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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?

Franz Schonwald
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  • Your approach to the problem looks good. How did you solve for $c$, though? –  Feb 24 '14 at 16:29

2 Answers2

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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. $$\mathrm{f}_x = \mathrm{f}_y = 0$$

Fly by Night
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When you solve these two lin systems of equations, you know that you get either one unique solution, or non e or infinite but not two. Go back to you two equations and verify that the first one is right

imranfat
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  • Note he did get a unique solution to the linear system of equations, so that isn't where the problem is. –  Feb 24 '14 at 16:31
  • @Hurkyl Not entirely. While his algebra to obtain $a$ and $b$ is certainly right, he went on to say "The problem is that two of the possible answers meet this condition: (2,−6,−52) and (2,−6,−26)" That wording "This condition" prompted me to reply wrt his system of 2 equations and verify what that means. – imranfat Feb 24 '14 at 16:35