technique for finding minima of quadratic surface with an $xy$-term? (without calculus)

Question

I am working thru the problems from Spivak's Calculus.

Problems 1-17 is a three part problem, asking for the smallest possible value first for a one-variable quadratic expression:

$2x^2-3x+4$

then for a two-variable quadratic expression:

$x^2 - 3x + 2y^2 + 4y + 2$

and finally for a two-variable quadratic expression with an $xy$-term:

$x^2 + 4xy + 5y^2 - 4x - 6y + 7$.

A hint is given for the first part, that one should "complete the square" by re-writing the expression as $2(x-\frac{3}{4})^2+?$.

I see that the first expression is a parabola, and the second expression is a parabolic surface. In both cases we can add a constant term $k$ so as to shift the curve/surface, turning it into a perfect square. These perfect squares can easily be solved for zero (i.e., their minimum). So the minimum of the original curve/surface is the negation of the shift $k$, and it occurs at the zero of the shifted curve/surface.

The third expression is also a parabolic surface – but it contains an $xy$ term, so as far as I can tell cannot be turned into a perfect square just by shifting.

Is there some algebraic technique used to find the minimum of a surface like this? (This problem is from chapter 1; no calculus has actually been introduced yet.)

Answer 1

Hint: You can use substitution $p = x + 2y$, $q = y$ to get rid of $xy$ term and solve it like the case before.

Answer 2

There is a standard way of rotating a conic which eliminates the $xy$ term. Given $Ax^2+Bxy +Cy^2 +Dx + Ey +F = 0$, solve $$\cot 2t = \frac{A-C}{B}.$$

In your example, $\cot 2t = -1$, so $t = 3\pi/8.$

Then the change of variables $$x = x'\cos t - y'\sin t$$ $$y=x'\sin t +y' \cos t$$

will eliminate the $xy$ term.