Find the absolute minimum of the function $ f(x,y) = x^2+y^2 $ subject to the constraint $x^2+ 2y^2 = 1 $.

Question

I tried this question on a site and found that none of the answers were the ones I got. I think my calculations were right but my understanding of applying the correct theorems may have been off.

I found the derivatives with respect to x and y and got:

$f_x = 2x \quad$ $f_{xx} = 2 \quad $ $ f_{y} = 2y \quad $ $ f_{yy}=2 \quad $ $ f_{xy} = 0 $

The critical values are when $ 2x = 2y $ which is the line x = y. the fancy formula:

$ D = f_{xx}*f_{yy} - (f_{xy})^2 $ gives us $D = 4$ always. So long as the point is a critical point, its a relative min. Since it is subject to the constraint, we need the values which satisfy $x^2+ 2y^2 = 1 $. Those are ones where $ x=y$ so we look at $x^2+ 2x^2 = 1 \implies 3x^2= 1 \implies x = \pm1/\sqrt{3}$. Either works so when I plug it in to the original function, since $ x=y$, I get $ 2/3 $. Thus my absolute minimum.

Is my reasoning correct? or am I missing tools or using the tools incorrectly from multivariable Calculus?

Answer 1

As we have

$x^2+2y^2=1$

we can put

$$x=\cos(t)$$

and

$$y=\frac{\sin(t)}{\sqrt{2}}$$

thus

$x^2+y^2=\frac{1}{2}(1+\cos^2(t))$

so, the absolute minimum is

$$\frac{1}{2}$$

Answer 2

From our constraint $x^2+2y^2=1$ we have $x^2=1-2y^2$ hence we can rewrite our function in terms of $y$:

$$f(y)=1-2y^2+y^2=1-y^2$$

With the constraint $y \in [-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}]$ because $x^2 \geq 0$ if $x \in \mathbb{R}$.

It should be clear that $f$ attains an absolute maximum at $1$ when $y=0$, and has absolute minimums of $\frac{1}{2}$ when $y=\pm \frac{\sqrt{2}}{2}$.

Answer 3

To do the Lagrange method let $$F=x^2+y^2+\lambda(x^2+2y^2-1)$$

Then $$F_x=(1+\lambda)x=0$$ $$F_y=(1+2\lambda)y=0$$ $$F_{\lambda}=x^2+2y^2-1=0$$

Thus $x=0$ or $y=0$ the first gives the minimum, the second the maximum.