maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$ without Lagrange multiplier What is the maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$? I wonder there is a precalculus method, without using the Lagrange multiplier.

First, let us substitute $x := u + v, y := u - v$ in the constraint to get
$$(u+v)^2 + (u+v)(u-v) + (u-v)^2 = 3u^2 + v^2 = 1.$$
(Note that this substitution is a standard one to try once you notice the fact that the constraint equation is symmetric with respect to swapping $x$ and $y$.  It is closely related to the operation of rotating the plane clockwise by $45^\circ$ to convert a curve symmetric about the line $y = x$ to a curve symmetric about the line $v = 0$.)
Thus, the constraint curve can be parameterized by $u = \frac{1}{\sqrt{3}} \cos \theta, v = \sin \theta$ for $0 \le \theta < 2\pi$, and so the parameterization in terms of $x, y$ is
$$(x, y) = \left(\frac{1}{\sqrt{3}} \cos\theta + \sin\theta, \frac{1}{\sqrt{3}} \cos\theta - \sin\theta \right).$$
Now, substituting into the objective function, we have
$$x^2 - 3xy - 2y^2 = 2 \sin^2\theta + 2\sqrt{3} \sin\theta \cos\theta - \frac{4}{3} \cos^2\theta = \\ \frac{1}{3} - \frac{5}{3} (\cos^2\theta - \sin^2\theta) + \sqrt{3} (2\sin\theta \cos\theta) = \frac{1}{3} - \frac{5}{3} \cos(2\theta) + \sqrt{3} \sin(2\theta).$$
From here, using the standard fact that the function $A \cos t + B \sin t$ has maximum value $\sqrt{A^2 + B^2}$, we see that $x^2 - 3xy - 2y^2$ has maximum value $\frac{1}{3} + \sqrt{(-5/3)^2 + (\sqrt{3})^2} = \frac{1}{3} + \frac{1}{3} \sqrt{52}$.
