<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] This solution uses no calculus.
By multiplying constrain with 4 we get $$(2x+y)^2+3y^2=4$$
Let $z= 2x+y$ then we have $z^2+3y^2=4\;\;(*)$ and we are searching for extreme points of $$L:= {1\over 4}(z^2-8zy-y^2)$$
Let $z= 2\cos t$ and $y= {2\sqrt{3}\over 3}\sin t$ be a parametrization of elipse $(*)$. Then we have $$L= {1\over 3}(3\cos ^2t -8\sqrt{3}\sin t\cos t - \sin^2t )$$

By further simplification you can get \begin{align} 3L &= 1+2\cos (2t) -4\sqrt{3} \sin(2t)\\
&= 1+\sqrt{52}\sin (2t+\phi)\\
&\leq 1+\sqrt{52}\end{align}

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