1. ## Directional Devirative question

Find values of the constant a, b, and c such that the directional derivative of f(x, y, z) = ax(y^2) + byz + c(z^2)(x^3) at the point (1, 2, -1) has a maximum value of 64 in a direction parallel to the z-axis.

What I tried doing was that since the direction is in the z-axis, this is a partial derivative. So I took D3f(x, y, z) and set it to less than or equal to 64. Then I'm stuck. Any ideas?

2. You may be misinterpreting the problem. It is not only telling you that the partial derivative with respect to z is 64 but also that that is the maximum of all directional derivatives.

The easiest way to do this is to use the fact that the gradient vector always points in the direction of fastest increase. That is, $\nabla f= (ay^2+ 3cx^2z^2)\vec{i}+ (2axy+ bz)\vec{i}+ (by+ 2czx^3)\vec{k}$, evaluated at x= 1, y= 2, z= -1, is just $64\vec{k}$.