Find min & max of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$ Problem: Find the maximum and minimal value of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$.
Since $x^2 > x$ (edit $x^2 \geq x$) for all $x \in \mathbb{R}$, $f$ is bowl-ish with a minimal value in the bottom.
This is a critical point which means that we can set partial derivatives of $f$ equal to $0$ and try to solve for $x$ and $y$ 
$\nabla f = (1+2x, 1+2y) = (0,0) \implies (x,y) = (\frac{-1}{2}, \frac{-1}{2})$
So we get the minimal value $f(\frac{-1}{2}, \frac{-1}{2}) = \frac{-1}{2} + \frac{-1}{2} + (\frac{-1}{2})^2 \frac{-1}{2})^2 = -\frac{1}{2}$
But how about the maximal value? How does $x^2 + y^2 = 1$ restrict $f$? 

$x^2 + y^2 =1$.  So $f(x,y) = x + y + 1$
Now consider $x + y = a ,$ Then  $y = x - a$
$x^2 + y^2 = 1$ So  $x^2 + (x-a)^2 = 1$
$2 x^2 -2 ax + a^2 -1 = 0$
The determinant must be non negative :  $4^2 - 8 a^2 + 8 >= 0 : a^2 <=2$ 
Which means maximum of $a$ is$\displaystyle \sqrt{2}$  and minimum is $\displaystyle -\sqrt{2}$
So $f(x,y)_{\bf{Min.}} = 1-\sqrt{2}$ and $f(x,y)_{\bf{Max.}} = 1+\sqrt{2}$
