# Finding Critical Points of a multivariable function

• arisachu
In summary, to find and classify critical points of f(x,y) = x2y- x2 - 2y2 - 2x4 - 3y4, you need to solve the system of equations 2xy - 2x - 8x3=0 and x2 - 4y - 12y3=0. This will give you the values of x and y for the critical points. Once you have these values, you can classify the critical points by plugging them into the original function and determining if they are local maxima, local minima, or saddle points.
arisachu

## Homework Statement

I'm supposed to find (and classify, which I'm really not sure what that means) critical points of f(x,y) = x2y- x2 - 2y2 - 2x4 - 3y4

## The Attempt at a Solution

I found the partial derivatives,
fx(x,y) = 2xy - 2x - 8x3
and
fy(x,y) = x2 - 4y - 12y3

I gather that x = 0 and x = sqrt(y-1/4)
I can't figure out how to solve for y, if I'm even supposed to.
I was under the impression to set both equations to 0 and solve, but I don't know if I am correct.
I guess my [STRIKE]main[/STRIKE] only question is if I am supposed to take the y solution and use it in the x? I want to solve it 100% on my own, so I only want broad answers, if that makes any sense?

Thanks to anyone who can help me! :)

Last edited:
Hi arisachu!

arisachu said:

## Homework Statement

I'm supposed to find (and classify, which I'm really not sure what that means) critical points of f(x,y) = x2y- x2 - 2y2 - 2x4 - 3y4

## The Attempt at a Solution

I found the partial derivatives,
fx(x,y) = 2xy - 2x - 8x3
and
fy(x,y) = x2 - 4y - 12y3

I gather that x = 0 and x = sqrt(y-1/4)
I can't figure out how to solve for y, if I'm even supposed to.
I was under the impression to set both equations to 0 and solve, but I don't know if I am correct.
I guess my [STRIKE]main[/STRIKE] only question is if I am supposed to take the y solution and use it in the x? I want to solve it 100% on my own, so I only want broad answers, if that makes any sense?

Thanks to anyone who can help me! :)

You'll need to know when fx and fy simultaniously become 0. So you'll need to solve the system

$$\left\{\begin{array}{c} f_x(x,y)=0\\ f_y(x,y)=0\\ \end{array}\right.$$

Thus

$$\left\{\begin{array}{c} 2xy - 2x - 8x^3=0\\ x^2 - 4y - 12y^3=0\\ \end{array}\right.$$

From your first equation, you have correctly deduced that x=0 or

$$x=\sqrt{\frac{y-1}{4}}$$

Now, what does that give you when you plug that in the second equation?

Thanks for you reply and help! :D

Ohhh, so it's as simple as that? :O
And then I solve for [STRIKE]x[/STRIKE] y from that? :)

Yes, solve for y in the second equation, and then you can solve for x.

micromass said:
Yes, solve for y in the second equation, and then you can solve for x.

Gosh, I had been thinking you solve for y in the second equation without the first equation's x, and then set those two solutions equal to each other and I couldn't for the life of me figure it out!
You're a life (and sanity haha) saver! Thank you so much!

## What is a critical point of a multivariable function?

A critical point of a multivariable function is a point where the partial derivatives of the function are equal to zero or do not exist. This means that the function may have a local maximum, minimum, or saddle point at that point.

## How do I find the critical points of a multivariable function?

To find the critical points of a multivariable function, you need to take the partial derivatives of the function with respect to each variable and set them equal to zero. Then, solve the resulting system of equations to find the values of the variables at the critical points.

## Are all critical points of a multivariable function local extrema?

No, not all critical points of a multivariable function are local extrema. Some critical points may be saddle points, where the function has neither a maximum nor a minimum value. Additionally, some critical points may not exist due to the function being undefined at those points.

## Can critical points exist at the boundary of a multivariable function's domain?

Yes, critical points can exist at the boundary of a multivariable function's domain. In this case, the critical point may not be a local extremum, but it can still provide valuable information about the behavior of the function near the boundary.

## Why are critical points important in multivariable functions?

Critical points are important in multivariable functions because they can help us identify local extrema and saddle points, which are essential for understanding the behavior of the function. They also play a crucial role in optimization problems, where we want to find the maximum or minimum value of a function.

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