Question

Among all points on the graph of z=12-x^2-y^2 that lies above the plane x+3y+6z=0, find the point farthest from the plane.

Answer 1

so we have a paraboliod and a plane

Answer 2

Yes

Answer 3

so z= (-1/6) ( x+3y)

so for z to be above the plane

x+3y<0

Answer 4

one constraint

Answer 5

so we need the distance between the two

Answer 6

so if we pick a general point (x , y , 12-(x^2+y^2 ) ) on the paraboliod

Answer 7

im not quite sure how to get the distance to the plane :|