Question

The paraboloid z = 3x2 + 2y2 + 1 and the plane 2x – y + z = 4 intersect in a curve C. Find the points on C that have a maximum and minimum distance from the origin.

The point on C is the maximum distance from the origin is (___ , ____ , ____).    The point on C is the minimum distance from the origin is (____ , ____ , ____).

So for this question I get the following:

f(x,y,z) = x2 +y2 +z2    ▽f(x,y,z) = <2x , 2y , 2z>    g(x,y,z) = 3x2 + 2y2 - z+ 1    h(x,y,z)= 2x – y + z -4 = 0    λ▽g(x,y,z) = λ<6x,4y,-1>    µ▽h(x,y,z) = µ<2, -1, 1>    The equations ▽f(x,y,z) = λ▽g(x,y,z) + µ▽h(x,y,z) as well as the constraint g(x,y,z) = 0 and h(x,y,z) = 0 form a system with 5 equations and 5 unknowns. This system is shown bellow. 2x = λ(6x)+2 µ 2y = λ(4y) - µ 2z = λ(-1) + µ 0 = 3x2 + 2y2 - z+ 1 0 = 2x – y + z -4

   THIS IS WHERE I NEED CHEGGS HELP

Use technology to find the two real solutions to this system of equations, rounding each coordinate to three decimal places.

SO I have a TI-89 Titanium Calculator, how would I put this into the calculator and solve to three decimal places??????    There should be six (6) answers with two per x,y,z. Thank you.

Answer 1