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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 $A=3x-2y$.

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    $\begingroup$ Are you familiar with Lagrange multipliers (for constrained optimization problems) or are you not supposed to use that method? $\endgroup$ – StackTD Feb 23 '18 at 16:01
  • $\begingroup$ I do not know this method, I'm at secondary school. $\endgroup$ – Mary Feb 23 '18 at 16:04
  • $\begingroup$ Have you done similar problems (in school) and if so; what approach or method are you supposed to use? $\endgroup$ – StackTD Feb 23 '18 at 16:19
  • $\begingroup$ This is a mathematical form that I have never encountered before. I know only some basic inequalities, for example $\frac{a+b}{2}\ge \sqrt{ab}\,\,\forall a,b>0$ or ${{\left( ax+by \right)}^{2}}\le \left( {{a}^{2}}+{{b}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}} \right)\,\,\forall a,b,x,y\in R$. Help me solve the problem by using basic inequalities. $\endgroup$ – Mary Feb 23 '18 at 16:32
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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 $$ \left\{ \matrix{ A = - 5 \hfill \cr y = 10/25 = 2/5 \hfill \cr x = \left( { - 5 + 4/5} \right)/3 = - 7/5 \hfill \cr} \right.\quad \left\{ \matrix{ A = 15 \hfill \cr y = 90/25 = 18/5 \hfill \cr x = \left( {15 + 36/5} \right)/3 = 37/5 \hfill \cr} \right. $$

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  • $\begingroup$ My way. We can safely focus on the discriminant only, ignoring the actual solution, right? $\endgroup$ – lab bhattacharjee Feb 23 '18 at 18:32
  • $\begingroup$ @labbhattacharjee: for sure. I preferred to express the whole solution for $y$ (at little extra effort), so to get from $A$ the corresponding $(x,y)$, to plug into the quadric and counter-check if null. $\endgroup$ – G Cab Feb 23 '18 at 18:56
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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.

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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 $$x=\frac{1}{5}(15-11\sqrt{6}),y=-\frac{2}{5}(-5+2\sqrt{6})$$

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  • $\begingroup$ Thank you but I did not learn the Lagrange Multiplier method. Please, help me solve the problem by using basic inequalities. $\endgroup$ – Mary Feb 23 '18 at 16:36
  • $\begingroup$ you can solve the equation $$x^2+5*y^2-4xy+2x-8y+1=0$$ for e.g. $x$ and plug this in the equation $$A=3x-2y$$ and you will get a Problem in only one variable. $\endgroup$ – Dr. Sonnhard Graubner Feb 23 '18 at 16:41

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