Find the point on the paraboloid z = 4x2 + 4y2 which is closest to the point
(4; 5; 6).

Using the method of Lagrange multipliers: the Lagrangian is

Fix λ > 0.

That is, we are minimizing the squared distance of a point (x, y, z) on the paraboloid to the point (4, 5, 6). We can use the squared distance in place of the proper Euclidean distance because both f(x) and √f(x) have critical points at the same point x ; plus the math works out much more easily.

Find the critical points of the Lagrangian:

Solving for λ in terms of x, y, or z gives

so right away, we get

Substitute these into the constraint and solve for x, then for y and z :

This has 3 roots, but we take the lone positive solution for x because otherwise λ would be negative. This root is x ≈ 0.7965.

Then it follows that the closest point on the paraboloid to (4, 5, 6) is the critical point (0.7965, 0.9956, 6.5028).

the bottom of the spinner is not flat.

step-by-step explanation:

i believe it is b and c. (side note, i got the same computer from k-12 lol i’m in 8th grade and this is my first year, will it get

step-by-step explanation: