Question

I need help! :( Thank you in advance. :)
f(x,y) = 2y^3 - 6xy + x^2 + 10
Locate where f has relative maximum/minimum value.

Answer 1

so...\[f(x,y)= 2y^{3}-6xy+x^{2} +10\]
and you need to find the max/min?

I would think you'd have to isolate y first so that you can find the vertex is, if any

Answer 2

But how? My only idea for now is the application of the second derivatives of x and y. How could that be?

Answer 3

Hmm I'm trying to remember how to do these.
You want to take the first partial derivatives of f WRT to x and then y.
Set those first partials equal to 0, and we'll be able to find critical points.

Answer 4

\[\large f_x=-6y+2x \qquad \qquad f_y=6y^2-6x\]This is what I'm getting for the first partials. Have you learned this stuff? Or this is just for fun? :D

Answer 5

First derivatives are for relative maximum/minimum while the second is for the concavity. Am I right? :)) I'm just confuse on how to solve it accordingly. Hmm. What are the implications of the critical points by the way?

Answer 6

When dealing with multivariable functions like this? Hmmm I'm not exactly sure what the implications are XD lol
I'm having trouble visualizing it :o

But yah, we can use the second derivatives test for functions like this to test each of our critical points and determine if they're max/min or saddle points.

I think they have to be one of those 3 if they're a critical point. Err maybe they can also potentially be asymptotic like in single variable calculus XD hmm

Answer 7

Hmmmmmm...... after equating the 1st derivatives to 0, what would be next? I know I can find it through graphing but I want to know first by solving.

Answer 8

We'll set our first derivatives equal to 0, to find a system of equations, and that will allow us to find critical points.\[\large 0=-6y+2x \qquad \qquad 0=6y^2-6x\]

Answer 9

Dividing both sides by 2 in the first equation, and 6 from the second gives us,
\[\large 0=x-3y \qquad \qquad 0=y^2-x\]
Rewriting the second equation, solving for x gives us,\[\large x=y^2\]We'll plug this into the first equation, giving us,\[\large 0=y^2-3y\]

Answer 10

From here we can solve for the y values of our critical points,\[\large 0=y(y-3)\]
After we get those values, we'll plug them into the other equation to get corresponding x values.

Answer 11

\[\large y=0 \quad \text{plugging into} \quad x=y^2 \quad \text{gives us} \quad x=0\]
So we have a critical point at f(0,0).
Understand how we found that? :D

Answer 12

This doesn't work out the same as in single variable calculus, if that's what you were hoping :c hehe

Answer 13

Yes. :) Then another x=9? :)

Answer 14

yah that sounds right! c: f(9,3) i think

Answer 15

If this is not the same with the single variable, then how?

Answer 16

So we first want to put them into the uhhh, second derivatives test for multivariable.. functions, or whatever it's called :D\[\large D(x,y)=f_{xx}(x,y)\cdot f_{yy}(x,y)-[f_{xy}(x,y)]^2\]We need to take the second partials in order to test these points.

Answer 17

Hopefully I'm remembering that formula correctly :3 one sec, i need to look it up to make sure lol

Answer 18

Yah looks good c: Ok soooo let's take our second partials now.

Answer 19

\[\large f_{xx}=2 \qquad \qquad f_{yy}=12y \qquad \qquad f_{xy}=-6\]This is what I'm coming up with, hopefully I didn't make any mistakes along the way c:

Answer 20

\[\large D(0,0)=f_{xx}(0,0)\cdot f_{yy}(0,0)-[f_{xy}(0,0)]^2\]We'll test our first point, filling in all the blanks gives us,\[\large D(0,0)=2\cdot 0-(-6)^2\]

Answer 21

The value that we get for the second derivative doesn't actually matter. We only care whether or not it's larger than 0.
If it's larger than 0, then it has the potential to be a max/min. We'll perform another quick test if that's the case.

If this test comes out negative, then the point is a saddle point.

Answer 22

So we've found that the point (0,0) is a saddle point.

Answer 23

\[\large D(9,3)=2\cdot (3\cdot 12)-(-6)^2\]\[\large D(9,3)>0\]

Answer 24

Grr sorry the equation tool keeps crashing me lol

Answer 25

It's okay. This is really a huge help for me. Just go on coz' I'm still following. :)

Answer 26

This point turned out to be greater than 0! So it might be a max or min!
Let's apply the other test.\[\large f_{xx}(x,y)<0 \quad \leftarrow \quad \text{This tells us that our point is a maximum.}\]\[\large f_{xx}(x,y)>0 \quad \leftarrow \quad \text{This tells us that our point is a minimum.}\]

Answer 27

\[\huge f_{xx}(9,3)=2\]This value is larger than 0! :o So we have found a local minimum! yay!

Answer 28

Assuming I didn't make any mistakes along the way, which is quite possible ^^

Answer 29

Oh and I suppose it might be a good idea to plug your critical point back into your original function\[\large f(9,3)=-17 \qquad \text{is a minimum.}\].

So this is what we can conclude about this function.
\[\large f(0,0)=10 \qquad \text{is a saddle point.}\]

Answer 30

woah that didn't paste correctly :O oh well hehe

Answer 31

Thank you so much and I'm sorry for this kinda late reply. My connection just slowed down. :D Hmm. Thank you so much again. I was following while you're solving and we did have the same answers on the calculations. So no worries. I'm now going to rewrite my homework. :))

Answer 32

Oh goodie \c:/ heh