6
$\begingroup$

Find the Maximum and Minimum of $$F=ax^2+2bxy+cy^2$$ when $$x^2+y^2=1$$ The variables a,b, and c are just real numbers.

I have attempted using partial differentiation in order to solve for the given maxima and minima, but I found the algebra used to solve for the variables just as complicated as using basic substitution. Is there a better method to approach this with? Any advice is helpful.

$\endgroup$
  • 2
    $\begingroup$ Do you know about Lagrange multipliers? $\endgroup$ – Git Gud Apr 14 '14 at 22:19
  • $\begingroup$ Yes but I was under the impression you needed 3 equations and 3 variables in order for that to be viable. $\endgroup$ – Amory Apr 14 '14 at 22:20
  • $\begingroup$ What if you just replace for $y = \pm\sqrt{1-x^2}$ $\endgroup$ – Hoda Apr 14 '14 at 22:21
  • 1
    $\begingroup$ @Amory No, any function with any numbers of variables subject to any number of constraints works. $\endgroup$ – Git Gud Apr 14 '14 at 22:21
  • 1
    $\begingroup$ $a \cos(t)^2 + 2b\sin(t)\cos(t) + c\sin(t)^2 $ with $t \in [0,2pi)$ $\endgroup$ – Alan Apr 14 '14 at 22:39
4
$\begingroup$

Let $A=\begin{bmatrix}a&b\\b&c\end{bmatrix}$, and $z=\begin{bmatrix}x\\y\end{bmatrix}$, then you have:

$$F=ax^2+2bxy+cy^2=z^TAz$$

$$x^2+y^2=1=z^Tz$$

So you can solve this problem instead:

$$\max~~~z^TAz ~~\text{subject to}~z^Tz=1 \tag{1}$$

and

$$\min~~~z^TAz ~~\text{subject to}~z^Tz=1\tag{2}$$

For (1), the value of $\max$ is equal to the largest eigenvalue of $A$, and $z=\begin{bmatrix}x\\y\end{bmatrix}$ is the corresponding eigenvector, and for (2), similarly, the value of $\min$ is equal to the smallest (second) eigenvalue of $A$, and the corresponding eigenvector is the solution for $z=\begin{bmatrix}x\\y\end{bmatrix}$.

Therefore, the only thing that you need to do, is forming the matrix $A=\begin{bmatrix}a&b\\b&c\end{bmatrix}$, and calculating its eigenvalues and eigenvectors.

$\endgroup$
  • $\begingroup$ Nice.${{{{}}}}$ $\endgroup$ – Git Gud Apr 14 '14 at 23:49

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.