Question

find the point on the graph z=12-x^2-y^2 above 5x+2y+3z=0 which is farthest from that plane

Answer 1

i understand how I would do it if it was from the function to a point but not in this case

Answer 2

prolly some hugely confounded distance formula contraption ...

Answer 3

first off, can we define the shape of the z equation?

Answer 4

x^2 + y^2 + z = 12

i think its a cylindar passing thru the xy plane riding along the z axis .... ill have to graph it to be sure tho

Answer 5

or, maybe a paraboloid

Answer 6

here it is with the plane thru it; is there an octant or other means of refining the ranges?

Answer 7

hummm do we need to do it that way? we are going over max/min problems at the moment and did one where we had to find the distance from z=sqrt(x^2+y^2) to (6,2,0) and got the equation to be d^2=((x-6)^2+(y-2)^2+x^2+y^2)

Answer 8

having a visual is always helpful.

Answer 9

nope thats the whole question i was given

Answer 10

the second setup shows that the parabaloid opens up indefinantly; and as it does gets further and further away form the plane; at least thats the way i see it

Answer 11

its the part that that is above the plane
so i think its the other part

Answer 12

its been awhile, but one idea i have is that the normal vector of the plane would be useful then