critical points: after first partials got non linear system

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eckimz

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[MATH]f(x,y)= x^3-y^3-2xy+6[/MATH]
First I did the partials

[MATH]fx= 3x^2-2y=0[/MATH][MATH]fy= -3y^2-2x=0[/MATH]
Ok, now on this system I got when [MATH]x= -3y^2/2 [/MATH] on the second equation.

So I put on the first one [MATH]3.(-3y^2/2)^2-2y=0[/MATH]Got: [MATH]y=+- (2/3)[/MATH]
Applied [MATH]y[/MATH] on [MATH]x= -3y^2/2 [/MATH] [MATH]x= -3(2/3)^2/2 [/MATH] [MATH]x= -3(-2/3)^2/2 [/MATH]Giving me [MATH]x= -2/3[/MATH]
So, first doubt is, does this math looks right? Seems pretty easy to mess it up on paper
Second doubt is: I can't clearly see what are my points, but I think its [MATH]P1(0,-2/3)[/MATH] and [MATH]P2(-2/3,0)[/MATH]
On the test of second derivative and to determe if its max/min I have no problem finishing... but need to know if my critical points are right, Idk
 
You're going to have to demonstrate how you managed y = -2/3.

Also, why didn't you find (x,y) = (0,0)?

Please demonstrate more steps so we can see where you wandered off.

[math]3\left(\dfrac{-3y^{2}}{2}\right)^{2} - 2y = 0[/math]
[math]\dfrac{27y^{4}}{4} - 2y = 0[/math]
[math]27y^{4} - 8y = 0[/math]
Very important. What, EXACTLY, did you do next?
 
tkhunny said:
You're going to have to demonstrate how you managed y = -2/3.

Also, why didn't you find (x,y) = (0,0)?

Please demonstrate more steps so we can see where you wandered off.

[math]3\left(\dfrac{-3y^{2}}{2}\right)^{2} - 2y = 0[/math]
[math]\dfrac{27y^{4}}{4} - 2y = 0[/math]
[math]27y^{4} - 8y = 0[/math]
Very important. What, EXACTLY, did you do next?
Click to expand...


Ok, I got a class right now, soon I'll type everything, sorry about it. But at that point u asked, I did:
[MATH]y(27y^3/4)-2=0[/MATH][MATH]27y^3=8[/MATH][MATH]y^3= (8/27)^(1/3) [/MATH] (sorry i dont know how to put cubic square cbrt?)
[MATH]y=+-2/3[/MATH]
And about the (0,0), thats one of my questions, I know should find two points. [MATH](0,0)[/MATH] and maybe [MATH](+-2/3 ; -2/3)[/MATH]but I can't see it clearly when I do the math, I mean, I know what to do, but not what I'm doing lol

Ahhh and after that I applied that y value on the x I found when I did x=0 on second equation
[MATH]x=−3y2/2[/MATH][MATH]x=(-3(2/3)^2)/2[/MATH]
 
Your first mistake was to divide by y. That assumes that it is not zero!

The right thing to do with [MATH]27y^{4} - 8y = 0[/MATH] is to factor out y to get [MATH]y(27y^{3} - 8) = 0[/MATH]. Then each factor leads to a solution.

Then you confused cube roots with square roots. Any number has two square roots, but only one cube root. There is no [MATH]\pm[/MATH].

You also forgot to check your solutions.

Give it another try.
 
that was like mindblowing! thank you. I will start from zero, this is hard level for me haha. I will try make is less messy aswell.
 
eckimz said:
[MATH]f(x,y)= x^3-y^3-2xy+6[/MATH]
First I did the partials

[MATH]fx= 3x^2-2y=0[/MATH][MATH]fy= -3y^2-2x=0[/MATH]
Ok, now on this system I got when [MATH]x= -3y^2/2 [/MATH] on the second equation.

So I put on the first one [MATH]3.(-3y^2/2)^2-2y=0[/MATH]Got: [MATH]y=+- (2/3)[/MATH]
Applied [MATH]y[/MATH] on [MATH]x= -3y^2/2 [/MATH][MATH]x= -3(2/3)^2/2 [/MATH][MATH]x= -3(-2/3)^2/2 [/MATH]Giving me [MATH]x= -2/3[/MATH]
So, first doubt is, does this math looks right? Seems pretty easy to mess it up on paper
Second doubt is: I can't clearly see what are my points, but I think its [MATH]P1(0,-2/3)[/MATH] and [MATH]P2(-2/3,0)[/MATH]
On the test of second derivative and to determe if its max/min I have no problem finishing... but need to know if my critical points are right, Idk
Click to expand...
Thank you for showing your work. to show my gratitude, I am going to be persnickety.

First , [MATH]f(x,\ y) = x^3 - y^3 - 2xy + 6 \implies f_x' = 3x^2 - 2y \text { and } f_y' = 3y^2 - 2x.[/MATH]
Derivatives are functions, not numbers. It may be important at times to remember that. What you should have said is

[MATH]f_x' = 0 \implies 3x^2 - 2y = 0 \text { and } f_y' = 0 \implies 3y^2 - 2x = 0.[/MATH]
You are trying to find where the derivatives are zero, not saying that the derivatives are zero everywhere.

