Why does the Lagrange multiplier $\lambda$ change when the equality constraint is scaled? Consider the problem 
$$\begin{array}{ll} \text{maximize} & x^2+y^2 \\ \text{subject to} & \dfrac{x^2}{25} + \dfrac{y^2}{9} = 1\end{array}$$
Solving this using the Lagrange multiplier method, I get
$$x = \pm5, \qquad y = 0, \qquad \lambda = 25$$
However, if I rewrite the constraint as $9x^2+25y^2=225$, I get a different value of $\lambda$, namely, $\lambda = \frac 19$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?
 A: If $x^\star$ minimizes $f(x)$ subject to the constraint that $g(x)=0$, then under mild assumptions there exists a Lagrange multiplier $\lambda$ that satisfies 
$$
\tag{1} \nabla f(x^\star) = \lambda \nabla g(x^\star).
$$ 
If $g$ is replaced with $c g$, then $x^\star$ is still a minimizer, but of course $\lambda$ no longer satisfies (1). We must correspondingly multiply $\lambda$ by $1/c$ in order for (1) to remain true.
Different but equivalent constraints have different Lagrange multipliers. The way you write the constraint matters.
A: Let $f(x,y)=x^2+y^2+\lambda(9x^2+25y^2-225).$
Thus, from $$\frac{\partial f}{\partial x}=2x+18\lambda x=0$$ and
$$\frac{\partial f}{\partial y}=2y+50\lambda y=0$$ we obtain two possibilities: $\lambda=-\frac{1}{9}$ or $\lambda=-\frac{1}{25}.$
The second gives a minimal value, wile the first gives a maximal value:
$$f(x,y)=x^2+y^2-\frac{1}{9}(9x^2+25y^2-225)=25-\frac{16}{9}y^2\leq25,$$ where the equality occurs for $y=0.$
If we consider $f(x,y)=x^2+y^2+\lambda\left(\frac{x^2}{25}+\frac{y^2}{9}-1\right)$ so we'll get $\lambda=-25$ 
and we'll get the same answer of course.
It happens because $-\frac{1}{9}\cdot225=-25$ and
$$x^2+y^2-\frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25\left(\frac{x^2}{25}+\frac{y^2}{9}-1\right).$$
A: Writing the equation $$\frac{x^2}{25}+\frac{y^2}{9}=1$$  in the form
$$y^2=9-\frac{9}{25}x^2$$ you will have the objective function
$$f(x)=\frac{16}{25}x^2+9$$
