5. Suppose that you are climbing the hill whose shape is given by the equation

z = 1000-0.01x² -0.02 y²

and you are at the point (60, 100, 764).

a) In what direction should you proceed initially in order to reach the top of the hill fastest.

b) If you keep climbing in the direction given in (a) how close

will you get to the top?

ascent. This is known to be

For this case

<f_x , f_y> = <-x/50, -y/25> = <-6/5, -4>

So, the direction of fastet ascent is

b) We assume the traveler starts and continues along the curve

Note that 20x(t) - 6y(t) = 1200-600 = 600, or 20x - 6y = 600. Thus, geometrically, the path of travel is the curve formed by the intersection of the plane 20x - 6y = 600 and the paraboloid

z = 1000-0.01x² -0.02 y² .

The distance from any point on the curve to the point (0,0, 1000) at the peak is given by

D(t) = (x(t)² + y(t)² + (z(t)-1000)²)^1/2

That is,

D(t) = (x(t)² + y(t)² + (0.01x(t)² + 0.02y(t)²)²)^1/2

We want to find the minimum of D(t). Graphing it we get

Thus, from the graph, the minimum distance is approximately 30, when t is about 112.