Can't find minimum using Lagrange multipliers

Question

I want to find the minimum of the function $f(x,y) = x + y^2$ with the constraint $2x^2 +y^2 = 1$.

Here are my partial derivatives:

$$f_x = 1$$ $$f_y = 2y$$ $$g_x = 4x$$ $$g_y = 2y$$

I have the following system of equations: \begin{align*} 1 = \lambda4x\\ 2y = 2y\lambda\\ 2x^2 + y^2 = 1 \end{align*}

and I found that $$\lambda = 1,\ x = 1/4,\ \text{and }y = +-(7/8)^{1/2}.$$ And Wolfram Alpha shows that the above is the maximum value, and that the minimum value is $$ x= -1/(2)^{1/2}\text{ and }y = 0.$$

How can I find this value? I missed something.

Answer 1

You correctly got the three equations for the Lagrange method. The 2nd equation tells you that $\lambda=1$ or $y=0$.

$\lambda=1$ gives you the stationary point you identified, which has $x=\frac{1}{4},y^2=\frac{7}{8}$ and hence $f(x,y)=\frac{9}{8}$.

$y=0$ gives $x^2=\frac{1}{2}$ and hence $x=\pm\frac{1}{\sqrt2}$ and $f(\pm\frac{1}{\sqrt2},0)=\pm\frac{1}{\sqrt2}$.

Note that the constraint limits $(x,y)$ to an ellipse which is a closed, bounded region, so one of these points must be the minimum. Hence the minimum is $-\frac{1}{\sqrt2}$.

Answer 2

To find the extreme values, you need to also check the conditions where $f_x=0$ and $f_y=0$, which is the way to find local extremas before using Lagrange.
In this case you can find through $f_y=0$, that $y=0$

Answer 3

There's no need to use a Lagrange multiplier. The ellipse

$$\{ (x,y) \in \mathbb{R}^2 \mid 2 x^2 + y^2 = 1 \}$$

is parametrized as follows

$$x (\theta) = \frac{\sqrt 2}{2} \, \cos (\theta) \qquad \qquad \qquad y (\theta) = \sin (\theta)$$

Hence,

$$g (\theta) := f (x (\theta), y (\theta)) = \frac{\sqrt 2}{2} \, \cos (\theta) + \sin^2 (\theta)$$

Differentiating,

$$g' (\theta) = -\frac{\sqrt 2}{2} \, \sin (\theta) + 2 \sin (\theta) \cos (\theta) = \left( 2 \cos (\theta) - \frac{\sqrt 2}{2} \right) \sin (\theta)$$

which vanishes when $\sin (\theta) = 0$ or $\cos (\theta) = \frac{\sqrt 2}{4}$. If $\sin (\theta) = 0$, then

$$x = \pm \frac{\sqrt 2}{2} \qquad \qquad \qquad y = 0 \qquad \qquad \qquad f = \pm \frac{\sqrt 2}{2}$$

If $\cos (\theta) = \frac{\sqrt 2}{4}$, then

$$x = \frac{1}{4} \qquad \qquad \qquad y = \pm \sqrt{\frac{7}{8}} \qquad \qquad \qquad f = \frac{9}{8}$$

Thus, the minimum is $- \frac{\sqrt 2}{2}$, which is attained at $(x,y) = \left(- \frac{\sqrt 2}{2}, 0\right)$.