# Thread: Method of tseepest descent to min. Help???

1. ## Method of tseepest descent to min. Help???

I have a question to solve and i really don't know how. if someone can help me please??????
Maria

2. Originally Posted by maria69
I have a question to solve and i really don't know how. if someone can help me please??????
Maria
What have you been covering in class, and what tool are you supposed to use for this.

(the main fact-oid that you need is that the direction of speepest decent from $\displaystyle (x,y)$ is $\displaystyle -\nabla f(x,y)/|\nabla f(x,y)|$

CB

3. i have an example but i really don't know how to use it to my specific problem. Can you please show me how? thank you very much!

4. The "method of steepest descent" is a numerical method. To find a minimum for f(x,y), starting at $\displaystyle (x_0, y_0)$, find the direction of $\displaystyle -\nabla f(x_0,y_0)$ and move a short distance in that direction. Repeat until $\displaystyle \nabla f(x_0, y_0)$ is short enough.

For this problem, $\displaystyle \nabla (x^2+ 1.1y^2)= 2x\vec{i}+ 2.2y\vec{j}$ and at (6, 3), that is $\displaystyle 12\vec{i}+ 6.6\vec{j}$. Just to make it "small" let's multiply that by, say, .1 to get $\displaystyle 1.2\vec{i}+ .22\vec{j}$. We want to move from (6, 3) in the opposite direction from that so subtract (1.2, .22) from (6, 3) to get (4.8, 2.78). Now repeat. Find $\displaystyle \nabla f$ at that point and subtract.

5. can you please tell me if my results are correct?
I found a=0.488 and x1=[0.144;-0.2208], a=0.466 and x2=[0.00979;0.00556]
, and a=0.4863 x3=[0.000268;-0.000388] can youi please just check theses results cause i am not sure. and i stop here i don't know if i must go on and do it again. Thank you