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

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)?
