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. \[ \begin{eqnarray*} f(1,-1)&=&13 \qquad {\color{#0066A7}{\hbox{Maximum value}}}\\ f(-1,1)&=&-3 \qquad {\color{#0066A7}{\hbox{Minimum value}}} \end{eqnarray*} \]