Critical Points and the Second Derivative Test for a Multivariable Function

In summary, the conversation discusses finding critical points and using the second derivative test to determine if they are local maxima, local minima, or saddle points for the function f(x,y) = x^4 + y^4 + 4xy. The individual's attempt at a solution involves taking the gradient and setting it equal to <0,0>, leading to the critical points (0,0), (1,-1), and (-1,1). However, there is a mistake in finding all the roots, as x^8-1 has two real roots. The Hessian is also mentioned, but the correct approach is to evaluate the determinant of the matrix of second derivatives at the critical points to determine the nature of the critical
  • #1
sheepcountme
80
1

Homework Statement



Find the critical points and use the second derivative test to decide if your critical points are local maxima, local minima, or saddle points.

f(x,y)=x4+y4+4xy


The Attempt at a Solution



so I took the gradient to get: <4x3+4y, 4y3+4x>

I know I need to set this equal to <0,0>..so,


4x3+4y=0 and 4y3+4x=0

but I'm stuck...I tried solving for y in the first one to get

y=-x3 and then plugging this into the next equation to get -x9+x=0

If I solve for x, I believe I get x=0 or x=1 and then plugging these into the first I get the points (0,0) and (1,-1)

Have I done this correctly?

And when we're talking about the second derivative test, is this the Hessian? And if so, I've gotten 144x2y2-12x2[tex]\lambda[/tex]-12y2[tex]\lambda[/tex]+[tex]\lambda[/tex]2-16=0

which seems awfully messy to be able to determine if the point is a maxima, etc.
 
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  • #2
That seems pretty ok, but what's wrong with (-1,1) as a critical point? I don't think you found all of the roots. And what are those lambdas doing the Hessian? Isn't it just the determinant of the matrix of second derivatives?
 
  • #3
sheepcountme said:

Homework Statement



Find the critical points and use the second derivative test to decide if your critical points are local maxima, local minima, or saddle points.

f(x,y)=x4+y4+4xy


The Attempt at a Solution



so I took the gradient to get: <4x3+4y, 4y3+4x>

I know I need to set this equal to <0,0>..so,


4x3+4y=0 and 4y3+4x=0

but I'm stuck...I tried solving for y in the first one to get

y=-x3 and then plugging this into the next equation to get -x9+x=0

If I solve for x, I believe I get x=0 or x=1 and then plugging these into the first I get the points (0,0) and (1,-1)
[itex]x^9- x= x(x^8- 1)= 0[/itex]
Yes, x= 0 is a root. But [itex]x^8- 1= 0[/itex] has two real roots.

Have I done this correctly?

And when we're talking about the second derivative test, is this the Hessian? And if so, I've gotten 144x2y2-12x2[tex]\lambda[/tex]-12y2[tex]\lambda[/tex]+[tex]\lambda[/tex]2-16=0

which seems awfully messy to be able to determine if the point is a maxima, etc.
Look at
[tex]\left|\begin{array}{cc}f_{xx} & f_{xy} \\ f_{xy} & f_{yy}\end{array}\right|= \left|\begin{array}{cc}12x^3 & 4 \\ 4 & 12 y^3\end{array}\right|[/tex]
Put in the values of x and y before evaluating the determinant and it is not at all complicated!
 
  • #4
Ah! I missed (-1,1), thank you!

And I got eigenvalues unnecessarily mixed up in all this (where the lamdas came from), thanks!
 

FAQ: Critical Points and the Second Derivative Test for a Multivariable Function

What is a critical point in a scientific context?

A critical point in a scientific context refers to a specific point in a phase diagram where a substance transitions from one state to another. This transition is usually between a liquid and a gas or between a liquid and a solid, and occurs at a specific combination of temperature and pressure.

What is the critical point problem?

The critical point problem is a mathematical and physical concept that focuses on understanding the behavior of substances at their critical points. This problem is also known as the critical point anomaly, which describes the unusual behavior of substances at their critical points.

Why is the critical point problem important?

The critical point problem is important because it helps scientists understand the unique properties and behaviors of substances at their critical points. This knowledge can be applied to various industries, such as in the development of new materials and in the design of industrial processes.

What are some real-world examples of the critical point problem?

Examples of the critical point problem can be found in various industries, such as in the production of supercritical fluids used in the extraction of essential oils and in the design of refrigeration systems that use carbon dioxide as a refrigerant. It also plays a role in the study of the behavior of Earth's atmosphere and in the development of new drugs.

How do scientists study the critical point problem?

Scientists use a combination of experimental and theoretical methods to study the critical point problem. This can involve conducting experiments at different temperatures and pressures, as well as using mathematical models and simulations to understand the behavior of substances at their critical points.

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