# Why does the method of Lagrange multiplier fail here?

$$\begin{array}{ll} \text{extremize} & f(x,y) := 2 y \sqrt{y^2 - x^2}\\ \text{subject to} & \pi x^2 y = 450\end{array}$$

WolframAlpha says that the minimum is $$0$$, but the Lagrange multipliers method fails. The Lagrangian is

$$L = 2y\sqrt{y^2-x^2}+\lambda\left(\pi x^2y-450\right)$$

and

\begin{aligned} \frac{\partial L}{\partial x} &= -\frac{2xy}{\sqrt{y^2-x^2}}+2\pi\lambda xy\\ \frac{\partial L}{\partial y} &= 2\sqrt{y^2-x^2}+\frac{2y^2}{\sqrt{y^2-x^2}}+\pi\lambda x^2\\ \frac{\partial L}{\partial \lambda} &= \pi x^2y-450 \end{aligned}

This system only has solutions, where the function is undefined.

• You should show where it fails according to your work on that.
– user
Sep 6, 2021 at 18:51
• What happens with the Lagrange multipliers method? Please share your work. Sep 6, 2021 at 18:51
• What's the domain of definition of $f$? Sep 6, 2021 at 19:21
• It is $y^2 \ge x^2$. Sep 6, 2021 at 20:01
• @Andrew Fount Notice Lagrange multipliers are a valid method for finding maxima in an open set. Is your domain open? Sep 6, 2021 at 20:51

Lagrange multipliers are useful for exploring open sets. In this case, your domain $$y^2 \geq x^2$$ is not open. Therefore the boundary $$y^2 = x^2$$ should be explored on its own. In those points you get: $$f(x,y) = 0$$ which is indeed the minimum. To find which point you want substitute in your extra equation $$x$$ by $$y^2$$

Given the lagrangian

$$L(x,y,\lambda) = f(x,y)+\lambda g(x,y)$$

the stationary solutions are the solutions to

$$\nabla f(x,y) +\lambda \nabla g(x,y) = 0$$

so at the solutions we should have $$\nabla f(x,y)\times \nabla g(x,y)=\vec 0$$. In our case

$$\cases{ \nabla f(x,y) = \left\{-\frac{2 x y}{\sqrt{y^2-x^2}},-\frac{2 \left(x^2-2 y^2\right)}{\sqrt{y^2-x^2}},0\right\}\\ \nabla g(x,y) = \left\{2 \pi x y,\pi x^2,0\right\} }$$

and $$\nabla f(x,y)\times \nabla g(x,y)=\left\{0,0,\frac{2 \pi x y \left(x^2-4 y^2\right)}{\sqrt{y^2-x^2}}\right\}$$ now to have $$-\epsilon\le \frac{2 \pi x y \left(x^2-4 y^2\right)}{\sqrt{y^2-x^2}}\le\epsilon$$. Making substitutions $$x\to r\cos\theta, y\to r\sin\theta$$ we have

$$-\epsilon\le \frac{\pi r^3 \sin (2 \theta ) (5 \cos (2 \theta )-3)}{2 \sqrt{-\cos (2 \theta )}}\le \epsilon$$

so to have small $$\epsilon$$ suffices a small $$r$$ giving small $$x$$ and small $$y$$ but then $$g(x,y) = \pi x^2y-450=0$$ can't be verified. Resuming, no real solutions to the stationary points determination.