610158…: “Opposite of the polite function”

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Use Lagrange multipliers to find the absolute maximum and minimum of $F(x,y) = x^2+2x+y^2+2y$ constrained to the circle $x^2+y^2=4$ . Start by setting $G(x,y) = \answer {x^2+y^2-4}$ .

Now compute: $\begin{align*} \grad F(x,y) &=\vector{\answer{2x+2},\answer{2y+2}}\\ \grad G(x,y) &=\vector{\answer{2x},\answer{2y}} \end{align*}$

Now solve the following system of equations for $x$ , $y$ , and $\lambda$ : $\begin{align*} \grad F(x,y) &= \lambda \cdot \grad G(x,y)\\ x^2+y^2 &= 4 \end{align*}$

The solutions will be the extrema (unless there are corners that need to be checked too). Write the extrema in order of increasing $x$ -values.

$\begin{align*} \vec{e}_1 &= \vector{-\sqrt{2},-\sqrt{2}}\\ \vec{e}_2 &= \vector{\sqrt{2},\sqrt{2}} \end{align*}$

The maximum value for $F$ when constrained to the set above is $\answer {4+4\sqrt {2}}$ . The minimum value for $F$ when constrained to the set above is $\answer {4-4\sqrt {2}}$ .