Find an equation of the plane tangent to the following surface at the given point. 8 xy plus 5 yz plus xzminus56equals0; (2 comma 2 comma 2 )The equation of the tangent plane at (2 comma 2 comma 2 )is nothingequals0.

The Question is :

Find an equation of the plane tangent to the following surface at the given point.

8xy + 5yz + xz - 56 = 0; (2, 2, 2 )

The equation of the tangent plane at (2, 2, 2 ) is 0.

The equation of the plane tangent to the surface

8xy + 5yz + xz - 56 = 0

at the point (2, 2, 2 )

is 9x + 13y + 6z - 56 = 0

Step-by-step explanation:

Given the equation

8xy + 5yz + xz - 56 = 0

and the point

(2, 2, 2 ).

To find the equation of the plane tangent to the surface, we first differentiate the given function with respect to x, y and z respectively.

F_x = 8y + z

F_y = 8x + 5z

F_z = 5y + x

At the point (2, 2, 2)

F_x = 18

F_y = 26

F_z = 12

The equation of the plane is given as

(F_x)(x - 2) + (F_y)(y - 2) + (F_z)(z - 2) = 0

18(x - 2) + 26(y - 2) + 12(z - 2) = 0

18x + 26y + 12z - 36 - 52 - 24 = 0

18x + 26y + 12z - 112 = 0

Divide through with 2

9x + 13y + 6z - 56 = 0

This is the equation we are looking for.

it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.