<s>[INST] 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.
 [/INST] The condition implies $xy = 1 - (x^2 + y^2)$, thus $x^2-3xy - 2y^2 = x^2 - 3 + 3(x^2 + y^2) - 2y^2$ $ = 4x^2 + y^2 - 3$.
Then as previously mentioned, the ellipse $x^2+xy+y^2 = 1$ can be parametrised as $(x,y) = \left(\frac{1}{\sqrt{3}}\cos t+\sin t,\frac{1}{\sqrt{3}}\cos t-\sin t\right)$. Substituting this into $4x^2 + y^2 - 3$ gives:
$$4 \left(\frac{1}{3} \cos^2 t + \frac{2}{\sqrt3} \cos t \sin t + \sin^2 t\right) + \frac{1}{3} \cos^2 t - \frac{2}{\sqrt3} \cos t \sin t + \sin^2 t - 3$$
$$= \frac{5}{3} \cos^2 t + \frac{6}{\sqrt3} \cos t \sin t + 5 \sin^2 t - 3 = \frac{10}{3} \sin^2 t + \sqrt{3} \sin(2t) - \frac{4}{3}$$
$$= -\frac{5}{3} \cos(2t) + \sqrt{3} \sin(2t) + \frac{1}{3}$$
and since the maximum value of $A \cos t + B \sin t$ is $\sqrt{A^2+B^2}$, the maximum value of the expression is $\sqrt{(-5/3)^2 + 3} - \frac{5}{6} = \frac{1 + \sqrt{52}}{3}$.
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