For real numbers $x,y$ satisfy the condition $x^2+5y^2-4xy+2x-8y+1=0$. Find the Maximum and Minimum Value of the expression $A=3x-2y$. For real numbers $x,y$ satisfy the condition $x^2+5y^2-4xy+2x-8y+1=0$.Find the Maximum and Minimum Value of the expression $A=3x-2y$.
 A: Using secondary school  "tools", you can replace
$$
x = \left( {A + 2y} \right)/3
$$
into the quadric equation, to get, after some simple simplifications
$$
25y^{\,2}  - \left( {8A + 60} \right)y + \left( {8A + 60} \right)^{\,2}  = 0
$$
whose solution is
$$
\eqalign{
  & y = {{2\left( {2A + 15} \right) \pm \sqrt {\left( {8A + 60} \right)^{\,2}  - \left( {10A + 30} \right)^{\,2} } } \over {25}} =   \cr 
  &  = {{2\left( {2A + 15} \right) \pm \sqrt {\left( {18A + 90} \right)\left( { - 2A + 30} \right)} } \over {25}} =   \cr 
  &  = {{2\left( {2A + 15} \right) \pm 3\sqrt {\left( {2A + 10} \right)\left( { - 2A + 30} \right)} } \over {25}} \cr} 
$$
For the solutions to be real, we must have
$$
0 \le \left( {2A + 10} \right)\left( { - 2A + 30} \right)\quad  \Rightarrow \quad  - 5 \le A \le 15
$$
corresponding to the $(x,y)$ values
$$
\left\{ \matrix{
  A =  - 5 \hfill \cr 
  y = 10/25 = 2/5 \hfill \cr 
  x = \left( { - 5 + 4/5} \right)/3 =  - 7/5 \hfill \cr}  \right.\quad \left\{ \matrix{
  A = 15 \hfill \cr 
  y = 90/25 = 18/5 \hfill \cr 
  x = \left( {15 + 36/5} \right)/3 = 37/5 \hfill \cr}  \right.
$$
A: The given equation defines a certain (not axis aligned) ellipse $E$ in the $(x,y)$-plane. The family of level lines $\ell_c: \ g(x,y)=c$ of the objective function
$$g(x,y):=3x-2y$$
covers the plane as a family of parallels. For certain values of $c$ the line $\ell_c$ will not intersect the ellipse, for other values of $c$ the line $\ell_c$ will intersect the ellipse in two points, and for the crucial values of $c$ the line $\ell_c$ will just be tangent to the ellipse at exactly one point. 
You therefore have to find the values $c$ for which $E\cap \ell_c$ consists of exactly one point, i.e., a certain equation has exactly one solution.
A: by the Lagrange Multiplier method we get $$A\le 5(1+\sqrt{6})$$ for $$x=\frac{1}{5}(15+11\sqrt{6})$$,$$y=\frac{10}{11}+\frac{4}{55}(15+11\sqrt{6})$$
and $$A\geq -5(-1+\sqrt{6})$$  for $$x=\frac{1}{5}(15-11\sqrt{6}),y=-\frac{2}{5}(-5+2\sqrt{6})$$
