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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.

Matthew Conroy
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Amory
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    Do you know about Lagrange multipliers? – Git Gud Apr 14 '14 at 22:19
  • Yes but I was under the impression you needed 3 equations and 3 variables in order for that to be viable. – Amory Apr 14 '14 at 22:20
  • What if you just replace for $y = \pm\sqrt{1-x^2}$ – Hoda Apr 14 '14 at 22:21
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    @Amory No, any function with any numbers of variables subject to any number of constraints works. – Git Gud Apr 14 '14 at 22:21
  • When I substituted y = +/- (1-x^2)^(1/2) into the equation and then took the derivative in terms of y and then set it equal to zero, I'm not sure where to proceed to solve for y, because the algebra is tedious. I was hoping for a less monotonous way of solving it. And -Git Gud, could you perhaps show me how you would solve for x and y in terms of lambda using Lagrange multipliers? That is the step I am stuck with using that method. – Amory Apr 14 '14 at 22:26
  • Set $\Lambda (x,y,\lambda)=ax^2+2bxy+cy^2+\lambda(x^2+y^2-1)$, for all $(x,y,\lambda)\in \mathbb R^3$. – Git Gud Apr 14 '14 at 22:37
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    $a \cos(t)^2 + 2b\sin(t)\cos(t) + c\sin(t)^2 $ with $t \in [0,2pi)$ – Alan Apr 14 '14 at 22:39
  • I know how to set up the problem -Git Gud but I am unsure after multiplying out the formula and I have 3 equations, I struggle on solving x and y in terms of lambda. – Amory Apr 14 '14 at 22:52
  • @Amory Me too. Sorry. – Git Gud Apr 14 '14 at 23:09

1 Answers1

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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.

Alt
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