# Critical points and partial differentiation

• dyn
In summary, you subtracted one equation from the other to obtain four solutions, but you may not have actually eliminated the second equation. You now have to check that each of these is a solution.

#### dyn

Homework Statement
Find the critical points of z(x,y) = xy ( x+y-3)
Relevant Equations
At the critical points the partial derivatives wrt x and y are both zero
zx = 2xy + y2 -3y = 0 and zy = 2xy + x2 - 3x = 0

Subtracting one equation from the other gives

y2 - 3y = x2- 3x ⇒ y (y-3) = x (x-3)

This leads to the following solutions ( 0 , 0) , (0 ,3) , (3 , 0) but the answer also gives ( 1, 1) as a solution. What have i done wrong to not get this solution ?
Thanks

dyn said:
Homework Statement:: Find the critical points of z(x,y) = xy ( x+y-3)
Relevant Equations:: At the critical points the partial derivatives wrt x and y are both zero

zx = 2xy + y2 -3y = 0 and zy = 2xy + x2 - 3x = 0

Subtracting one equation from the other gives

y2 - 3y = x2- 3x ⇒ y (y-3) = x (x-3)

This leads to the following solutions ( 0 , 0) , (0 ,3) , (3 , 0) but the answer also gives ( 1, 1) as a solution. What have i done wrong to not get this solution ?
Thanks
I don't what else to say except that you just didn't get it. You have:
$$y(y -3) = x(x-3)$$
That equation is satisfied by ##x = y##, isn't it?

WWGD, Delta2 and etotheipi
Well actually, all ##(a,b)## with ##a=b## satisfy ##y(y-3) = x(x-3)##, so it doesn't really help you much. Try another way!

Delta2
But if you substitute y=x in the pair of partial derivatives they are only satisfied for x=y=1

NB This is in reply to #3

Delta2
dyn said:
But if you substitute y=x in the pair of partial derivatives they are only satisfied for x=y=1

Exactly!

When you reduce the equations from 2 to 1 by subtraction that is equivalent to requiring the polynomials to be equal. This is a weaker constraint than both being equal to zero. Consider the point x=y=2, for example.

Delta2
DaveE said:
When you reduce the equations from 2 to 1 by subtraction that is equivalent to requiring the polynomials to be equal. This is a weaker constraint than both being equal to zero. Consider the point x=y=2, for example.
I'm not sure what you mean by this as x=y=2 is not a solution to the pair of partial derivative equations. But i have to admit that simultaneous equations don't normally cause me problems but there was something about this example that caused me problems

dyn said:
I'm not sure what you mean by this as x=y=2 is not a solution to the pair of partial derivative equations. But i have to admit that simultaneous equations don't normally cause me problems but there was something about this example that caused me problems
But x=y=2 does satisfy the equation you derived in your solution, y2 - 3y = x2- 3x ⇒ y (y-3) = x (x-3). That is precisely my point. You are not solving your system correctly.
2xy + y2 -3y = 0 and 2xy + x2 - 3x = 0 is more restrictive than y2 - 3y = x2- 3x; for example, when x=y=2.

Delta2
DaveE said:
But x=y=2 does satisfy the equation you derived in your solution, y2 - 3y = x2- 3x ⇒ y (y-3) = x (x-3). That is precisely my point. You are not solving your system correctly.
2xy + y2 -3y = 0 and 2xy + x2 - 3x = 0 is more restrictive than y2 - 3y = x2- 3x; for example, when x=y=2.
So ; how do you solve the original pair of equations to get all 4 solutions ? And what have i done wrong in my method ?

Your conclusion that ##y=x## is correct but incomplete:
1. we have to accompany ##y=x## by the initial equations that ##2xy+y^2-3y=0## and ##2xy+x^2-3x=0## and solve the system of those 3 equations
2. from ##y^2-3y=x^2-3x## we can infer ##y=x## as just one case. What about the case ##y=0\neq x=3## this case also satisfies ##y^2-3y=x^2-3x##.

dyn said:
So ; how do you solve the original pair of equations to get all 4 solutions ? And what have i done wrong in my method ?
In your method you have a one-way implication: IF ##(x, y)## is a solution THEN ##(x, y) = (0,3), (3,0)## or ##x = y##. You know that these are the only possible solutions, but you don't know that they actually are solutions.

