Find the absolute maximum and absolute minimum of $f(x,y)=x^{2}+y^{2}+4x-4y+3$ subject to $x^{2}+y^{2}\leq 2$.

Solution From the Extreme Value Theorem, $f$ has both an absolute maximum and an absolute minimum on the closed, bounded set $x^{2}+y^{2}\leq 2.$ We begin by finding any critical points. $$\begin{equation*} \begin{array}{rcl@{\qquad}rrcl} f_{x}=2x+4&=&0 & f_{y}=2y-4&=&0\\ x&=&-2 & y&=&2 \end{array}\end{equation*}$$

The point $( -2,2)$ lies outside the disk $x^{2}+y^{2}\leq 2,$ so $f$ has no relevant critical points. The extrema must occur on the boundary of the domain, that is, on the curve $x^{2}+y^{2}=2$. To find the extrema, we use Lagrange multipliers.

Step 1 Using $g(x,y)=x^{2}+y^{2}-2=0$ as the constraint, the problem is expressed in the required form.

Step 2 The functions $f$ and $g( x,y) =x^{2}+y^{2}-2=0$ each have continuous partial derivatives.

Step 3 We solve the system of equations ${\nabla }f( x,y) =\lambda {\nabla }g( x,y)$ and $g( x,y) =0.$ $$\left\{ \begin{array}{rcl@{\qquad}l} 2x+4&=&2\lambda x \qquad \enspace{\color{#0066A7}{\hbox{\(f_{x}(x,y) =\lambda g_{x}(x,y)\)}}}\\ 2y-4&=&2\lambda y \qquad \enspace{\color{#0066A7}{\hbox{\(f_{y}(x,y) =\lambda g_{y}(x,y)\)}}}\\ x^{2}+y^{2}&=&2 \qquad \enspace \enspace \enspace \enspace {\color{#0066A7}{\hbox{\(g(x,y) =0\)}}} \end{array}\right.$$

We eliminate $\lambda$ from the first two equations. $$\begin{eqnarray*} 2x+4&=&2x\lambda \qquad \enspace 2y-4&=&2y\lambda\\ \lambda &=& \dfrac{2x+4}{2x} & \lambda &=&\dfrac{2y-4}{2y}\\ \lambda &=& 1+\dfrac{2}{x} & \lambda &=& 1-\dfrac{2}{y} \end{eqnarray*}$$

Then $1+\dfrac{2}{x}=1-\dfrac{2}{y}$, so $y=-x.$

If we substitute $y=-x$ into the third equation $x^{2}+y^{2}=2,$ we find $2x^{2}=2.$ Then $x=\pm 1$ and $y=\mp 1.$ There are two test points: $(1,-1)$ and $(-1,1)$.

Step 4 We evaluate $z=f( x,y)$ at each test point. $$\begin{eqnarray*} f(1,-1)&=&13 \qquad {\color{#0066A7}{\hbox{Maximum value}}}\\ f(-1,1)&=&-3 \qquad {\color{#0066A7}{\hbox{Minimum value}}} \end{eqnarray*}$$