From the second equation you correctly got

[MATH]x = -\ \dfrac{3y^2}{2}.[/MATH]
Then you correctly substituted into the first equation.

[MATH]3 * \left ( -\ \dfrac{3y^2}{2} \right )^2 - 2y = 0 \implies 3 * \dfrac{9y^4}{4} - 2y = 0 \implies \dfrac{27y^4}{4} - 2y = 0[/MATH]
What you next did was to divide by y, whether you formally factored or not.

So, the second point to remember is that before you ever divide by an unknown, you must first consider whether that unknown may be zero. in this case, zero is a possible solution because

[MATH]\dfrac{27 * 0^4}{4} - 2 * 0 = 0 - 0 = 0.[/MATH]
Thus, you have a critical point when x = 0 and y = 0.

[MATH]y \ne 0 \text { and } \dfrac{27y^4}{4} - 2y = 0 \implies \dfrac{27y^3}{4} - 2 = 0 \implies y^3 = \dfrac{8}{27} \implies \sqrt[3]{y^3} = \text {WHAT?}[/MATH]
My third point is that a real number has either two or no real even roots and always has one real odd root.

You may not have known this important fact.

Lots to learn from this little problem.

Now finish the problem up.
 
Dr.Peterson said:
Then you confused cube roots with square roots. Any number has two square roots, but only one cube root. There is no [MATH]\pm[/MATH].
Click to expand...

Amazing info, wish I had better basic math!! The rest of your comment, I think I fixed it, I'm still doing it. Gonna add JeffM comments into my math.
 
JeffM said:
Thank you for showing your work. to show my gratitude, I am going to be persnickety.

First , [MATH]f(x,\ y) = x^3 - y^3 - 2xy + 6 \implies f_x' = 3x^2 - 2y \text { and } f_y' = 3y^2 - 2x.[/MATH]
Derivatives are functions, not numbers. It may be important at times to remember that. What you should have said is

[MATH]f_x' = 0 \implies 3x^2 - 2y = 0 \text { and } f_y' = 0 \implies 3y^2 - 2x = 0.[/MATH]
You are trying to find where the derivatives are zero, not saying that the derivatives are zero everywhere.

From the second equation you correctly got

[MATH]x = -\ \dfrac{3y^2}{2}.[/MATH]
Then you correctly substituted into the first equation.

[MATH]3 * \left ( -\ \dfrac{3y^2}{2} \right )^2 - 2y = 0 \implies 3 * \dfrac{9y^4}{4} - 2y = 0 \implies \dfrac{27y^4}{4} - 2y = 0[/MATH]
What you next did was to divide by y, whether you formally factored or not.

So, the second point to remember is that before you ever divide by an unknown, you must first consider whether that unknown may be zero. in this case, zero is a possible solution because

[MATH]\dfrac{27 * 0^4}{4} - 2 * 0 = 0 - 0 = 0.[/MATH]
Thus, you have a critical point when x = 0 and y = 0.

[MATH]y \ne 0 \text { and } \dfrac{27y^4}{4} - 2y = 0 \implies \dfrac{27y^3}{4} - 2 = 0 \implies y^3 = \dfrac{8}{27} \implies \sqrt[3]{y^3} = \text {WHAT?}[/MATH]
My third point is that a real number has either two or no real even roots and always has one real odd root.

You may not have known this important fact.

Lots to learn from this little problem.

Now finish the problem up.
Click to expand...

Yes, lots to learn.
But first things first, thank you very much for your time and help, it was very useful and detailed. I hope my work goes as that aswell.

I'm kind of new here, I can't do much in on TeX yet, like those squares and fractions, no idea how yet. My final test is today, but in the next semester I will get better and frequent here, math is going inside my brain somehow! haha

So replying your last question, when
[MATH]y≠0 [/MATH][MATH] 3√y3= 2/3[/MATH]
So I have [MATH]y= 0[/MATH] and [MATH]y= 2/3.[/MATH]
Next thing I did was to apply this values of y on
[MATH]x=−3y^2/2[/MATH]
When [MATH]y= 0, x= 0.[/MATH]Confirming what you had told me before.
And when [MATH]y= 2/3, x=-2/3.[/MATH]
So I have two points [MATH](0,0)[/MATH] and [MATH](-2/3, 2/3).[/MATH]
I applied on
[MATH]F'xx= 6x[/MATH];
[MATH]F'yy= -6y[/MATH];
[MATH]F'xy= -2[/MATH].

[MATH]For (0,0)[/MATH];
[MATH]F'xx= 6x = 0[/MATH];
[MATH]F'yy= -6y = 0[/MATH];
[MATH]F'xy= -2 = -4[/MATH];
[MATH]D<0 = saddle[/MATH].

[MATH]For (-2/3, 2/3)[/MATH];
[MATH]F'xx= 6x = -4[/MATH];
[MATH]F'yy= -6y = -4[/MATH];
[MATH]F'xy= -2 = -2[/MATH];
[MATH]D= (-4)*(-4)-(-2)^2[/MATH];
[MATH]D= 12[/MATH];
[MATH][ D < 0 ; F'xx < 0 ] = local - maxima[/MATH].

My test in few hours, hope it's good haha
 
A good way to start learning LaTeX is to hit the reply button: it will let you see the actual code when LaTeX is used in a post.

And good luck on the test.
 
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