Max/Min Values of $f(x,y)=xy$ subject to $g(x,y)=x^2+2y^2-1$

Question

Find the maximun and minimum values of the function $f(x,y)=xy$ on the ellipse $g(x,y)=x^2+2y^2-1$.

I realised this problem relied on the use of Legrange Multipliers. So i let $\nabla f=t\nabla g$. Solving this, i arrived at $y=\pm \frac{1}{\sqrt{2}}$. I thought of substituting this in for $y$ to get the corresponding $x$ values, but this would not give me the correct max/min after substituting these points into $f(x,y)$. I solution to this problem would be very helpful.

Answer 1

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

Since $f$ and $g$ are functions of $2$ variables, we can rewrite this as

$$\frac{f_x}{g_x}=\frac{f_y}{g_y}$$

$$\frac{y}{2x}=\frac{x}{4y}$$

Then solving for $x$

$$2x^2=4y^2$$

$$x=\pm y\sqrt{2}$$

subject to

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

$x$ is squared so the result will be the same if we use $x$ or $-x$, but the points are still there

$$4y^2=1$$

$$y=\pm\frac{1}{2}$$

Then the critical points are $(\frac{\sqrt{2}}{2},\frac{1}{2}),(\frac{\sqrt{2}}{2},\frac{-1}{2}),(\frac{-\sqrt{2}}{2},\frac{1}{2}),(\frac{-\sqrt{2}}{2},\frac{-1}{2})$

All of which only differ by sign, and $f(x,y) = xy$

Then the maxima are $\left\{(\frac{\sqrt{2}}{2},\frac{1}{2}),(\frac{-\sqrt{2}}{2},\frac{-1}{2})\right\}$

and the minima are $\left\{(-\frac{\sqrt{2}}{2},\frac{1}{2}),(\frac{\sqrt{2}}{2},\frac{-1}{2})\right\}$

Answer 2

Alt. hint:  by the RMS-GM means inequality the following holds, with equality iff $\,|x|=\sqrt{2}\,|y|\,$:

$$ \sqrt{|x| \cdot \sqrt{2} |y|} \le \sqrt{\frac{x^2+2y^2}{2}} = \frac{1}{\sqrt{2}} \;\;\iff\;\; |xy| \le \frac{1}{2 \sqrt{2}} $$