Finding the maximum value of a function on an ellipse Let $x$ and $y$ be real numbers such that $x^2 + 9 y^2-4 x+6 y+4=0$. Find the maximum value of $\displaystyle \frac{4x-9y}{2}$.
My solution: the given function represents an ellipse. Rewriting it, we get $\displaystyle (x-2)^2 + 9(y+\frac{1}{3})^2=1$. To find the maximum of $\displaystyle \frac{4x-9y}{2}$, $x$ should be at its maximum and $y$ at its minimum. Solving the equations, I get that $x = 3$ and $\displaystyle y = -\frac{2}{3}$, but the answer i get is wrong. What am I doing wrong?

HINT : Letting $k=\frac{4x-9y}{2}$, this represents a line $y=\frac{4}{9}x-\frac{2}{9}k.$
Then, find the max $k$ such that the line is tangent to the ellipse.
You can set $y=\frac{4}{9}x-\frac{2}{9}k$ in the equation of the ellipse, then you'll get an equation of $x$. Then, find the condition that the equation has only one real solution. This gives you two $ks$. Then, the bigger $k$ is what you want.
