# Relative extrema of multivariables: f(x,y)= (xy)/7 find critical points and relative extrema, given an open region.

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

2020-10-28
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.

### Relevant Questions

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b) Find the function's range.
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