Minimise this (if you can)!

@Otto Bretscher – Exercise Question : Find the minimum/maximum value of $W(x,y,z) = 2x^2 + 4y^2 + 6z^2 + 8xy + 10xz + 12yz + 14x + 16y + 18z + 20$ (if it exists).

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Your method : To prove that it has no maximum value

Let $W (\mathbf v) = \mathbf v ^T Q\mathbf v + \mathbf b^T \mathbf v + c$ , where $Q = \begin{bmatrix} 2 & 4 & 5 \\ 4 & 4 & 6 \\ 5 & 6 & 6 \end{bmatrix}$ , $\mathbf b^T = [14 \enspace 16 \enspace 18 ]$ and $c=20$ .

Because $\det(Q) = 2(4\cdot6 - 6\cdot6) - 4(4\cdot6 - 6\cdot5) + 5(4\cdot6 - 4\cdot5) = 20 > 0$ , so that $Q$ has a positive eigenvalue; let $\mathbf x$ be a unit leading eigenvector with eigenvalue $\lambda$ . Then $W(k x) = \lambda k^2 + (\mathbf b^T \mathbf x)k + c$ is quadratic function of $k$ with a positive leading coefficient, so $W(x,y,z)$ fails to be bounded above. In other words, there is no maximum value of $W(x,y,z)$ .

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My method : To prove that is has no minimum value either.

Let's find the type of second-order algebraic surface , referencing this, we have

$e = Q = 20 > 0 , \quad \Delta E = \begin{bmatrix} 2&4&5&7 \\ 4&4&6&8 \\ 5 &6&6&9 \\ 7 & 8 & 9 & 20 \end{bmatrix} = 140, \rho_3 = \text{rank } e = 3, \rho_4 = \text{rank } E = 4$

To find the roots of $k_1,k_2,k_3$ of $\begin{bmatrix} 2-x & 4 & 5 \\ 4 & 4 -x & 6 \\ 5 & 6 & 6-x \end{bmatrix} = 0$ so that we can determine the value of $k$ such that $k := \begin{cases} 1 \quad \text{ if the signs of non-zero }k\text{'s are the same} \\ 0 \quad \text{ otherwise} \end{cases}$ .

and the roots to the equation below satisfy the condition $-x^3+12 x^2+33 x+20 = 0$ . By Descartes' rule of signs , there is exactly one positive root to this cubic equation only (or just use WolframAlpha ). So $k = 0$ . Comparing it with value in the Mathworld link as given above , we have

$\rho_3 = 3 ,\quad \rho_4 = 4, \quad \quad \Delta E = 140 > 0 \Rightarrow \text{sgn} \Delta(E) \equiv + ,\quad \quad k = 0$

to get a hyperboloid of one sheet. So both the domain and range of $W(x,y,z)$ is $\mathbb R^3$ . Thus, there is no maximum nor minimum value for this function.

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Follow-up questions :

( 1 ) Is my method correct too? (This is a rhetorical question)

( 2 ) Using your method, how do you prove that $W(x,y,z)$ has no minimum value either?

( 3 ) Is it possible to use partial derivatives to show that $W(x,y,z)$ has no minimum nor maximum value too? Because I know how to solve it for 2-dimensional functions , but I don't know any method for 3-dimensional functions or higher. If I recall correctly, there is something to do with Hessian matrix , but I'm not sure how to proceed. Thoughts? (Don't ask me to read that wikipedia page, it explains nothing to me)