Answer:
Required equation of tangent plane is [tex]z=\frac{6}{5}(x-5y-11)[/tex].
Step-by-step explanation:
Given surface function is,
[tex]f(x,y)=6-\frac{6}{5}(x-y)[/tex]
To find tangent plane at the point (5,-1,1).
We know equation of tangent plane at the point $(x_0,y_0,z_0)[/tex] is,
[tex]z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)\hfill (1)[/tex]
So that,
[tex]f(x_0,y_0)=6-\frac{6}{5}(5+1)=-\frac{6}{5}[/tex]
[tex]f_x=-\frac{6}{5}y\implies f_x(5,-1,1)=\frac{6}{5}[/tex]
[tex]f_y=-\frac{6}{5}x\implies f_y(5,-1,1)=-6[/tex]
Substitute all these values in (1) we get,
[tex]z=\frac{6}{5}(x-5)-6(y+1)-\frac{6}{5}[/tex]
[tex]\therefore z=\frac{6}{5}(x-5y-11)[/tex]
Which is the required euation of tangent plane.
