Question

I have a question about a method for solving a critical points problem.

Answer 1

It will take me a minute to type it.

Answer 2

Find and classify all critical points of \[f(x,y) = xy - x^2y - xy^2\]

Answer 3

So I found that:

\[f_{x} = y-2xy-y^2\]
\[f _{y} = x-x^2-2xy\]

Answer 4

I now need to solve for x and y.

I know I could solve for x by setting each equation equal to 0 and using the substitution method. But what if I set each equation equal to each other like this.

\[y^2+2xy-y=x^2+2xy-x\]

Then both of the 2xy's would cancel and I would be left with \[y^2 - y = x^2 -x\]
By that I can see that y=x

so if I substitute in x for y in the first equation I would get \[x^2 + 2x^2 - x = 3x^2-x = x(3x-1) = 0\]

Thus x = \[\frac{ 1 }{ 3 }, 0\]

Answer 5

and y = 1/3, 0 also

Answer 6

So I'm just wondering if that approach works correctly, or would it only work for positive values of x and y?

Answer 7

Also, I realize I still need to use the second derivative test to determine the min and max

Answer 8

@phi

Answer 9

I thought we solve for each partial equals 0 ?

Answer 10

Pretty sure I need to solve for x and y using the partial equations

Answer 11

and then I find f(x, y) to determine the min and max

Answer 12

At least that is how my teacher was showing, lol

Answer 13

the critical points are where \( f_x = f_y = 0 \)
it is easier to first solve for when each =0, and after that, solve for x and y

Answer 14

CRITICAL POINTS or the minimum and maxium are when the slope of the tangent line in a curved equation is zero

Answer 15

Here is an example of what I am suppose to do in another problem

Answer 16

Right, that makes sense

Answer 17

I understand how to solve the problem when solving for x and y. I'm just trying to figure out if the method of setting the two equations equal to each other to solve for x and y would give me the correct results

Answer 18

Like if you look at the example problem I attached, you can see that they solve for x and y buy substituting

Answer 19

setting them equal is not the same as setting each equal to 0
they could be equal to each other , but not equal to 0

Answer 20

So would this solution here be incorrect then?

Answer 21

they start by setting each equation = 0, and then go from there.
they did not start by setting the equations equal to each other

Answer 22

Right right, I did that also if you look above at my solution. but I was just trying to figure out if my solution only worked for positive values of x and y

Answer 23

Well I didn't actually write it above, but both of the equations could be set to 0 I guess

Answer 24

it looks like it works (you are getting the correct critical points)

Answer 25

Hmm ok, y shouldn't be negative?