Maximising directional derivative of a polynomial in 3 variables

Question

I am a beginner in multivariable calculus and had started reading Apostol. I have solved all the exercises in the portions I have covered, except one problem.

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

I am not sure that I understand the wording of the problem. I know the definition of directional derivative at a point $a$ in the direction of a vector $y$: it is the derivative $f'(a;y)$ where $y$ is a unit vector. So what I think is, since $y$ is a unit vector parallel to $z$-axis in our case, we must have $y=(0,0,1)$.

Therefore, we need to obtain the partial derivative $D_zf(x,y,z)=by+2cx^3z$, at $(1,2,-1)$ hence $D_zf(1,2,-1)=2(b-c)$.

How do I maximize this quantity (surely the supremum is infinity?)? I believe I am not understanding the problem at all.

Help is appreciated.

Answer 1

Directional derivatives are maximized when they are taken in the direction of the gradient (to prove this, use Cauchy-Schwarz). We have: $$\nabla f(x,y,z) = (ay^2 + 3cx^2 z^2, 2axy + bz, by + 2cx^3z),$$ $$\nabla f(1,2,-1) = (4 a+3 c,4 a-b,2 b-2 c)$$ We want this to be parallel to $(0,0,1)$, i.e. we require $$(4 a+3 c,4 a-b,2 b-2 c) = (0,0,k)$$ for some $k \in \mathbb{R^+}$. That is, we require \begin{align*} 4a + 3c &= 0, \\ 4a - b &= 0, \\ b - c &= k', \end{align*} where $k' = k/2$ (the constant doesn't matter). Solving, $c = -k'/4$, $b = 3k'/4$, and $a = 3k'/16$. It remains to find $k'$. We also know the maximum directional derivative is $|| \nabla f(1,2,-1) ||$ (this also follows from the proof I mentioned in the first line). But this is easy, since we set the gradient equal to $(0,0,k) = (0,0,2k')$. Then $$|| \nabla f(1,2,-1) || = ||(0,0,2k')|| = 2k' = 64$$ so $k' = 32$. Plugging this into the relations we found earlier, $$\boxed{(a,b,c) = (6, 24, -8)}.$$

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Thm. Let $f: \mathbb{R}^n \to \mathbb{R}$ be differentiable function. The directional derivative $D_{\vec{u}}f$ at $x$ is maximized when $\vec{u}$ is in the direction of the gradient $\nabla f$. In this case, $D_{\vec{u}}f = || \nabla f(x) ||.$

Proof. Since $f$ is differentiable, for any unit vector $\vec{u}$, we can write $$D_\vec{u} f = \nabla f \cdot \vec{u}.$$ By the Cauchy-Schwarz inequality, $$D_\vec{u} f = \nabla f \cdot \vec{u} \leq ||\nabla f||\;||\vec{u}|| = ||\nabla f||$$ with equality if and only if $\nabla f = k \vec{u}$ for some $k \in \mathbb{R^+}.$

Answer 2

The directional derivative is maximized in the direction of greatest change, i.e. the gradient, given by $\operatorname{grad} f=\left(ay^2+3cx^2z^2,2axy+bz,by+2cx^3z\right)$. Now, at $(1,2,-1)$ the gradient is as follows: $$(\operatorname{grad} f)\big|_{(1,2,-1)}=\left(4a+3c,4a-b,2b-2c\right)$$So we expect the $4a+3c=4a-b=0$ while $2b-2c=64$. This gives $b+3c=0,b-c=32$ so $4c=-32\implies c=-8$. This ultimately gives $b=c+32=24$ and $4a=24\implies a=6$.