6
$\begingroup$

Here we are more interested in the method to minimize the function, rather than what the actual result is.

The function I currently have, which I may need to change, is:

$$b / d + (1/6 \log{(b)})(1+\log{(b)})(1+2\log{(b)})m/(gd)$$

with:

$$b=g \log_2(\log_2{(g)}+d\log_2{(2n^2a^2)}+\log_2{(nm)})$$

Here $a$, $m$, and $n$ are given and known. I'm interested in knowing the method to find the $d$ and $g$ which minimize the function. For example, can I simply take the derivative to find the minimum values?

My question is, what method can I use to find the minimum values of a function of this type? Can it be as simple as taking the derivative, assuming that finding the derivative is easy?

$\endgroup$
14
$\begingroup$

In general, if you have a function of two variables, $f(x,y)$, to find the critical points you need to take partials and set them equal to zero $$\frac {\partial f}{\partial x}=0$$ $$\frac {\partial f}{\partial y}=0$$ The values of $x$ and $y$ which satisfy these equations will be either minima, maxima, or saddle points. You can plug them into the function to see which is bigger and compare them.
If you do not want to manually plug these values into the function, you can instead use the second derivative test. Let $D=f_{xx}f_{yy}-f_{xy}^2$, evaluating $D$ and all second partials at the critical points you have four options:
If $D>0$ and $f_{xx}>0$ you have a local minimum.
If $D>0$ and $f_{xx}<0$ you have a local maximum.
If $D<0$ you have a saddle point.
If $D=0$ you need to use a third order test to determine the nature of the critical point, although from the nature of your function, I believe it will be laborious enough to compute the first derivative, let alone the third partials. I would instead resort to a computer. But partial derivatives are the way to go with multivariate functions when looking for maxima and minima.

$\endgroup$
4
  • $\begingroup$ Thank you for your help on such a basic question. I've forgotten almost everything on multivariate calculus, and this refreshed my memory and gave me a way to proceed. Much appreciated, +1 and accept! $\endgroup$
    – Matt Groff
    Dec 14 '16 at 13:50
  • $\begingroup$ You're welcome, I'm glad I could help. $\endgroup$ Dec 14 '16 at 13:56
  • $\begingroup$ What is the meaning of D? Geometrical meaning or why $D = f_{xx} f_{yy} - f_{xy}^2$? $\endgroup$
    – SHREE6174
    Oct 11 '18 at 5:13
  • $\begingroup$ @SHREE6174 $D$ is the determinant of the Hessian matrix of $f$. See this page for more info: en.wikipedia.org/wiki/Second_partial_derivative_test $\endgroup$
    – on-pasta
    Sep 29 '19 at 4:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.