To find $a,b,c$ as the directional derivative of $f(x,y,z)=axy^2+byz+cz^2x^3$ , at $(1,2-1)$ , is atmost $64$ in a direction parallel to $z$-axis?

Question

How to find $a,b,c$ such that the directional derivative of $f(x,y,z)=axy^2+byz+cz^2x^3$ , at $(1,2-1)$ , has a maximum value of $64$ in a direction parallel to $z$-axis ? I think I have to equate $\nabla f(1,2,-1) . (1,2,\alpha)=64\sqrt{1^2+2^2+\alpha^2}$ , but I am not sure and I don't even know how to proceed from here even if it is correct . Please help . Thanks in advance .

Answer 1

The directional derivative in the z-direction is just $\partial f/\partial z$ (or in the opposite direction, which would just be the negative of that). So you just need to compute that, evaluate it at the desired point, and find the conditions on the constants which ensure it is less than 64.

Answer 2

Find grad of $f$ and put $(1,2,-1)$ in it. You will get $(4a+3c) i +(4a-b)j +(2b-2c)k$. Since parallel to $y$ axis, $4a+3c=0$ and $2b-2c=0$, From here we get $b=c=-4a/3$. Put the value of $b$ in $4a-b=64$ as given in question, and you will get $a=12, b=c=-16$. The answer will be $-20$.