You are given the following function maxz x1 22 x1 x21 P
Solution
http://www.math.uri.edu/~bkaskosz/flashmo/graph3d/
or https://academo.org/demos/3d-surface-plotter/
(a) Use one of the online plotters (key in the function -(x-2)2 -x-y2 , say for x between 0 and 3 and y between -5 and 5)
(b)Let x and y be the variables x1 and x2.
Set F(x,y) =-(x-2)2 -x-y2
Stationary point is obtained by setting the partial derivatives equal to 0. So we have
-2(x-2) -1 =0 , x =1.5
-2y=0 , y=0
Now Fxx =-2, Fyy = -2 and Fxy =0. So this point is a maximum .
(c) Start from the point (2.5,1.5)
The gradient at thi2eepest ascent, set
G(t) = F(2.5-2t,1.5-3t) = -(0.5-t)2 -(2.5-2t)-(1.5-3t)2
So G\'(t) = 4(0.5-2t)+2t+6(1.5-3t)
So G\'(t) =0 implies t =0.5
The new point (approximation is (1.5,0), at which the gradient vanishes.
So the method of steepest ascent terminates at (1.5,0) in one step.