You now have to check that each of these is a solution.

In general, you ought to be aware when algebraic steps involve two-way implication. E.g. $$2x + 5 = 17 \ \Leftrightarrow \ 2x = 12 \ \Leftrightarrow \ x = 6$$ In which case you know you have precisely the one solution. And be aware when the algebraic steps involve a one-way implication. E.g. in your OP you explicitly have ##\Rightarrow## only, which means that you may not have ##\Leftarrow## and you cannot assume that all the solutions to the final equation are solutions to the original.

Delta2
I don't fully understand what you mean by 1 or 2 way implications. If i start with

2xy + y2 -3y = 0 and 2xy + x2 - 3x = 0

what is the best way to solve these equations ? I eliminated ( 2xy ) by subtracting one equation from the other. Is this wrong ?

dyn said:
I don't fully understand what you mean by 1 or 2 way implications. If i start with

A one-way implication is IF, THEN or ##\Rightarrow##. A two-way implication is IF and ONLY IF or ##\Leftrightarrow##.

Subtracting the equations was an IF, THEN. Nothing wrong with that. It means that any solution must have the properties you deduce; but, not everything with those properties is a solution. A crude eaxmple is: $$x = 2 \ \Rightarrow \ x^2 = 4 \ \Rightarrow \ x = \pm 2$$ By squaring the equation you add a second solution (##x = -2)## that does not satisfy the original equation.

In your case you deduced that ##x = y## was a necessary condition. But, you had to go back and check for precisely which ##x = y## the original equation was true.

Note that when you are learning algebra this point is often not highlighted. Technically, unless all your steps are reversible, then you need to check any "solutions" you get really are solutions to the original equation(s).

dyn
OK, consider this:

First, just to save me a bunch of typing, let's define P(x,y)≡2xy+y2-3y

[P(x,y)=0] & [P(y,x)=0]
[P(x,y) -0=0] & [P(y,x)=0]
[P(x,y) -P(y,x)=0 IFF P(y,x)=0] & [P(y,x)=0]
etc...

Then you can simplify P(x,y) -P(y,x)=0, as you did, but you can't ignore the constraint IFF P(y,x)=0.

You can subtract the equations (like this), but you have to maintain all of the constraints that accumulate as you do it.

Here's another example to elaborate what @PeroK and @DaveE have been saying.

Solve the system
x + y = 0
2x - y = 0

Using the technique of post #1, we have
##x + y = 2x - y \Rightarrow 2y = x## or ##y = \frac 1 2 x##
This equation represents a line through the origin with slope m = 1/2. IOW, we have found an infinite number of solutions. Unfortunately, most of them don't satisfy the original system.

If we solve the system of equations using, say, adding the equations, we get 3x = 0, so x = 0, and by substitution, y = 0. This is a unique solution.

By setting x + y equal to 2x - y, we have lost the condition that both expressions also have to be zero. The fact that both x + y and 2x - y are zero, places an additional constraint on the system that is lost when we merely set the two expressions equal to each other.

dyn
dyn said:
2xy + y2 -3y = 0 and 2xy + x2 - 3x = 0
I subtracted one equation from the other to remove 2xy which lead to the issues above.

Is there a better way to solve the above pair of equations ?

dyn said:
I don't fully understand what you mean by 1 or 2 way implications. If i start with

2xy + y2 -3y = 0 and 2xy + x2 - 3x = 0

what is the best way to solve these equations ? I eliminated ( 2xy ) by subtracting one equation from the other. Is this wrong ?
The way to solve a two-variable systen of eequations is to eliminate one variable. You eliminated xy but neither x nor y. You got a single two-variable equation with infinite many solutions.

Delta2 and PeroK
dyn said:
I subtracted one equation from the other to remove 2xy which lead to the issues above.

Is there a better way to solve the above pair of equations ?
Factor out y from the first equation .
As it is zero, teither y must be zero, or the other factor. Work with both cases separateli, and substitute what you got into the secomd equation. Solve.

DaveE said:
OK, consider this:

First, just to save me a bunch of typing, let's define P(x,y)≡2xy+y2-3y

[P(x,y)=0] & [P(y,x)=0]
[P(x,y) -0=0] & [P(y,x)=0]
[P(x,y) -P(y,x)=0 IFF P(y,x)=0] & [P(y,x)=0]
etc...

