 # Relative extrema of multivariables: f(x,y)= (xy)/7 find critical points and relative extrema, given an open region. sibuzwaW 2020-10-27 Answered
Relative extrema of multivariables:
$f\left(x,y\right)=\frac{xy}{7}$
find critical points and relative extrema, given an open region.
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We have, $f\left(x,y\right)=\frac{xy}{7}$
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
${f}_{x}\left(x,y\right)=\frac{y}{7}=0$
$⇒y=0$
Differntiate f(x,y) with respect to y and equate to zero
${f}_{y}\left(x,y\right)=\frac{x}{7}=0$
$⇒x=0$
Hence, the critical point is (0,0)
Now we have to find the relative extrema.
Use:
$D\left(x,y\right)={f}_{xx}\left(x,y\right){f}_{yy}\left(x,y\right)-{\left({f}_{xy}\left(x,y\right)\right)}^{2}$
Find D(0,0):
Now
${f}_{xx}\left(0,0\right)=0$
${f}_{yy}\left(0,0\right)=0$
${f}_{xy}\left(0,0\right)=\frac{1}{7}$
Therefore,
$D\left(0,0\right)=0×0-{\left(\frac{1}{7}\right)}^{2}=\frac{1}{49}<0$
Hence, by the ${2}^{nd}$ derivative test:
f(x,y) neither max. nor min at (0,0) which means (0,0) is saddle point.