Finding Absolute Extrema

Find the absolute maximum and the absolute minimum of \[ \begin{equation*} z=f( x,y) =x^{2}+y^{2}-2x+2y-5 \end{equation*} \]

whose domain is the disk \(x^{2}+y^{2}\leq 9\).

Solution  The function \(f\) is continuous on its domain, a closed, bounded set. The Extreme Value Theorem guarantees that \(f\) has an absolute maximum and an absolute minimum on its domain. To find them, we find the critical points of \(f,\) namely the solutions of the system of equations \[ \left\{\begin{array}{l} f_{x}( x,y) =2x-2=0\\[3pt] f_{y}( x,y) =2y+2=0 \end{array}\right. \]

The only solution is \(x=1,\) \(y=-1,\) which is an interior point of the domain of \(f.\) The value of \(f\) at the critical point is \(f(1,-1) =-7\).

The boundary of the domain of \(f\) is the circle \(x^{2}+y^{2}=9.\) We express the boundary using the parametric equations \(x=x(t) =3\,\cos t\), \(y=y(t) =3\,\sin t,\) \(0\leq t\leq 2\pi .\) Next we evaluate \(f\) on its boundary. \[ \begin{eqnarray*} z=f( x,y) &=&f( 3\,\cos t,3\,\sin t)\\[4pt] &=&9\,\cos ^{2}t+9\,\sin^{2}t-6\,\cos t+6\,\sin t-5\\[4pt] &=&4-6\,\cos t+6\,\sin t \end{eqnarray*} \]

Now we find where \(\dfrac{dz}{dt}=0\). \[ \begin{eqnarray*} \dfrac{dz}{dt}&= & 6\,\sin t+6\,\cos t=0 \\[4pt] \tan t&=&-1 \\[4pt] t&=&\dfrac{3\pi }{4}\qquad\hbox{or}\qquad t=\dfrac{7\pi }{4} \end{eqnarray*} \]

At \(t=\dfrac{3\pi }{4},\) the value of \(f\) is \[ \begin{eqnarray*} f\left( 3\,\cos \dfrac{3\pi }{4},3\,\sin \dfrac{3\pi }{4}\right) &=&4-6\,\cos \dfrac{ 3\pi }{4}+6\,\sin \dfrac{3\pi }{4}=4-6\left( -\dfrac{\sqrt{2}}{2}\right) +6\left( \dfrac{\sqrt{2}}{2}\right)\\[4pt] &=&4+6\sqrt{2} \end{eqnarray*} \]

At \(t=\dfrac{7\pi }{4},\) the value of \(f\) is \[ \begin{eqnarray*} f\left( 3\,\cos \dfrac{7\pi }{4},3\,\sin \dfrac{7\pi }{4}\right) &=&4-6\,\cos \dfrac{ 7\pi }{4}+6\,\sin \dfrac{7\pi }{4}=4-6\left( \dfrac{\sqrt{2}}{2}\right) +6\left( -\dfrac{\sqrt{2}}{2}\right)\\[4pt] &=&4-6\sqrt{2} \end{eqnarray*} \]

886

The absolute maximum value of \(f\) is \(4+6\sqrt{2}\approx 12. 485;\) the absolute minimum value of \(f\) is \(-7.\)