If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$. If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$.
I did a little bit of manipulation and got $4(x+1)(x-2) + (y+1)^2=0$. I then got $x=2$ and $y=-1$ which means the maximum must be $4$, but the answer key says it's $16$. How come?
 A: $$4x^2-4x + 1+y^2 + 2y +1 =9\implies (2x-1)^2+(y+1)^2=9$$
So a pair $(x,y)$ is on elipse: $${(x-{1\over 2})^2\over {9\over 4}}+{(y+1)^2\over 9}=1$$ so you can write $x= {3\over 2}\cos t +{1\over 2}$ and $y= 3\sin t -1$ 
So your expresion is now $${15\over 2}\cos t +18\sin t -{7\over 2}$$ 
This you can write as $${39\over 2}\sin (t+\varphi) -{7\over 2}$$ for some phase $\varphi$. So the maximum is when $\sin (t+\varphi)=1$ and it is $16$ and minimum is when $\sin (t+\varphi)=-1$ and it is $-23$.
A: Let $5x+6y=k$.
Thus, $y=\frac{k-5x}{6}$ and the following equation has real roots:
$$4x^2+\left(\frac{k-5x}{6}\right)^2=7+4x-2\cdot\frac{k-5x}{6}$$ or
$$165x^2-2(5k+102)x+k^2+12k-252=0,$$ which gives
$$(5k+102)^2-169(k^2+12k-252)\geq0$$ or
$$(k-16)(k+23)\leq0$$ or
$$-23\leq k\leq16.$$
For $k=16$ the equality occurs for
$$x=\frac{5\cdot16+102}{169}=\frac{14}{13}$$ and from here $$y=\frac{23}{13},$$ which says that we got a maximal value.
A: We can rewrite the equation as follows:
$$4x^2 - 4x + y^2 + 2y - 7 = 0$$
By completing the squares, we obtain:
$$4\left (x^2 - x + \color{red}{\frac 1 4} \right ) + (y^2 + 2y + \color{blue}1) - 7 - \color{red}1 - \color{blue}1 = 0$$
that is,
$$\frac {\left (x - \frac 1 2 \right )^2} {\left ( \frac 3 2 \right )^2} + \frac {(y + 1)^2} {3^2} = 1$$
which represents the ellipse with center $(\frac 1 2, -1)$ and semiaxes $\frac 3 2, 3$.
For a given $k \in \mathbb R$, the equation $5x + 6y = k$ represents a straight line parallel to the line of equation $5x + 6y = 0$. We want to find the greatest $k$ such that the line $5x + 6y = k$ intersects the ellipse at a point $(x, y)$.
As $k$ increases, the line will intersect the ellipse in either $0$, $1$ or $2$ points (see the picture below). It should be clear that the greatest $k$ is such that there is only one point of intersection. Therefore, we want to find $k$ such that the system
$$\begin{cases} 5x + 6y = k \\ 4x^2 - 4x + y^2 + 2y - 7 = 0 \end{cases}$$
has exactly one solution.
By substituting $y = -\frac 5 6 x + \frac 1 6 k$ in the second equation, we get
$$169 x^2 - (10k + 204) x + (k^2 + 12 k - 252) = 0$$
which has exactly one solution if and only if its discriminant with respect to $x$ is $0$:
$$\Delta = (10 k + 204)^2 - 4 \cdot 169 \cdot (k^2 + 12 k - 252) = 0$$
This last equation has solutions $k = -23$ and $k = 16$.
Therefore, the maximum value of $5x + 6y$ is $16$.
The solution is represented in the following picture:

A: You've observed that $x\in[-1,2]$, but your mistake is that the maximum need not be at one of the endpoints. Once you know that 
$$4(x+1)(x-2) + (y+1)^2=0,$$
you see that any solution $(x,y)$ must have $-1\leq x\leq 2$. Then
$$y=-1\pm2\sqrt{-(x+1)(x-2)},$$
and so you want to find the maximum of
$$5x+6y=5x-1\pm2\sqrt{-(x+1)(x-2)},$$
for $x\in[-1,2]$, which I'm sure you can do by standard methods.
A: Also, we can use C-S here:
$$19.5=\sqrt{6.5^2\cdot9}=\sqrt{(2.5^2+6^2)((2x-1)^2+(y+1)^2)}\geq$$
$$\geq2.5(2x-1)+6(y+1)=5x+6y+3.5,$$
which gives $$5x+6y\leq16.$$
The equality occurs for $$(2.5,6)||(2x-1,y+1),$$ which says that $16$ is a maximal value.
A: This can be solved as a tangency problem.
Given the non degenerated conic $C(x,y) = 0$ determine the value for $\lambda$ such that $a x + b y = \lambda$ is tangent to $C(x,y)=0$ so we proceed as follows:
first we substitute $y = \frac 1b(\lambda-a x)$ into $C(x,y) = 0$ giving the relationship
$$
C\left(x, \frac 1b(\lambda-a x)\right)=0
$$
second, we solve for $x$ giving the condition
$$
x^* = \frac{1}{169} \left(102\pm12 \sqrt{368-\lambda^2-7 \lambda}+5 \lambda\right)
$$
but at tangency $x^*$ is unique so follows the condition
$$
368-\lambda^2-7 \lambda = 0
$$
which solved gives
$$
\lambda^* = \{-23, 16\}
$$
hence the answer is $\lambda^* =16$
