<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] Let $f(x,y)=x^2-3xy-2y^2$ and $g(x,y)=x^2+xy+y^2-1$. Then,
$$(f_x’, f_y’)= (2x-3y, -3x-4y),\>\>\>\>\>( g_x’, g_y’)= (2x+y, x+2y)$$
At the extrema, the two curve are tangential to each other, i.e. the extreme points satisfying
$$\frac{-3x-4y}{2x-3y}= \frac{x+2y}{2x+y}$$
along with $ g(x,y)=0$.
Solve to obtain $x^2= \frac1{26\pm 7\sqrt{13}}$, $\frac yx = 3\pm \sqrt{13}$.
Then, the extreme values are
$$f_m(x,y)= x^2(1-3\frac yx -2\frac {y^2}{x^2})= 
\frac13(1\pm 2 \sqrt{13})$$
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