Simple Lagrange Multiplyers Problem Can anyone please help me with the following:
Find the stationary values of $u=x^2+y^2$ subject to the constraint $t(x,y)  = 4x^2 + 5xy + 3y^2 = 9$.
The answer is given as $u = 9$ and $x = \pm 3/\sqrt2$ and $y = \mp 3/\sqrt2$
I do the following:
Let $g(x,y) = 4x^2 + 5xy + 3y^2 - 9 = 0$
$$\begin{align*}
\partial u/\partial x &= 2x \\
\partial u/\partial y &= 2y \\
\partial g/\partial x &= 8x + 5y \\
\partial g/\partial y &= 5x + 6y
\end{align*}$$
So solve the system:
$$\begin{align*}
2x + k(8x + 5y) &= 0\\
2y + k(5x + 6y) &= 0\\
4x^2 + 5xy + 3y^2 - 9 &= 0
\end{align*}$$
And this system does not solve to give the stated answer.
Where have I erred, or is the book answer (or question?) wrong?
Thanks,
Mitch.
 A: I haven't worked out a numerical answer by hand:
if I eliminate k from:
$$\begin{align*}
2x + k(8x + 5y) &= 0\\
2y + k(5x + 6y) &= 0\\
\end{align*}$$
I get:
$$\begin{align*}
2y - \frac{2x(5x+6y)}{8x+5y} &= 0\\
\end{align*}$$
And if one substitutes in
$$\begin{align*}
x=\frac{3}{\sqrt{2}}\\
y=\frac{-3}{\sqrt{2}}
\end{align*}$$
One would expect to get zero (if the supplied answers are correct) but one gets
$$\begin{align*}
-2\sqrt{2}
\end{align*}$$
Mitch.
A: I do believe your book answer is wrong, or the question has a typo. 
Your set up is correct. From my habit, I used $-\lambda$ in the following, but it should not give any difference:
\begin{align*}
2x -\lambda (8x + 5y) &= 0\\
2y - \lambda (5x + 6y) &= 0\\
4x^2 + 5xy + 3y^2 - 9 &= 0
\end{align*}
So from 1, $\lambda = \frac{2x}{8x+5y}$. Plug this into 2 gives
$$2y-\frac{2x}{8x+5y}(5x+6y)=0\implies 5x^2-2xy-5y^2=0$$
You can easily check that the denominator cannot be zero to eliminate that case.
Now this is a quadratic equation, we can solve it by quadratic formula with $y$ in the solution. This gives us
$$(x-\frac{1+\sqrt{26}}{5}y)(x-\frac{1-\sqrt{26}}{5}y)=0$$ 
Plug this into 3 and simplify:
$$y^2=\frac{225}{208+33\sqrt{26}}$$
I put square root of this into wolframalpha and it doesn't give me a good number. You can find $x$ value from this. There must be a typo somewhere but this is one way to solve this type of problem.
