Simple Lagrange Multiplyers Problem

Question

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.

Answer 1

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.

Answer 2

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.