Question

Relative extrema of multivariables:
\(\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{{x}{y}}}{{7}}\)
find critical points and relative extrema, given an open region.

Answer 1

We have, \(\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{{x}{y}}}{{7}}\)
To find the critical point:
Find the partial derivative and equate to zero:
Now,
Differntiate f(x,y) with respect to x and equate to zero
\(\displaystyle{{f}_{{x}}{\left({x},{y}\right)}}=\frac{{y}}{{7}}={0}\)
\(\displaystyle\Rightarrow{y}={0}\)
Differntiate f(x,y) with respect to y and equate to zero
\(\displaystyle{{f}_{{y}}{\left({x},{y}\right)}}=\frac{{x}}{{7}}={0}\)
\(\displaystyle\Rightarrow{x}={0}\)
Hence, the critical point is (0,0)
Now we have to find the relative extrema.
Use:
\(\displaystyle{D}{\left({x},{y}\right)}={{f}_{{\times}}{\left({x},{y}\right)}}{{f}_{{{y}{y}}}{\left({x},{y}\right)}}−{\left({{f}_{{{x}{y}}}{\left({x},{y}\right)}}\right)}^{{2}}\)
Find D(0,0):
Now
\(\displaystyle{{f}_{{\times}}{\left({0},{0}\right)}}={0}\)
\(\displaystyle{{f}_{{{y}{y}}}{\left({0},{0}\right)}}={0}\)
\(\displaystyle{{f}_{{{x}{y}}}{\left({0},{0}\right)}}=\frac{{1}}{{7}}\)
Therefore,
\(\displaystyle{D}{\left({0},{0}\right)}={0}\times{0}-{\left(\frac{{1}}{{7}}\right)}^{{2}}=\frac{{1}}{{49}}{<}{0}\)</span>
Hence, by the \(\displaystyle{2}^{{{n}{d}}}\) derivative test:
f(x,y) neither max. nor min at (0,0) which means (0,0) is saddle point.