2.9: Maximum And Minimum Values - Mathematics LibreTexts

We are putting quotation marks around the word “Proof”, because we are not going to justify the fact that it suffices to analyse the quadratic approximation in equation \((*)\text{.}\) Let's temporarily suppress the arguments \((a,b)\text{.}\) If \(f_{xx}(a,b)\ne 0\text{,}\) then by completing the square we can write

\[\begin{align*} &f_{xx}\,\Delta x^2 +2f_{xy}\,\Delta x\Delta y + f_{yy}\,\Delta y^2\\ &\hskip0.5in=f_{xx}\left(\Delta x +\frac{f_{xy}}{f_{xx}}\Delta y\right)^2 +\left(f_{yy}-\frac{f_{xy}^2}{f_{xx}}\right)\,\Delta y^2\\ &\hskip0.5in=\frac{1}{f_{xx}}\left\{\big(f_{xx}\,\Delta x+f_{xy}\,\Delta y\big)^2 +\big(f_{xx}f_{yy}-f_{xy}^2\big)\,\Delta y^2\right\} \end{align*}\]

Similarly, if \(f_{yy}(a,b)\ne 0\text{,}\)

\[\begin{align*} &f_{xx}\,\Delta x^2 +2f_{xy}\,\Delta x\Delta y + f_{yy}\,\Delta y^2\\ &\hskip0.5in=\frac{1}{f_{yy}}\left\{\big(f_{xy}\,\Delta x+f_{yy}\,\Delta y\big)^2 +\big(f_{xx}f_{yy}-f_{xy}^2\big)\,\Delta x^2\right\} \end{align*}\]

Note that this algebra breaks down if \(f_{xx}(a,b)=f_{yy}(a,b)=0\text{.}\) We'll deal with that case shortly. More importantly, note that

• if \(\big(f_{xx}f_{yy}-f_{xy}^2\big) \gt 0\) then both \(f_{xx}\) and \(f_{yy}\) must be nonzero and of the same sign and furthermore, whenever \(\Delta x\) or \(\Delta y\) are nonzero,

  \[\begin{align*} \left\{\big(f_{xx}\,\Delta x+f_{xy}\,\Delta y\big)^2 +\big(f_{xx}f_{yy}-f_{xy}^2\big)\,\Delta y^2\right\} & \gt 0\quad\text{and}\\ \left\{\big(f_{xy}\,\Delta x+f_{yy}\,\Delta y\big)^2 +\big(f_{xx}f_{yy}-f_{xy}^2\big)\,\Delta x^2\right\} & \gt 0 \end{align*}\]

  so that, recalling \((*)\text{,}\)

  • if \(f_{xx}(a,b) \gt 0\text{,}\) then \((a,b)\) is a local minimum and
  • if \(f_{xx}(a,b) \lt 0\text{,}\) then \((a,b)\) is a local maximum.
• If \(\big(f_{xx}f_{yy}-f_{xy}^2\big) \lt 0\) and \(f_{xx}\) is nonzero then

  \[ \left\{\big(f_{xx}\,\Delta x+f_{xy}\,\Delta y\big)^2 +\big(f_{xx}f_{yy}-f_{xy}^2\big)\,\Delta y^2\right\} \nonumber \]

  is strictly positive whenever \(\Delta x\ne 0\text{,}\) \(\Delta y= 0\) and is strictly negative whenever \(f_{xx}\,\Delta x+f_{xy}\,\Delta y=0\text{,}\) \(\Delta y\ne 0\text{,}\) so that \((a,b)\) is a saddle point. Similarly, \((a,b)\) is also a saddle point if \(\big(f_{xx}f_{yy}-f_{xy}^2\big) \lt 0\) and \(f_{yy}\) is nonzero.
• Finally, if \(f_{xy}\ne 0\) and \(f_{xx}=f_{yy}=0\text{,}\) then

  \[ f_{xx}\,\Delta x^2 +2f_{xy}\,\Delta x\,\Delta y + f_{yy}\,\Delta y^2 =2f_{xy}\,\Delta x\,\Delta y \nonumber \]

  is strictly positive for one sign of \(\Delta x\,\Delta y\) and is strictly negative for the other sign of \(\Delta x\,\Delta y\text{.}\) So \((a,b)\) is again a saddle point.