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Find the equation of the tangent plane to the surface z=54−5x^2−4y^2 at the point (3,1,5)

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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

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