find the equation of the tangent plane to the surface ( z = 8x^{2}+y^{2}+2xy ) at the point ( (0,3,9) ).

Explanation:

Step1: Find the partial derivatives

Let (F(x,y,z)=8x^{2}+y^{2}+2xy - z). The partial derivative with respect to (x): (F_{x}=\frac{\partial F}{\partial x}=16x + 2y). Evaluate at ((0,3)): (F_{x}(0,3)=16\times0+2\times3 = 6). The partial derivative with respect to (y): (F_{y}=\frac{\partial F}{\partial y}=2y+2x). Evaluate at ((0,3)): (F_{y}(0,3)=2\times3+2\times0 = 6). The partial derivative with respect to (z): (F_{z}=\frac{\partial F}{\partial z}=- 1).

Step2: Use the tangent - plane formula

The equation of the tangent plane is (F_{x}(x_{0},y_{0})(x - x_{0})+F_{y}(x_{0},y_{0})(y - y_{0})+F_{z}(x_{0},y_{0})(z - z_{0})=0). Here (x_{0}=0,y_{0}=3,z_{0}=9). Substitute the values: (6(x - 0)+6(y - 3)-1(z - 9)=0). Expand: (6x+6y-18 - z + 9=0). Rearrange to get (z = 6x+6y - 9).

Answer:

(z = 6x+6y - 9)