I am given the surface $z = x^{2} – 7xy – y^{2} – 46x + 2y$ and have to find the point, among four options, at which the tangent plane to the surface is horizontal.
Now my reasoning is this: if we write $F (x, y, z) = x^{2} – 7xy – y^{2} – 46x + 2y – z,$ the tangent plane to the surface at the point $(a, b, c)$ is $$(2a – 7b – 46)(x – a) + (-7a – 2b + 2) (y – b) + (-1) (z – c) = 0,$$ and since horizontal planes have the equation $z = k$, for some constant $k$, we need find the point which makes $2a – 7b – 46$ and $-7a – 2b + 2$ zero. The problem is that two of the possible answers meet this condition: $(2, -6, -52)$ and $(2, -6, -26)$.
So my question is, am I doing something wrong?
Best Answer
HINT
Your surface is given by $\mathrm{f}(x,y,z)=0$, where $$\mathrm{f}(x,y,z) = x^2−7xy−y^2−46x+2y-z$$ The gradient vector $\nabla\mathrm{f} = \left(\mathrm{f}_x,\mathrm{f}_y,\mathrm{f}_z\right)$ is, if non-zero, perpendicular to the the tangent plane. The tangent plane is horizontal if, and only if, it is perpendicular to the $z$-axis. Hence, we need $$\nabla\mathrm{f} \propto (0,0,1)$$ We need to find the partial derivatives $\mathrm{f}_x$, $\mathrm{f}_y$ and $\mathrm{f}_z$ and check when $\nabla\mathrm{f} \propto (0,0,1)$, i.e. $$\mathrm{f}_x = \mathrm{f}_y = 0$$