Optimizing Quadratic Functions with the Second Derivative Test

In summary, the second derivative test guarantees that the function has a local maximum at (0,0,0) if the Hessian matrix is positive-definite, and a local minimum at (0,0,0) if the Hessian matrix is negative-definite.
  • #1
Zyuke
2
0
Under what conditions on the constants a and b does the second derivative test guarantee that the function

g(x,y,z)=ax^2+2axz+by^2-2byz+z^2

has a local maximum at (0,0,0)? a local minimum at (0,0,0)?

well, i used the Hessian matrix to compute the eigenvalues to set them above zero. but the computation is so complicated that i used Mathematica to solve them and it turned out to be some messy stuff. i was convinced that this is not the way to do it. but then how?
 
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  • #2
[STRIKE]Just looking at the problem makes me want to try Lagrange multipliers.[/STRIKE]
 
Last edited:
  • #3
Jufro said:
[STRIKE]Just looking at the problem makes me want to try Lagrange multipliers.[/STRIKE]

Won't work here.

They ask about the second derivative test, so you'll need to calculate the Hessian. It's what they ask for.
 
  • #4
Zyuke said:
Under what conditions on the constants a and b does the second derivative test guarantee that the function

g(x,y,z)=ax^2+2axz+by^2-2byz+z^2

has a local maximum at (0,0,0)? a local minimum at (0,0,0)?

well, i used the Hessian matrix to compute the eigenvalues to set them above zero. but the computation is so complicated that i used Mathematica to solve them and it turned out to be some messy stuff. i was convinced that this is not the way to do it. but then how?

Why would you try to set the eigenvalues to zero? The Hessian is positive-definite if all the eigenvalues are > 0 (so the origin is a global minimum---not just local!)

The Hessian cannot be negative-definite, so the origin cannot be a local maximum. (Note: a matrix A is negative-definite if and only if its negative B = -A is positive-definite, and since B(3,3) = -2 < 0, B cannot be positive-definite---no fancy tests needed here!)

To test for positive-definiteness of the Hessian H, just perform Cholesky decomposition; see, eg., http://en.wikipedia.org/wiki/Cholesky_decomposition . The Cholesky algorithm is easy to carry out in this case.
 
  • #5
thanks a lot.
 

Related to Optimizing Quadratic Functions with the Second Derivative Test

What is the "Second Derivative Test"?

The Second Derivative Test is a mathematical method used to determine the nature of a critical point in a function by analyzing the concavity of its graph. It helps determine whether the critical point is a local maximum, minimum, or neither.

How does the "Second Derivative Test" work?

The Second Derivative Test involves taking the second derivative of a function at a critical point and evaluating the resulting value. If the second derivative is positive, the critical point is a local minimum; if it is negative, the critical point is a local maximum; and if it is zero, the test is inconclusive.

When is the "Second Derivative Test" used?

The Second Derivative Test is used when finding the critical points of a function and determining their nature. It is a useful tool in optimization problems and can also be used to analyze the behavior of a function near its critical points.

What are the limitations of the "Second Derivative Test"?

The Second Derivative Test can only be applied to functions with continuous second derivatives. It also cannot determine the global maximum or minimum of a function, only the local ones. Additionally, it may give inconclusive results for certain functions, such as those with inflection points at their critical points.

How is the "Second Derivative Test" related to the "First Derivative Test"?

The First Derivative Test and the Second Derivative Test are both used to analyze the behavior of a function at its critical points. The First Derivative Test looks at the sign of the first derivative to determine if a critical point is a local maximum or minimum, while the Second Derivative Test looks at the concavity of the graph to make the same determination. Both tests are useful and can be used together to get a more complete understanding of a function's critical points.

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