Convert This Equation Into An Inequality

First we rewrite the condition $x^2+y^2+2(3y-2x)-4=0\Leftrightarrow (x-2)^2+(y+3)^2=17$

This suggests that we should look at it in terms of $(x-2)$ and $(y+3)$ . Hence, if we want to maximize $Q = x+4y$ , it is better to try and maximize $(x-2) + 4(y+3) = x + 4y + 10 = Q + 10$ .

By Cauchy - Schwarz we have $(Q+10)^2\leq (1^2+4^2)[(x-2)^2+(y+3)^2]=17^2\Rightarrow -17\leq Q+10\leq 17$ $\therefore -27\leq Q\leq 7$ So the sum is $-20$ and $\begin{cases} Q_{min}=-27\Leftrightarrow (x,y)=(1;-7) \\ Q_{max}=7\Leftrightarrow (x,y)=(3;1)\end{cases}$

Basic geometry to the rescue! The given equation represents a circle described by $(x-2)^2+(y+3)^2=17$ . The circle must be tangent to the parallel lines $x+4y=m$ and $x+4y=n$ . By symmetry, the line that is equidistant to both of these lines (and also parallel to them) must pass through the center $(2,-3)$ , and this line can be described by $x+4y=\frac{m+n}{2} (*)$ . Substituting $x=2, y=-3$ and solving gives $m+n=\boxed{-20}$ .

A justification of $(*)$ is to just look at ratios. Let the lines described be

• $a: x+4y=m$

• $b: x+4y=t$

• $c: x+4y=n$

with $y$ -intercepts of $A,B,C$ respectively, where $b$ is equidistant to $a$ and $c$ . Then by projection ratios $AB=BC\Rightarrow t-m=n-t\Rightarrow t=\frac{m+n}{2}$ .

$\begin{aligned} x^2 + y^2 +2(3y-2x) - 4 & = 0 \\ (x^2 - 4x +{\color{#3D99F6}{4}}) + (y^2 + 6x + {\color{#3D99F6}{9}}) -{\color{#D61F06}{4}}-{\color{#D61F06}{9}}-4 & = 0 \\ (x-2)^2 + (y+3)^2 & = 17 \\ \frac {(x-2)^2}{17} + \frac {(y+3)^2}{17} & = 1 \end{aligned}$

Now, we can use the identity $\sin^2 \theta + \cos^2 \theta = 1$ . Therefore, we have:

$\begin{cases} \dfrac {x-2}{\sqrt{17}} = \cos \theta & \implies x = \sqrt{17} \cos \theta + 2 \\ \dfrac {y+3}{\sqrt{17}} = \sin \theta & \implies y = \sqrt{17} \sin \theta - 3 \end{cases}$

And

$\begin{aligned} A & = x + 4y \\ & = \sqrt{17} \cos \theta + 2 + 4\sqrt{17} \sin \theta - 12 \\ & = \sqrt{17} \cos \theta + 4\sqrt{17} \sin \theta - 10 \\ & = \sqrt{17}\cdot \sqrt{17}\left(\frac 1{\sqrt{17}} \cos \theta + \frac 4{\sqrt{17}} \sin \theta \right) - 10 \\ & = 17 \sin \left( \theta + \tan^{-1} \frac 14 \right) - 10 \end{aligned}$

$\begin{aligned} \implies A_{max} & = 17-10 = 7 & \small \color{#3D99F6}{\text{when }\sin \left( \theta + \tan^{-1} \frac 14 \right) = 1} \\ A_{min} & = -17-10 = -27 & \small \color{#3D99F6}{\text{when }\sin \left( \theta + \tan^{-1} \frac 14 \right) = -1} \end{aligned}$

$\implies A_{max} + A_{min} = 7 - 27 = \boxed{-20}$

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Alternative solution

The problem can also be solved using Lagrange multipliers as follows:

$\begin{aligned} F(x,y,\lambda) & = x+4y -\lambda (x^2 + y^2 +6y-4x - 4) \\ \frac {\partial F}{\partial x} & = 1 - \lambda (2x -4) \\ \frac {\partial F}{\partial y} & = 4 - \lambda (2y + 6) \\ \frac {\partial F}{\partial \lambda} & = -x^2 - y^2 -6y+4x + 4 \end{aligned}$

Putting $\displaystyle \frac {\partial F}{\partial x} = \frac {\partial F}{\partial y} = \frac {\partial F}{\partial \lambda} = 0$ , we have $\dfrac 1{2x-4} = \dfrac 2{y+3} = \lambda$ $\implies y = 4x - 11$ , and that

$\begin{aligned} x^2 + y^2 +6y-4x - 4 & = 0 \\ x^2 + (4x - 11)^2 + 6(4x - 11) - 4x - 4 & = 0 \\ 17x^2 - 68x + 51 & = 0 \\ x^2 - 4x + 3 & = 0 \\ (x-1)(x-3) & = 0 \end{aligned}$

Let us use Hessian matrix $\text H$ to check for the maximum and minimum.

