The function $f(x, y, z) = x^2 + y^2$ is constrained by $5x^2 + 6xy + 5y^2 = 1$. Using Lagrange, maximise and minimise this function.
My original Lagrange equation was
$$L = x^2 + y^2 + \lambda(1 – 5x^2 – 6xy – 5y^2).$$
After partially differentiating I got:
$$\begin{aligned}
L_x &= 2x – 10\lambda x – 6\lambda y ,\\
L_y &= 2y – 6\lambda x – 10\lambda y , \\
L_\lambda &= 1 – 5x^2 – 6xy – 5y^2 .
\end{aligned}$$
As I’m attempting to find a minimum and maximum, I set all the equations equal to $0$. Rearranging the partial derivative w.r.t. $x$ gave me
$$\lambda = \frac{2x}{10x + 6y}$$
and rearranging the partial derivative w.r.t. $y$ gave me $\lambda = 2y/(10y + 6x)$. Putting these two equations equal to one another gives $$\frac{2y}{10y + 6x} = \frac{2x}{10x +6y}.$$ Multiplying both sides out led to $20xy + 12y^2 = 20xy + 12x^2$. As such I could equate $x$ to $y$.
Subbing in $x$ for $y$ into the constraint gave me $16x^2 = 1$ and as such $x$ (and $y$) $= \pm \frac14$. However, I don’t think either $\left(\frac14, \frac14\right)$ or $\left(-\frac14, -\frac14\right)$ are the minimum and maximum points. If I have done something wrong please correct me but as of right now I am unsure what to do. Also, $z$ being listed in the function but actually having no part in any of the question confused me too. Was just hoping someone could help me understand.
Best Answer
I think that I can safely assume that $f$ is a function of $2$ variables, rather than $3$.
Let $g(x,y)=5x^2+6xy+5y^2$. Note that $$ E=\{(x,y)\in\mathbb{R}^2\,:\,g(x,y)=1\} $$ is a compact set (it is an ellipse), and therefore the restriction of $f$ to that set must have a maximum and a minimum. The points at which they are attained are such that $\nabla g=(0,0)$ or that, for some $\lambda\in\mathbb{R}$, $(x,y,\lambda)$ is a solution of the system
$$ \left\{\begin{array}{l} 2\bigl((1-5\lambda)x-3\lambda y\bigr)=0\\ 2\bigl(-3\lambda x+(1-5\lambda)y\bigr)=0\\ 5x^2+6xy+5y^2=1. \end{array}\right.\label{sys}\tag{1} $$
But $\nabla g(x,y)=(10x+6y,6x+10y)$, which is $(0,0)$ only at the origin, which does not belong to $E$. On the other hand, the first two equations of the system \eqref{sys} form a system of two linear equations in two variables depending upon a parameter $\lambda$. The matrix of the coefficients of this system is $$ \begin{pmatrix} 1-5\lambda&-3\\ -3&1-5\lambda \end{pmatrix}, $$ whose determinant is $0$ if and only if $\lambda=\frac12$ or $\lambda=\frac18$. If the determinant is not $0$, then the only solution of the system which consists of the first two equation is $x=y=0$, which is not a solution of the third equation.
Now, suppose that $\lambda=\frac12$. Then the system becomes $$ \left\{\begin{array}{l} -\frac32x-\frac32y=0(\iff x+y=0)\\ 5x^2+6xy+5y^2=1, \end{array}\right. $$ whose only solutions are $\pm\left(\frac12,-\frac12\right)$. And if $\lambda=\frac18$, then the system becomes $$ \left\{\begin{array}{l} \frac38x-\frac38y=0(\iff x-y=0)\\ 5x^2+6xy+5y^2=1, \end{array}\right. $$ whose only solutions are $\pm\left(\frac14,\frac14\right)$. But $f\left(\pm\left(\frac12,-\frac12\right)\right)=\frac12$ and $f\left(\pm\left(\frac14,\frac14\right)\right)=\frac18$. So, the maximum is $\frac12$ and the minimum is $\frac18$.