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?

$$4x^2-4x + 1+y^2 + 2y +1 =9\implies (2x-1)^2+(y+1)^2=9$$
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 $-23$.