$\begin{aligned} H & = \Delta^2 F(x,y,\lambda) = \begin{vmatrix} \frac {\partial^2 F}{\partial x^2} & \frac {\partial^2 F}{\partial x \partial y} & \frac {\partial^2 F}{\partial x \partial \lambda} \\ \frac {\partial^2 F}{\partial y \partial x} & \frac {\partial^2 F}{\partial y^2} & \frac {\partial^2 F}{\partial y \partial \lambda} \\ \frac {\partial^2 F}{\partial \lambda \partial x} & \frac {\partial^2 F}{\partial \lambda \partial y} & \frac {\partial^2 F}{\partial \lambda^2} \end{vmatrix} = \begin{vmatrix} -2\lambda & 0 & -2x+4 \\ 0 & -2\lambda & -2y-6 \\ -2x+4 & -2y-6 & 0 \end{vmatrix} = \begin{vmatrix} -2\lambda & 0 & -\frac 1\lambda \\ 0 & -2\lambda & -\frac 1\lambda \\ -\frac 1\lambda & -\frac 1\lambda & 0 \end{vmatrix} = - \frac 4\lambda \end{aligned}$

$\begin{cases} x = 1 & y = 4(1)-11 = -7 & \lambda = - \frac 12 & x+4y = -27 & \text H = 2 > 0 & \text{minimum} \\ x = 3 & y = 4(3)-11 = 1 & \lambda = \frac 12 & x+4y = 7 & \text H = -2 < 0 & \text{maximum} \end{cases}$

Therefore, the required answer $7-27 = \boxed{-20}$ .

The given equation represents a circle with centre at $(2,-3)$ and radius $r$ .

Now , we will use polar coordinates on the circle which are

$(x,y)$ = $(2 + r\cos\theta , -3 + r\sin\theta)$ . And now we have to maximize and minimize the new $x + 4y$ ,which is given by $-10 + \sqrt{17r^{2}}$ and $-10 - \sqrt{17r^{2}}$ respectively.

Adding these two we get result as $\boxed{-20}$ .

$x^2+y^2+2(3y-2x)-4=0\Leftrightarrow (x-2)^2+(y+3)^2=17.\\ \text{So the condition is that both x and y are on this circle Center (2,-3).}\\ \text{For the circle the highest and the lowest points on it are above and below the center (2,-3).}\\ \text{So for both, x=2 on the above circle, }y_{max}=-3+ a, \ \ or\ \ y_{min}=-3 - a, \text{ a is some number .}\\ \therefore\ 4*y_{max}+2 \ + \ 4*y_{min}+2=\Huge\ \ \ \ \color{#D61F06}{20}.$

As the real solutions of $x^2+y^2-4x+6y-4=0 \Rightarrow (x-2)^2+(y+3)^2=17$ make a circle with radius $\sqrt{17}$ , so we have: $m=min(x+4y)=2-\sqrt{17}+4(-3-\sqrt{17})=-10-5\sqrt{17}$ $n=max(x+4y)=2+\sqrt{17}+4(-3+\sqrt{17})=-10+5\sqrt{17}$ So, summing up we find: $m+n=-20$

$From\quad the\quad equation,\quad y=\quad \pm \sqrt { 13-{ x }^{ 2 }+4x } -3\\ \therefore \quad f\left( x \right) =\quad x+4y=\quad \pm 4\sqrt { 13-{ x }^{ 2 }+4x } -12+x\\ \Rightarrow \quad f^{ ' }\left( x \right) =\quad \frac { \pm \left\{ 16-8x \right\} }{ 2\sqrt { 13-{ x }^{ 2 }+4x } } +\quad 1\quad \\ For\quad extrema,\quad f^{ ' }\left( x \right) =0\\ \Rightarrow \frac { \pm \left\{ 16-8x \right\} }{ 2\sqrt { 13-{ x }^{ 2 }+4x } } =-1\Rightarrow { x }^{ 2 }-4x+3=0\Rightarrow \quad x=1\quad or\quad 3\\ We\quad thus\quad consider\quad the\quad points:\\ (1,1)\longrightarrow \quad 5\\ (1,-7)\longrightarrow \quad \boxed { -27 } \quad Minima\\ (3,1)\longrightarrow \quad \boxed { 7 } \quad Maxima\\ (3,-7)\longrightarrow \quad -25\quad \\ \therefore \quad \boxed { -27\le f\left( x \right) \le 7 }$

Answer would be twice the coordinate of centre of the circle. Centre of circle is 2,-3. Hence answer is -20. If you are unable to understand draw the figure of circle and a variable line x+4 y=c. Then you would understand.