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.
 A: 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}

A: 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 $\frac{1}{3} + \sqrt{(-5/3)^2 + (\sqrt{3})^2} = \frac{1}{3} + \frac{1}{3} \sqrt{52}$.
A: 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
$$f_m(x,y)= x^2(1-3\frac yx -2\frac {y^2}{x^2})= 
\frac13(1\pm 2 \sqrt{13})$$
A: One approach is to exploit the fact that both functions are homogeneous:
$$x^2-3xy-2y^2 = \frac{x^2-3xy-2y^2}{1} = \frac{x^2-3xy-2y^2}{x^2+xy+y^2}$$
Now you can substitute $t=x/y$ and reduce this to a single-variable problem:
$$...=\frac{t^2-3t-2}{t^2+t+1}$$
Which can then be solved without calculus by finding the largest $f$ such that the following quadratic in $t$ has a non-negative discriminant:
$$t^2 - 3t - 2 = f\times(t^2+t+1)$$
A: $$x^2+xy+y^2=1\implies y_\pm=-\frac{1}{2} \left(x \pm\sqrt{4-3 x^2}\right)$$
Using $y_-$ gives
$$f=x^2-3xy_m-2y_m^2=\frac{7 x^2}{2}+\frac{1}{2} \sqrt{4-3 x^2} x-2$$ Compute the derivative, set it equal to $0$ and solve for $x$. You should get
$$x_1=\sqrt{\frac{1}{39} \left(26+7 \sqrt{13}\right)} \qquad x_2=-\sqrt{\frac{1}{39} \left(26-7 \sqrt{13}\right)}$$ Compute the coreesponding $y_1$ and $y_2$.
Do the same with $y_+$.
A: 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 $\sqrt{(-5/3)^2 + 3} - \frac{5}{6} = \frac{1 + \sqrt{52}}{3}$.
