Step 1

We first find the partial derivatives

\(\displaystyle{f}={8}{x}^{{{2}}}-{2}{x}{y}^{{{2}}}\)

\(\displaystyle{f}_{{{x}}}={16}{x}-{2}{y}^{{{2}}}\)

\(\displaystyle{f}_{{{y}}}=-{4}{x}{y}\)

At (5,4)

\(\displaystyle{f}_{{{x}}}={16}{x}-{2}{y}^{{{2}}}={16}{\left({5}\right)}-{2}{\left({4}\right)}^{{{2}}}={48}\)

\(\displaystyle{f}_{{{y}}}=-{4}{x}{y}=-{4}{\left({5}\right)}{\left({4}\right)}=-{80}\)

So the direction vector of the normal is = (48,-80)

Step 2

So the equation of the tangent plane is:

\(\displaystyle{z}={\left({8}{\left({5}\right)}^{{{2}}}{02}{\left({5}\right)}{\left({4}\right)}^{{{2}}}\right)}+{48}{\left({x}-{5}\right)}+{\left(-{80}\right)}{\left({y}-{4}\right)}\)

z=40+48x-240-80y+320

z=48x-80y+120

Answer: B

We first find the partial derivatives

\(\displaystyle{f}={8}{x}^{{{2}}}-{2}{x}{y}^{{{2}}}\)

\(\displaystyle{f}_{{{x}}}={16}{x}-{2}{y}^{{{2}}}\)

\(\displaystyle{f}_{{{y}}}=-{4}{x}{y}\)

At (5,4)

\(\displaystyle{f}_{{{x}}}={16}{x}-{2}{y}^{{{2}}}={16}{\left({5}\right)}-{2}{\left({4}\right)}^{{{2}}}={48}\)

\(\displaystyle{f}_{{{y}}}=-{4}{x}{y}=-{4}{\left({5}\right)}{\left({4}\right)}=-{80}\)

So the direction vector of the normal is = (48,-80)

Step 2

So the equation of the tangent plane is:

\(\displaystyle{z}={\left({8}{\left({5}\right)}^{{{2}}}{02}{\left({5}\right)}{\left({4}\right)}^{{{2}}}\right)}+{48}{\left({x}-{5}\right)}+{\left(-{80}\right)}{\left({y}-{4}\right)}\)

z=40+48x-240-80y+320

z=48x-80y+120

Answer: B