Final answer:
The equation of the tangent plane to the surface z=54−5x^2−4y^2 at the point (3,1,5) is -30(x - 3) - 8(y - 1) - (z - 5) = 0
Step-by-step explanation:
To find the equation of the tangent plane to the surface z = 54 - 5x^2 - 4y^2 at the point (3,1,5), we need to determine the partial derivatives and evaluate them at that point. The partial derivatives are:
∂z/∂x = -10x
∂z/∂y = -8y
Evaluating these at (3,1,5), we get:
∂z/∂x = -10(3) = -30
∂z/∂y = -8(1) = -8
Using the point-normal form of the equation of a plane, where the normal vector is (∂z/∂x, ∂z/∂y, -1) and the point is (3,1,5), we can write the equation of the tangent plane as:
-30(x - 3) - 8(y - 1) - (z - 5) = 0