Problem: Find the direction of greatest increase at $P$.
$$f(x,y)=4x^2+y^2+2y$$
$$P=(1,2,12)$$
Solution: The greatest increase in $f(x,y)$ at $P$ can be attained by moving in the direction of $$\nabla f(1,2)= \langle 8, 6 \rangle$$
Can someone explain how they got this solution? I know it will involve partial derivatives, but I don't know where to go from there.
Thanks.
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$\begingroup$(I am using the notation $x=(x_1,x_2)$.)
If we pick some direction $h$, then the rate of change of $f$ in the direction $h$ at a point $x$ is given by $df(x,h) = \lim_{t\downarrow 0} { f(x+th)-f(x) \over h}$.
It is easy to show that if $\lambda \ge 0$, then $df(x,\lambda h) = \lambda df(x,h)$, so we can just look at the case where $\|h\| = 1$.
A (unit) direction $\hat{h}$ gives the greatest descent if $df(x,\hat{h}) \ge df(x,h)$ for all other unit length directions $h$.
If the function is differentiable, we can show that $df(x,h) = \langle \nabla f(x), h \rangle$, and the Cauchy-Bunyakovsky-Schwarz-Bieber inequality gives $|\langle \nabla f(x), h \rangle| \le \|h\| \|\nabla f(x)\|$. Furthermore, if we let $\hat{h} = {1 \over \|\nabla f(x)\|} \nabla f(x)$, we have $\langle \nabla f(x), \hat{h} \rangle = \| \nabla f(x) \|$, and so $df(x,h) \le df(x,\hat{h})$ for all other unit length $h$.
The particular length doesn't matter since we are only interested in the direction, so we can just pick the direction $h=\nabla f(x)$.
In your example, $\nabla f(x) = \binom{8x}{2y+2}$, so $\nabla f((1,2)) = \binom{8}{6}$.
$\endgroup$ 2 $\begingroup$So we define the gradient of a function to be: $$ \nabla f(x,y,z)=\space\left\langle\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right\rangle $$ (in three dimensions)
So to answer your question, you need to compute these derivatives, and evaluate the gradient at the point in question to obtain the direction of greatest increase.
It also appears as if, since your function is independent of z, your book is ignoring the z component of the gradient.
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