Question

minimize f(x)=(x1-2)^2+(x2-5)^2 subject to g1=-x1-x2+10<=0 and g2=-2x1+3x2-10<=0

Answer 1

It's under the single objective optimization problem

Answer 2

I dont get you

Answer 3

Why? is the question not clear?

Answer 4

Can anyone help me?

Answer 5

I need the steps o solve the problem

Answer 6

Make a coordinate axis "thing" with vertical axis as "x2" and horizontal as "x1".
For each of the two constraints, simplify so that you solve for x2 on one side, and the form will be like x2 >= mx1 + b so that you can sketch the inequality on your graph. Shade the appropriate region based on the inequality.

So this will give you two constraint regions...

I think, then, the process is somewhat of an iterative trial and error of looking at x1 and x2 combinations that are "allowable' (meaning they fall into the overlapping regions of the 2 constraints), and choosing the x1,x2 pair that creates the minimum f(x) result.

Answer 7

there may be a straight analytical method, but what I have seen in the last few minutes points toward iterative approaches. Not my favorite, but maybe that's the right idea here.

Answer 8

isn't the f(x) result a single number or a coordinate point (x,y)?

Answer 9

yes, I think... it is a single number result based on the inputs (x1, x2)

Answer 10

you mean the final number will be the intersetion between the two lines?

Answer 11

No, the f(x) value is the single value that results from choosing x1 and x2 inputs from the shaded region above, plugging them into the f(x) expression, and just getting the number result.

But different choices of x1 and x2 give different f(x) results... you need to find the x1 and x2 that give the SMALLEST (i.e. minimum) f(x) result.

Answer 12

just for fun, it looks like x1 = 20, x2 = 0 falls in the allowed region.
So...
f(x)=(x1-2)^2+(x2-5)^2
f(x) = (20-2)^2 + (0-5)^2 = 18^2

Or you could try x = 21, x2 = 0 -->> f(x) = 19^2 so that's bigger, and not the minimum... better try searching smaller values of x1 and x2, but you can only choose from within the constraint region.

Answer 13

the constraint regionis the shaded part on the right?

Answer 14

yes, the constraint region is the shaded area that satisfies both g1 and g2 constraint inequalities

Another point I just checked is the point where the two constraint lines hit the x2 vertical axis is (x1,x2) = (0,10)

f(x) at that point is (0-2)^2 + (10-5)^2 = 4 + 25 = 29 this is much better than 19^2, but still may not be the minimum f(x)

Answer 15

so do I just guess some points for f(x) to get the smallest number that fits in the shaded area?

Answer 16

well, that's the part that I don't really like, but yeah, I think that's the approach. But don't just guess randomly... think about closing in on the target... like a kid's game of finding a lost object with "hotter, colder' responses... adjust your next guess to take advantage of the results of your last guess.

My intuition is that the minimum f(x) will result from an x1, x2 pair that is very close to the left side of the shaded region. However, because the two inputs cause different "contributions" to the overall f(x), it could be that choosing the smallest x1 is NOT the best choice... maybe choosing a bigger x1 than 0 will allow a smaller x2 choice which will then lead to a lower overall f(x) value.

Answer 17

So (0,10) is one guess, but also try some points where x1 is not 0... small steps to the right, and maybe downward from (0,10), like (1,9)... just be sure to stay inside the shaded region