Question

A surface is represented by the following multivariable function,
\(\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{1}}{{3}}{x}^{{3}}+{y}^{{2}}-{2}{x}{y}-{6}{x}-{3}{y}+{4}\)
a) Calculate \(\displaystyle{f}_{{x x}},{f}_{{{y}{x}}},{f}_{{{x}{y}}}{\quad\text{and}\quad}{f}_{{{y}{y}}}\)
b) Calculate coordinates of stationary points.
c) Classify all stationary points.

Answer 1

Since, we have been given a surface that is represented by the following multivariable function given below:
\(\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{1}}{{3}}{x}^{{3}}+{y}^{{2}}-{2}{x}{y}-{6}{x}-{3}{y}+{4}\) (i)
To calculate:
a) \(\displaystyle{f}_{{x x}},{f}_{{{y}{x}}},{f}_{{{x}{y}}}{\quad\text{and}\quad}{f}_{{{y}{y}}}\)
Solution \(\displaystyle\Rightarrow{f}_{{x}}=\frac{{3}}{{3}}{x}^{{2}}-{2}{y}-{6}\)
\(\displaystyle\Rightarrow{f}_{{x x}}={2}{x}\)
\(\displaystyle{f}_{{y}}={2}{y}−{2}{x}−{3}\)
\(\displaystyle{f}_{{{y}{y}}}={2}\)
\(\displaystyle{f}_{{{x}{y}}}=−{2}\)
\(\displaystyle{\quad\text{and}\quad}{f}_{{{y}{x}}}=−{2}\)
b) \(\displaystyle{f}_{{x}}={x}^{{2}}−{2}{y}−{6}\)
\(\displaystyle{\quad\text{and}\quad}{f}_{{y}}={2}{y}−{2}{x}−{3}\)
\(\displaystyle{f}_{{x}}={0}\Rightarrow{x}^{{2}}−{2}{y}={6}\) (ii)
\(\displaystyle{\quad\text{and}\quad}{f}_{{y}}={0}\Rightarrow{2}{y}−{2}{x}={3}\) (iii)
Adding equation (ii) and (iii), we have:\(\displaystyle−{x}^{{2}}−{2}{x}={9}\)
Solving for x, we have:
\(\displaystyle{x}=\frac{{-{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{2}}{a}\Rightarrow{x}=\frac{{{2}\pm\sqrt{{{4}+{36}}}}}{{2}}\Rightarrow{x}=\frac{{{2}\pm\sqrt{{10}}}}{{2}}\Rightarrow{x}={1}\pm\sqrt{{10}}\)
Since, \(\displaystyle{2}{y}−{2}{x}={3}\Rightarrow{2}{y}={2}{x}+{3}\Rightarrow{y}=\frac{{{2}{x}+{3}}}{{2}}=\frac{{{2}{\left({1}\pm\sqrt{{10}}\right)}+{3}}}{{2}}=\frac{{{2}\pm{2}\sqrt{{{10}}}+{3}}}{{2}}\)
\(\displaystyle\Rightarrow{y}=\frac{{1}}{{2}}{\left({5}\pm{2}\sqrt{{10}}\right)}\)
So, stationary points are:
\(\displaystyle{\left({1}-\sqrt{{10}},\frac{{{5}-{2}\sqrt{{10}}}}{{2}}\right)},{\left({1}+\sqrt{{10}},\frac{{{5}-{2}\sqrt{{10}}}}{{2}}\right)},{\left({1}-\sqrt{{10}},\frac{{{5}+{2}\sqrt{{10}}}}{{2}}\right)},{\left({1}+\sqrt{{10}},\frac{{{5}+{2}\sqrt{{10}}}}{{2}}\right)}\)
c) \(\displaystyle{\left({1}+\sqrt{{10}},\frac{{{5}+{2}\sqrt{{10}}}}{{2}}\right)}\) is minimal and \(\displaystyle{\left({1}-\sqrt{{10}},\frac{{{5}-{2}\sqrt{{10}}}}{{2}}\right)}\) is a saddle point.