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 $f$? 
 A: Your method to get the minimum gets in the entire plane, but you want the minimum on the unit circle $x^2+y^2=1$.
You can use Lagrange multipliers. You could also parameterize the unit circle on one variable then find the extrema on that variable. Two parametrizations are
$$x=\cos\theta,\quad y=\sin\theta,\quad 0\le\theta<2\pi$$
or
$$y=\pm\sqrt{1-x^2},\quad -1\le x\le 1$$
In your problem it looks like the cosine/sine parametrization would be easier.
A: If $x^2 + y^2 = 1$ then
$$
f(x,y) = 1 + x + y.
$$
Now, let $x=\cos t$, $y=\sin t$. So,
$$
f(t) = 1 + \cos t + \sin t = 1 + \sqrt2 \sin(t + \pi/4).
$$
Could you proceed?
A: Use the lagrange multiplier !
minimize the function:
 $g(x,y,\lambda) = f(x,y)+\lambda(x^2+y^2-1)$
A: By $MA\geq MG$ we have $2xy\leq x^2+y^2=1$. Note that $(x+y)^2-(x^2+y^2)=2xy$ and this implies that $(x+y)^2\leq 2$ . So, $-\sqrt{2}\leq x+y\leq \sqrt{2}$. Then $f(x,y)\leq 1+\sqrt{2}$ with equality for $x=y=\frac{\sqrt{2}}{2}$ and $f(x,y)\geq 1-\sqrt{2}$ with equality for $x=y=-\frac{\sqrt{2}}{2}$.
A: $x^2 + y^2 =1$.  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 $f(x,y)_{\bf{Max.}} = 1+\sqrt{2}$
A: $$2(x^2+y^2)-(x+y)^2=(x-y)^2\ge0\iff(x+y)^2\le2(x^2+y^2)=?$$
A: Think of a unit circle centred at the origin, and consider another circle centred at $(-\frac12,-\frac12)$. Expand the latter circle from a small size until it just touches the unit circle, at $(-\frac12\sqrt2,-\frac12\sqrt2)$. Then expand it further until it just touches the unit circle again, at $(\frac12\sqrt2,\frac12\sqrt2)$. From this, the corresponding values of $(x+\frac12)^2+(y+\frac12)^2-\frac12$ are $1\pm\sqrt2$. 
