# Thread: Points on horizontal tangent plane?

1. ## Points on horizontal tangent plane?

This is a question from my calculus 3 class.

"Find all points on the surface at which the tangent plane is horizontal."
z = x^2 - xy + y^2 - 2x + 4y

Okay, so I'm not sure where to begin. I know that the partial derivatives z_x and z_y are both zero. And so I've thought of making a tangent equation but I got stuck.

Tangent plane = (z - z0) ?

I'm not sure if I have to solve for x and y in the given equation, or really what I have to do in general.

Can anyone give me a push in the right direction? Maybe I'm over-thinking this problem.

2. Like you said, $\dfrac{\delta z}{\delta x} = 0$ and $\dfrac{\delta z}{\delta y} = 0$, which is a system of equations. Solutions to the system of equations are the points on the surface at which the tangent plane is horizontal.

3. This might be a silly question but how do I go about solving this? I know how to solve systems of equations, but all I see is dz/dx = 0 and dz/dy = 0. What am I not seeing?

4. First, find $\dfrac{\delta z}{\delta x}$ given $z = x^2 - xy + y^2 - 2x + 4y$ like so:

$\dfrac{\delta z}{\delta x} = 2x - y - 2$

Then, find $\dfrac{\delta z}{\delta y}$ like so:

$\dfrac{\delta z}{\delta y} = 2y - x + 4$

Now, we know:

$2x - y - 2 = 0$
$-x + 2y + 4 = 0$

I assume you can solve the systems of equation from here. If not, then feel free to post again.

5. If a moderator would delete this double post, then I would be much obliged. :P

6. Oooooh okay I see. Thanks!

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# tangent plane is horizontal

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