Find constants so that the directional derivative of $f(x,y,z) = axy^2+byz+cx^3z^2$ has maximum value $32$ in point $P$ given the direction

I am asked to find $a$, $b$ and $c$ so that the directional derivative of $$f(x,y,z) = axy^2+byz+cx^3z^2$$ has maximum value of $32$ in the point $P(1,2,-1)$ and in the direction $\overrightarrow{u} = (0,0,1)$.

What I have so far is

$$\frac{\partial f}{\partial x} = ay^2 + 3cx^2z^2\\ \frac{\partial f}{\partial y} = 2axy+bz\\ \frac{\partial f}{\partial z} = by+2cx^3z\\ \nabla f(1,2,-1) = (4a+3c,4a-b,2b-2c)$$

$$D_{u} f(1,2,-1) = (4a+3c, 4a-b, 2b-2c) \cdot (0,0,1) = 32 \therefore b-c = 16$$

and I'm not sure how to proceed.

Answer: $$a = 3\\ b = 12\\ c = -4$$

• Try using systems of equations Commented Apr 6, 2016 at 0:52
• @Red I have one equation. What would be the other two? Thank you. Commented Apr 6, 2016 at 0:52
• The max the of directional derivative occurs when the gradient and $u$ have the same direction. Commented Apr 6, 2016 at 0:55
• The requirement that grad$f$ have the same direction as $u$ gives some equations. Commented Apr 6, 2016 at 0:56
• @spinoza thank you for the input. But that would just say that $4a+3c=0$ and $4a-c=0$. Am I making a mistake here? I'll make a small edit on the post in one minute. Commented Apr 6, 2016 at 1:08

1. Find values of the constants $$a, b, c$$ such that the directional derivative of $$f(x, y, z) = axy^2 + byz +cz^2x^3$$ at the point $$(1, 2, −1)$$ has a maximum value of $$32$$ in the direction parallel to the $$z$$-axis. First compute the gradient: $$\nabla f = (ay^2 + 3cz^2x^2)i + (2axy + bz)j + (by + 2czx^3)k$$ and $$\nabla f(1, 2, −1) = (4a + 3c)i + (4a − b)j + (2b − 2c)k$$ We know that the gradient vector points in the direction of the greatest rate of change. So, if $$f$$ attains a maximum value in a direction parallel to the $$z$$-axis, this means that $$\nabla f(1, 2, −1)$$ points in a direction parallel to the $$z$$-axis (but we don’t know if it points up or down). This gives the equations $$4a + 3c = 0, \quad 4a − b = 0, \quad 2b − 2c = 32$$ where the final equation follows from the fact that we know that the maximum value attained by $$f$$ is $$64$$, and the fact that the maximum value of the gradient vector equals the length of the gradient vector. Solving this, we get $$(a, b, c) = (3, 12, −4) \text{ or } (a, b, c) = (−3,12,-4)$$