16. So, for any A there are two values of x! Surely this means that we can get two sets of (x,y) values for each A? Well, no. There's something we're forgetting

So let's say we chose A=18 (we know a solution set for this is (x,y)=(6,3) ), if we plug this into the equation we get
x=\frac{18\pm\sqrt{18^2-16\cdot 18}}{4}

And simplifying,
=\frac{18\pm\sqrt{18(18-16)}}{4}

=\frac{18\pm\sqrt{18\cdot 2}}{4}

=\frac{18\pm\sqrt{36}}{4}

=\frac{18\pm 6}{4}

x=6, 3

Notice that we found both the values 6 and 3 for x. This is because we haven't specified any added restrictions on x such as the length is always bigger than the width or anything similar. We could've fooled ourselves into thinking that there were two solution sets when in fact the two solution sets we were getting would be (6,3) and (3,6), but we consider both these rectangles to be equivalent, so we don't count that as separate solutions.

So in fact, that equation is giving us the solution sets for both length and its corresponding width (you can choose whether length is going to be the + and width - or vice versa).

And so finally, since the restricted domain (let's assume the length x is longer than the width)

x=\frac{A+\sqrt{A^2-16A}}{4}

is a function, then your question is answered.

Statement 2: Given the shape of a rectangle, if you let x = any positive value, there will always be some corresponding positive y value, such that the dimensions (x,y) satisfy area = perimeter.

Yes, because xy=2x+2y thus y=\frac{2x}{x-2} is a continuous function (within its domain, so we exclude the value x=2). This means that for any x-value you choose, there will be some y-value to go with it. There are no holes in the function.

Statement 3: The ratio between height and length of a rectangle can be used to rewrite xy = 2x + 2y into an equation in terms of x, (such as x = y so that x(x) = 2x + 2(x)). So that the x dimension of the described rectangle can be found, which can then be used to find the specific y value that let's the equation area = perimeter be true.

I actually touched on the length to width ratios in one of my previous posts. If we want to restrict ourselves to rectangles where the length is three times the width, then we use y=3x and substitute that into the equation, giving us x(3x)=2x+2(3x)
3x^2=2x+6x

With this quadratic, if you solve for x you'll be given the value of the length such that Area=Perimeter when the length is three times the width. It only happens at one value (solve the quadratic and toss out unwanted values, such as x=0).

As I also added before, you can try the same procedure for all possible ratios, which we can denote as m or anything else you prefer. You'll find that when you get to your answer, it will in fact be a function.

In summary, yes. And just to add, when you say ratio x=y that means you're looking for the specific case when the rectangle is a square (x=y means length=width).

Statement 4: And in the example of ratio x = y, [x(x) = 2x + 2x] ==> [x^2 = 4x] where x = 4 and y = 4. You can not use the fact that the sides are of a specific proportion in order to assume that any value of x will satisfy Area = Perimeter. In other words, just because the sides are x = y, it doesn't mean that 5 = x and 5 = y will satisfy Area = Perimeter. As a second example, just because (6,3) satisfies [y = 2x] and thus [x(2x) = 2x + 2(2x)] = [Area = Perimeter], it doesn't mean that (8,4), (10,5), (12,6) are going to satisfy Area = Perimeter.

Yes, you're right. This is the corollary to all the points we've covered before. We've shown that for any given area, there is only one possible rectangle that can satisfy area=perimeter, and also shown that given any length (except 2) there is only one value of width that'll satisfy area=perimeter, thus, we can't just choose any combination of values and expect it to satisfy area=perimeter.

It always has to satisfy xy=2x+2y

Statement 5: I can be given a random Area (greater than 16) and by way of algebra, find the two dimensions that satisfy Area = Perimeter.

Yes, that's the same as statement 1.

Statement 6: I can start with a value representing x, and from there, use algebra to determine the a y value such that area = perimeter, even if I haven't yet determined the specific value of area. And once I have found the y value corresponding to the x value, a simple multiplication will tell me the area (and of course perimeter).

Yes. What you have there are 3 unknowns: length, width, area. Once you know 2 of them, you can solve the last one. So this means that you can have length and area, and correspondingly find width.

Statement 7: Given [Area = xy] and [Perimeter = 2x + 2y] Area and Perimeter do not have to be equal to each other in order to substitute one into the other. Confirmed by a sample case where if I have a rectangle with area 50 and give it dimensions x = 5 and y = 10, the corresponding perimeter will be 30, and since 50 ≠ 30, then Area ≠ Perimeter. However, I can still arrange Area = 10(5) into Area/10 = 5 and substitute into [Perimeter = 2x + 2y] to get [Perimeter = 2x + 2(Area/10)] or [30 = 2(10) + 2(50/10)] ==> [30 = 30]

Area and perimeter are hardly ever equal to each other. This is obvious from the fact that xy and 2x+2y are considerably different functions from each other (although for some specific values, they do intersect, which gives us (x,y) when area=perimeter).
You can solve for anything you want though. You can find when area is twice the perimeter, so xy=2(2x+2y) or when the perimeter is k times more than area, so kxy=2x+2y. Just because we equate these equations doesn't mean that for any value of x,y they're going to be equal. When we equate them, we're assuming they're equal, so if you chose values that don't satisfy the equality, then you'll get something that doesn't equate, such as 1=0.

Also, you really aren't asking a question here, you're just confirming something that you've already come to understand on your own. So I guess I can say, yes, you're right

Good luck trying to take all that in!