2 cal 3 problems?

Question

1) Find the point on the curve of intersection of the paraboloid x^2+y^2+2z = 4 and the plane x-y+2z = 0 that is closest to the origin.

2) Find the point on the curve of intersection of the paraboloid x^2+y^2+2z = 4 and the plane x-y+2z = 0 that is farthest from the origin.


I dont necessarily need the answers , i can solve them myself if someone can tell me how in the world am I supposed to solve them. I just need an explanation cause i don't know where to start.

in my book its chapter 13. right before ch.14 which is multiple integration. my book name is CALCULUS : EIGHT EDITION by Larson, Hostetler, and Edwards

Answer 1

The intersection point is the common point, satisfies the given equations. Find z in terms of x and y
1) find distance and its min value use derivative

2) find distance and its max value use derivative

Example:Given the intersection of the parboloid surface z=x^2+y^2 and the plane z+y+x=1 is an ellipse, find the maximum and minimum distance from origin (0,0) to the points of this ellipse.

Solution: This problem can be formulated as:
d(x,y,z) = x^2+y^2+z2,
subject to g(x,y,z ) = x^2+y^2 -z
and h(x, y, z)=z+y+x=1.

Method I: We can eliminate the variable z in this problem by replacing z =1-x-y.
Then it reduces to
d(x,y) = x^2+y^2+(1-x-y)^2
subject to 1-x-y=x^2+y^2.

Method II:
We use the modified Lagrange method as follows:
grad d = a grad g + b grad h,
for some unknown constants a and b.
(2x, 2y, 2z) =a(2x, 2y -1) +b( 1, 1, 1). ie.
x=b/(1-2a),
y =b/(1-2a),
z=(b-a)/2.
Set x=y=t=b/(1-2a), z=s=(b-a)/2.
Then we have 2t2=s, and s+2t=1. Eliminating s, we got 2t2=1-2t.