Answer
a) The top of the hill is 3 miles north, 2 miles west, of South Hadley.
b)Height of the hill, h=720ft.
c)The slope at the point is 200$\sqrt{2}$ ft/mile in the northwest direction.
Work Step by Step
a)$\nabla{h}=(\large\hat{x}\frac{\partial}{\partial{x}}+\hat{y}\frac{\partial}{\partial{y}}+\hat{z}\frac{\partial}{\partial{z}})\normalsize(h(x,y))$
$\hspace{1cm}=10(2y-6x-18)\hat{x}+(2x-8y+28)\hat{y}$
At the top of hill $\nabla{h}=0$, so (by comparison)
$2y-6x-18=0\hspace{1cm}$ --(1)
$2x-8y+28=0\hspace{1cm}$ --(2)
eq(1)+3$\times$eq(2) gives
$-22y+66=0$ ; $\Rightarrow y=3$
Putting y=3 in eq(1) ,
$\Rightarrow x=-2$
Since x is the distance(in miles) east, and y the distance north, of South Hadley, thus the top is located at 3 miles north, 2 miles west, of South Hadley.
b)Height of the hill, $h(-2,3)=10(-12-12-36+36+84+12)=720$ feet or ft.
c)The slope at x=1 and y=1 (slope at a point 1 mile north and 1 mile east of South Hadley) is given by,
$\hspace{4cm}\nabla{h}\Big|_{x=1,y=1}=10(2-6-18)\hat{x}+(2-8+28)\hat{y}$
$\hspace{6.1cm}=220(-\hat{x}+\hat{y})$
$|\nabla{h}|=200\sqrt{2}$
The direction of this steepest slope at the point (x=1,y=1) is $(-\hat{x}+\hat{y})$, which corresponds to northwest.
