# Homework Help: Setting area = perimeter of rectangle, does it have any meaning?

1. Mar 23, 2012

### 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.

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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?

2. Mar 23, 2012

### checkitagain

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.

3. Mar 23, 2012

### phinds

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

x=4, y=4

4. Mar 23, 2012

### LearninDaMath

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?

5. Mar 23, 2012

### LearninDaMath

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?

6. Mar 23, 2012

### phinds

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 !

7. Mar 23, 2012

### phinds

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.

8. Mar 23, 2012

### 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.

Last edited: Mar 23, 2012
9. Mar 23, 2012

### checkitagain

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?

Last edited: Mar 23, 2012
10. Mar 23, 2012

### 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.

11. Mar 23, 2012

### phinds

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.

12. Mar 23, 2012

### LearninDaMath

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.

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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?

13. Mar 23, 2012

### phinds

OOPS

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

14. Mar 24, 2012

### 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?

15. Mar 24, 2012

### Mentallic

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.

16. Mar 24, 2012

### phinds

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

17. Mar 24, 2012

### checkitagain

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

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

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."

18. Mar 24, 2012

### LearninDaMath

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?

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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?

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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.

19. Mar 25, 2012

### checkitagain

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.

20. Mar 25, 2012

### phinds

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?

21. Mar 25, 2012

### LearninDaMath

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?

22. Mar 25, 2012

### phinds

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.

23. Mar 25, 2012

### 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.

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 =]

24. Mar 25, 2012

### HallsofIvy

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.

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

25. Mar 25, 2012

### LearninDaMath

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?