Maximizing quadratic form subject to quadratic constraint

Question

How can I find the maximum and minimum value of the following quadratic form

$$Q(x) = x_1^2+3x_2^2+10x_1x_3+25x_3^2$$

subject to the equality constraint $\|x\|_2 = 3$? The norm is the Euclidian one.

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Normally if I was given the constraint $\|x\|_2 = 1$ I would find the matrix A:

$$ A = \begin{bmatrix} 1 & 0 & 5\\ 0 & 3 & 0\\ 5 & 0 & 25 \end{bmatrix} $$

Where I would calculate the eigenvalues of $A$ , which are $\lambda_1 = 26, \lambda_2 = 3, \lambda_3 = 0.$ Then the maximum value would be 26, and and the minimum value would be 0. But how am I supposed to do this with a constraint such as $\|x\|_2 = 3$ ?

Any guidance will be much appreciated.

Answer 1

This problem is homogeneous then calling $x_2 = \lambda x_1, x_3 = \mu x_1$ we arrive at

$$ \min(\max) \left(x_1^2\left(1+2\lambda^2+10\mu+25\mu^2\right)\right), \ \ \ \text{s. t.}\ \ \ x_1^2(1+\lambda^2+\mu^2)=k^2 $$

so we can follow without the constraint as

$$ \min(\max)_{\lambda,\mu} \frac{\left(1+2\lambda^2+10\mu+25\mu^2\right)k^2}{1+\lambda^2+\mu^2} $$

and deriving we find the conditions

$$ \cases{ 2 \lambda (10 \mu + 23 \mu^2-1) k^2=0\\ 2 ( 5 \mu^2-5 - 5 \lambda^2 - 24 \mu - 23 \lambda^2 \mu) k^2=0 } $$

giving the real solutions $(\lambda=0,\mu = -\frac 15)$ and $(\lambda=0,\mu = 5)$ with respective minimum and maximum $(0, 26k^2)$

Answer 2

If the norm of $\mathbf x=(x_1,x_2,x_3)$ is $3$, then the norm of $\frac13\mathbf x$ is $1$. Thus, if we multiply $Q(x)$ by $\frac19$ to get $R(x)$, the maximum and minimum of $R(x)$ with $\|\mathbf x\|=1$ translates to a maximum and minimum of $Q(x)$ with $\|\mathbf x\|=3$, by multiplying the solution vector for $R(x)$ by $3$ (and hence the extrema by $9$).