Finding the highest point for a 3d function The height of something at point $(x,y)$ is given by a function $z(x,y)$
$z(x,y)=z_0exp(-\alpha (x^2+y^2))$   where  $z_0$ and $\alpha$ are positive constants
a) At what location in the plane is the height the greatest.
My solution:
$grad (z) = -2az_0 xexp(-\alpha (x^2+y^2)) \hat{e}_x-2az_0 yexp(-\alpha (x^2+y^2)) \hat{e}_y =0 $
Then solve for values of x and y. However i cannot do that as we have two unknown variables and only one equation. We i graph the function, i can see the function is height at (0,0)
b) Find the maximum slope at the point $(x,y)=(3,4)$ and find the unit vecotr in the xy-plane that points in the direction of maximum upward slope.
My solution:
In this part you just plug the values of y and x in to grad(z)  but i was wondeing how we find the unit vecotr that points in the direction of maximum upwards slope
 A: a) You do not have simply one equation and two unknowns. You have one vector equation that can be written as two scalar equations with two unknowns. Observe that $\nabla z = \vec 0$ directly implies that
$$\frac{\partial z}{\partial x} = 0$$
$$\text{and}$$
$$\frac{\partial z}{\partial y} = 0$$
individually. You can thus solve the system
$$\begin{align} -2az_0 x\exp(-\alpha (x^2+y^2))  &= 0 \\-2az_0 y\exp(-\alpha (x^2+y^2)) &= 0 \end{align}$$
which is trivially $$\begin{align}x&=0 \\ y&=0 \;. \end{align}$$
b) To find the maximal slope, you need to compute the directional derivative with respect to the direction of maximum increase. The gradient already points in the direction of maximal increase thus you can dot the gradient vector with itself and divide by the magnitude of the gradient vector. This can be easily shown to be equal to the magnitude of the gradient vector.
The unit vector in the $x\text{-}y$ plane that points in the direction of maximal increase is the normalized gradient vector. Compute the gradient vector at your point and divide by it's magnitude.
Let me know if you need any more clarification.
A: Here is a solution taking into account that you have a surface of revoultion by converting the equation of your function in cylindrical coordinates (a generalization of polar coordinates).
$$\begin{cases}r \\ \theta\\ z=z_0 \exp(-\alpha r^2)\\\end{cases}$$
The maximum of $z=z_0 \exp(-\alpha r^2)$ is clearly for $r=0$, otherwise said for $(x,y)=(0,0)$.
b) Let us write the equation under the form:
$$f(r,\theta,z)=z-z_0 \exp(- \alpha r^2)$$
In point $(3,4)$ (whose cylindrical coordinates are $(r=5,\theta=arctan(4/3),z=z_0 \exp(-\alpha 25))$, the gradient is (see here):
$$({\partial f \over \partial r}, {1 \over r}{\partial f \over \partial \theta},{\partial f \over \partial z})=(-z_0 5 \exp(-\alpha 25),0,1)$$
which is in cylindrical coordinates the orthogonal vector to the tangent plane.
Can you take it from here (it remains the conversion into cartesian coordinates)?
