The plane 2y + 4z = 5 meets the cone z^2 = 4(x^2 + y^2) in a curve. use the method of Lagrange multipliers to find the point on this curve nearest the origin.
I had to refresh myself on the method of Lagrange multipliers. The following theorem is paraphrased from a problem in Spivak's "Calculus on Manifolds", an excellent calculus text for mathematicians.
Let A be a subset of R^n. If g:A -> R^m is differentiable and has rank m on the kernel of g, and if f:R^n -> R is also differentiable and the extremum of f on the kernel of g occurs at a, then there exist m real multipliers L1, ..., Lm such that df(a) = (L1)d(g1)(a) + ... + (Lm)d(gm)(a). (d is the differential operator)
The proof is not that difficult, but a concise one would require the introduction of too much machinery, so we'll leave it as something to look up.
If g has only one component function (if g:A -> R), then the above is very simple with only one multiplier. If n=2, the kernel of g is just a curve in R^2 and the graph of f is a surface in R^3. In this case, you can visualize the constraint on the surface as a curve in the surface of f whose projection on R^2 considered as a subset of R^3 is the kernel of g.
In your case, we want to minimize the distance function f(x, y, z) = Sqrt[x^2 + y^2 + z^2] over the two surfaces above, or equivalently the kernel of the differentiable function g(x, y, z) = (2y + 4z - 5, 4(x^2 + y^2) - z^2) (g is thus a map from R^3 -> R^2, so m = 2. The kernel of g is trivially the intersection of the two surfaces. You should show that the rank of g is 2 whenever g(x, y, z) = 0 to validate the use of Lagrange multipliers).
We simplify this by noting that the points corresponding to the minimum of f over the kernel of g coincides with the points corresponding to the minimum of [f(x, y, z)]^2 = F(x, y, z) over the kernel of g which is easier to deal with. The theorem tells us we will have to deal with two Lagrange multipliers, one for each component of g. We thus have the system of 5 equations:
2x = (L1)*0 + (L2)*8x
2y = (L1)*2 + (L2)*8y
2z = (L1)*4 + (L2)*(-2z)
g(x, y, z) = (2y + 4z - 5, 4(x^2 + y^2) - z^2) = (0, 0)
These are 5 equations in 5 unknowns, so it is easily solved. Note that the actual values of the multipliers are seldom important.
In case you are not familiar with the differential operator, if f(x, y, z) = x^2 + y^2 + z^2 + xy, then df(x, y, z) is a convector (actually a cotangent bundle in more formal terms) such that df(x, y, z) = 2x dx + 2y dy + 2z dz. It is similar to notation for implicit differentiation but has a much more rigorous foundation and is used in differential geometry. The dx, dy and dz terms are treated as basis vectors, thus the above is a 3-dimensional vector, normally called the gradient of f in multivariable calculus, where it is treated as if it were a vector in R^3: (2x, 2y, 2z). You should now see where the first 3 equations come from.