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

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  • $\begingroup$ Try using systems of equations $\endgroup$
    – Bunny
    Commented Apr 6, 2016 at 0:52
  • $\begingroup$ @Red I have one equation. What would be the other two? Thank you. $\endgroup$
    – bru1987
    Commented Apr 6, 2016 at 0:52
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    $\begingroup$ The max the of directional derivative occurs when the gradient and $u$ have the same direction. $\endgroup$
    – user310540
    Commented Apr 6, 2016 at 0:55
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    $\begingroup$ The requirement that grad$f$ have the same direction as $u$ gives some equations. $\endgroup$
    – user310540
    Commented Apr 6, 2016 at 0:56
  • $\begingroup$ @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. $\endgroup$
    – bru1987
    Commented Apr 6, 2016 at 1:08

1 Answer 1

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  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 $$(a, b, c) = (3, 12, −4) \text{ or } (a, b, c) = (−3,12,-4)$$
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  • $\begingroup$ For some basic information about writing mathematics at this site see, e.g., here, here, here and here. $\endgroup$ Commented Feb 21, 2020 at 13:59

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