# Find the equation of the tangent plane to the graph of f(x,y)=8x^{2}-2xy^{2} at the point (5,4). A)z=23x-15y+42 B)z=48x-80y+120 C)0=48x-80y+120 D)0=23x-15y+42

Derivatives
Find the equation of the tangent plane to the graph of $$\displaystyle{f{{\left({x},{y}\right)}}}={8}{x}^{{{2}}}-{2}{x}{y}^{{{2}}}$$ at the point (5,4).
A)z=23x-15y+42
B)z=48x-80y+120
C)0=48x-80y+120
D)0=23x-15y+42

2021-05-23
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