Equation of the height of the hill z = 2xy - 3x^2 -4y^2 -18x + 28y +12 z: Height of the hill x: Distance East y: distance South Question: In which compass direction is the slope at x = y = 1 steepest? My question: What does this suppose to mean? The vector that is tangent to that point pointing to the top of the hill? If so how do I do that? Note: I have already calculated the critical point of the hill and the angle between the normal vector of the hill at x=y=1 and the z-axis. Thanks!