Please solve it by( the method of Lagrange multipliers )! Suppose you are diming

Question

Please solve it by( the method of Lagrange multipliers )!

Suppose you are diming a hill whose shape is given by the equation z = 1000 - O.O1x2 - 0.02y2 where x, y and z are measured in meters. Here, x and y are horizontal coordinates while z denotes elevation above sea level. Assume that the earth is flat and the vector i points to the north and j points to the east. Suppose you start off from the point whose x and y coordinates are (50, 80). If you walk due south, will you ascend, or will you descend? At what slope? In what direction should you start out in order to go downhill as steeply as possible? If you walk northwest at a rate of 2 meters per second, how fast would you be ascending/descending?

Explanation / Answer

z = 1000 - 0.01x^2 - 0.02y^2
grad(z) = (-0.02x, - 0.04y)

at (50,80,847),
grad(z) = (-0.02*50, - 0.04*80)
= (-1,-3.2) (this vector points in the direction of the steepest uphill slope)

a) The unit vector for "south" is (0,-1), so (-1,-3.2) . (0,-1) = 3.2 > 0. So a walk south from the point (50,80,847) is uphill with gradient 3.2

b) The unit vector for "northwest" is (-1/sqrt(2), 1/sqrt(2)), so (-1,-3.2) . (-1/sqrt(2), 1/sqrt(2)) = 1/sqrt(2) * (1 - 3.2) = -2.2/sqrt(2) = -11*sqrt(2)/10 < 0. So a walk northeast from (50,80,847) is downhill with gradient -11*sqrt(2) / 10.

c) The slope is the largest (i.e. most uphill) in the direction of grad(z), and has gradient |grad(z)|. The direction of steepest ascent at (50,80,847) is in the direction of the vector (-1,-3.2) with gradient sqrt(1^2 + 3.2^2) = 3.35.

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