# Finding constants that maximise a certain directional derivative

Find values $a, b, c$ such that the maximal directional derivative of $$f(x,y,z) = a x y^2 + b y z + c z^2 x^3$$ at point $(1,2-1)$ is $64$ in a direction parallel to the $z$ axis.

My work so far:

Claim 1: I know the directional derivative can only be in the z direction because to must be parallel to the z axis. so out of the vector $<i,j,k>$ only k will be used

I also know that the maximum directional derivative will concern the unit vector, but thiis means that the length of the vector in the direction of j will be 1 ( I think this is right? )

so $\frac{\partial f}{\partial z} = by +2cx^3$ and given that the unit vector should be <0,0,1>

I get the equation:

$64$ = $\frac{\partial f}{\partial z} * 1$

$64 = (by +2cx^3) * 1$ and this is at the point (1,2,-1)

so my equation becomes $64 = 2b-2c$ so I know that the equality $32=b-c$ exists for the values a,b,c

but I cannot get further than that.

• You’re going astray when you say that “only $k$ will be used.” What the condition really means is that $i$ and $j$ must be zero, which is rather different.
– amd
Sep 4 '17 at 23:22

The formula for directional derivative of $f$ in the direction of $\vec u$ is $$D_{\ \vec u} \ f = \vec \nabla f \ \bullet \frac{\vec u}{|\vec u|}$$
where $\vec \nabla f := <f_x,f_y,f_z> \$ (the vector of partial derivatives) and "$\bullet$" is the dot product.
Observe that you can choose $\vec u$ in whathever length you want, because it will be divided by its length. In other words, if $\vec u$ is unit, $D_{\ \vec u} \ f = \vec \nabla f \ \bullet \ \vec u$.
To maximize the directional derivative, $\vec \nabla f$ must be parallel to $\vec u$.
So, $f_x$ and $f_y$ must be equal to zero at the point $(1,2,-1)$. And $f_z$ can be equal to anything.
In addition to the equality you have already found, solving $\ f_x \big|_{(1,2,-1)} =0$ and $\ f_y \big|_{(1,2,-1)}=0$ will give you all equations you need.