# Max and min with Lagrange multipliers, question about getting a single extrema point back

I just did a problem with Lagrange multipliers and found the two extrema points, which gave the same results which according to the book were minimums. It was $f(x,y)=x^2+y^2$ with constraint $xy=1$ with mins $(1,1)=2$ and $(-1,-1)=2$. How do I know to call these minimums instead of maximums? Are they just always minimums if they both equal the same thing or there is just one point? Or could they have been maximums and there's some way to tell which?

• Plug a random point say $(1/2, 2)$ into the function and compare with the value at the extreme point. Oct 19, 2016 at 4:07
• Hmm Alright. Thanks! Oct 19, 2016 at 4:47

Suppose the same problem with contraint $xy=a\implies y=\frac a x$; so the function is $$f(x)=x^2+\frac{a^2}{x^2}\implies f'(x)=2 x-\frac{2 a^2}{x^3} \implies f''(x)=2+\frac{6 a^2}{x^4}>0$$ So,if $a>0$, in the real domain the derivative cancels for $x=\pm \sqrt a$ and $f(\pm \sqrt a)=1+a^2$. The second derivative being always positive, then two minimum points.