Then you can simplify P(x,y) -P(y,x)=0, as you did, but you can't ignore the constraint IFF P(y,x)=0.

You can subtract the equations (like this), but you have to maintain all of the constraints that accumulate as you do it.
I didn't ignore the constraint because after eliminating 2xy i set

y2- 3y = x2 -3x = 0

I tried the other method of substituting x from one equation into the other. This way involves a lot more algebra and still involves going back to the original equations

dyn said:
I didn't ignore the constraint because after eliminating 2xy i set

y2- 3y = x2 -3x = 0
Subtracting the two equations, you got y2- 3y-(x2 -3x )=0 . That does not mean y2- 3y = x2 -3x = 0.
dyn said:
I tried the other method of substituting x from one equation into the other. This way involves a lot more algebra and still involves going back to the original equations
Show what you did. That method should work.

dyn said:
I didn't ignore the constraint because after eliminating 2xy i set

y2- 3y = x2 -3x = 0

I tried the other method of substituting x from one equation into the other. This way involves a lot more algebra and still involves going back to the original equations
You need to pull yourself together. You made one small oversight in your OP. Accept that and move on.

DaveE
BTW, there is an easy way to do this that doesn't involve subtracting those two equations. You start by factoring, then a bit of "[a or b] and [c or d]" style logic to generate multiple solutions.

I think I'm done explaining about constraints when manipulating equations, maybe you instructor or TA can help with that.

ehild said:
Subtracting the two equations, you got y2- 3y-(x2 -3x )=0 . That does not mean y2- 3y = x2 -3x = 0.

Show what you did. That method should work.

I agree with your 1st comment. The other method does work but involves more work

PeroK said:
You need to pull yourself together. You made one small oversight in your OP. Accept that and move on.
My original method worked and contained no errors as you stated in #5 and was a lot quicker. It may not have suited mathematical purists but it worked !

DaveE said:
BTW, there is an easy way to do this that doesn't involve subtracting those two equations. You start by factoring, then a bit of "[a or b] and [c or d]" style logic to generate multiple solutions.

I think I'm done explaining about constraints when manipulating equations, maybe you instructor or TA can help with that.
I don't have an instructor or TA. I self-study

dyn said:
I agree with your 1st comment. The other method does work but involves more work

My original method worked and contained no errors as you stated in #5 and was a lot quicker. It may not have suited mathematical purists but it worked !
OK, here's something you need to consider. You came to this website to ask for help understanding a problem. A website that is populated with people who volunteer their time and effort to help people like you with questions like this. We don't have to respond to these posts, and you don't have to listen to our advice. Advice that is well intentioned and usually has some value, although incorrect responses aren't impossible.

If you choose to argue with freely offered advice, you are not providing a lot of incentive for us to continue the dialog.

I, for one, am done; there are other posts to occupy my time. Good luck with your studies.

ehild
dyn said:
I subtracted one equation from the other to remove 2xy which lead to the issues above.

Is there a better way to solve the above pair of equations ?
What you did to there was fine, but you need to understand exactly what you had established.
Essentially, you made two errors in managing the information. One lost constraints while the other created constraints that did not exist.

1. You had two equations, so you had two "degrees of freedom" nailed down. In producing a third from them, you can discard one of the original pair but not both. By only using this new equation you are losing constraints, hence allowing extra solutions.

2. In handling this third equation, you spotted certain solutions but did not search for all solutions.
y (y-3) = x (x-3)
y2-x2-3y+3x=0
(y-x)(y+x)+3(x-y)=0
(y-x)(y+x-3)=0
y=x or y+x=3
Now you can plug these two possibilities, in turn, into one of your pair of equations.

dyn
DaveE said:
OK, here's something you need to consider. You came to this website to ask for help understanding a problem. A website that is populated with people who volunteer their time and effort to help people like you with questions like this. We don't have to respond to these posts, and you don't have to listen to our advice. Advice that is well intentioned and usually has some value, although incorrect responses aren't impossible.

If you choose to argue with freely offered advice, you are not providing a lot of incentive for us to continue the dialog.

I, for one, am done; there are other posts to occupy my time. Good luck with your studies.

There are 2 possibilities that exist ; either i am arguing with freely offered advice or the advice offered has not really addressed my question or explained it clearly and i am seeking further clarification .
Both possibilities exist !