Using the method of Lagrange multipliers: the Lagrangian is
[tex]L(x,y,z,\lambda) = (x-4)^2+(y-5)^2+(z-6)^2 - \lambda (z-4x^2-4y^2)[/tex]
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:
[tex]\dfrac{\partial L}{\partial x} = 2(x-4) + 8\lambda x = 0 \\\\ \dfrac{\partial L}{\partial y} = 2(y-5) + 8\lambda y = 0 \\\\ \dfrac{\partial L}{\partial z} = 2(z-6) - \lambda = 0 \\\\ \dfrac{\partial L}{\partial\lambda} = -z+4x^2+4y^2 = 0[/tex]
Solving for λ in terms of x, y, or z gives
[tex]\displaystyle \lambda = \frac1x-\frac14 = \frac5{4y} - \frac14 = 2z-12[/tex]
so right away, we get
[tex]\dfrac5{4y} - \dfrac14 = \dfrac1x - \dfrac14 \implies y=\dfrac{5x}4 \\\\ 2z-12 = \dfrac1x-\dfrac14 \implies z = \dfrac1{2x} + \dfrac{47}8[/tex]
Substitute these into the constraint and solve for x, then for y and z :
[tex]\dfrac1{2x} + \dfrac{47}8 = 4x^2 +4\left(\dfrac{5x}4\right)^2 \implies 82x^3 - 47x - 4 = 0[/tex]
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).
