Find the critical points of the following functions.Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, a local minimum,or a saddle point. If the Second Derivative Test is inconclusive,determine the behavior of the function at the critical points. f(x,y)=(4x-1)^2+(2y+4)^2+1

Question
Analyzing functions
asked 2020-11-08
Find the critical points of the following functions.Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, a local minimum,or a saddle point. If the Second Derivative Test is inconclusive,determine the behavior of the function at the critical points.
\(\displaystyle{f{{\left({x},{y}\right)}}}={\left({4}{x}-{1}\right)}^{{2}}+{\left({2}{y}+{4}\right)}^{{2}}+{1}\)

Answers (1)

2020-11-09
Given function \(\displaystyle{f{{\left({x},{y}\right)}}}={\left({4}{x}-{1}\right)}^{{2}}+{\left({2}{y}+{4}\right)}^{{2}}+{1}={16}{x}^{{2}}+{4}{y}{2}-{8}{x}+{16}{y}+{18}\)
Critical points:
\(\displaystyle{f}_{{x}}=\frac{{\partial{f}}}{{\partial{x}}}={32}{x}-{8}\)
\(\displaystyle{f}_{{y}}=\frac{{\partial{f}}}{{\partial{y}}}={8}{y}+{16}\)
Setting these equal to zero.
\(\displaystyle{32}{x}-{8}={0}\Rightarrow{x}=\frac{{1}}{{4}}\)
\(\displaystyle{8}{y}+{16}={0}\Rightarrow{y}=-{2}\)
so, the critical point is \(\displaystyle{\left(\frac{{1}}{{4}},-{2}\right)}.\)
Now find the second derivatives,
\(\displaystyle{f}_{{\times}}={32},{f}_{{{y}{y}}}={8},{f}_{{{x}{y}}}={0}\)
So, \(\displaystyle{D}={f}_{{\times}}\cdot{f}_{{{y}{y}}}-{\left({f}_{{{x}{y}}}\right)}^{{2}}\)
For \(\displaystyle{\left(\frac{{1}}{{4}},-{2}\right)},\)
\(\displaystyle{D}={32}\cdot{8}-{0}={256}{>}{0}\)
Since D>0 and \(\displaystyle{f}_{{\times}}={32}{>}{0}\), so the obtaines critical point is local minimum.
The local minimum at \(\displaystyle{\left(\frac{{1}}{{4}},-{2}\right)}\) is 1.
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