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fa272
 3 years ago
minimize f(x)=(x12)^2+(x25)^2 subject to g1=x1x2+10<=0 and g2=2x1+3x210<=0
fa272
 3 years ago
minimize f(x)=(x12)^2+(x25)^2 subject to g1=x1x2+10<=0 and g2=2x1+3x210<=0

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fa272
 3 years ago
Best ResponseYou've already chosen the best response.1It's under the single objective optimization problem

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1Why? is the question not clear?

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1I need the steps o solve the problem

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0Make 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.

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0there 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.

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1isn't the f(x) result a single number or a coordinate point (x,y)?

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0yes, I think... it is a single number result based on the inputs (x1, x2)

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1you mean the final number will be the intersetion between the two lines?

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0No, 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.

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0just for fun, it looks like x1 = 20, x2 = 0 falls in the allowed region. So... f(x)=(x12)^2+(x25)^2 f(x) = (202)^2 + (05)^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.

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1the constraint regionis the shaded part on the right?

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0yes, 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 (02)^2 + (105)^2 = 4 + 25 = 29 this is much better than 19^2, but still may not be the minimum f(x)

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1so do I just guess some points for f(x) to get the smallest number that fits in the shaded area?

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0well, 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.

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0So (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

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0If you are struggling at all with what this optimization means, an easy example might be: f(x) is my # of questions missed on the exam. I want to minimize f(x). f(x) is some function that depends on "hours studied" and "hours of sleep". I can maximize my study hours if I study 24 hours a day, which then makes sleep necessarily zero, but this may or may not optimize (meaning minimize) my "# of questions missed" score. Then again, I could sleep 24 hours a day and study none, and I might score well, but I doubt it. I need to choose the optimum study/sleep balance to minimize my missed questions.

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1I got the idea ... I just want to make sure I understand the steps: first I take the constraints and make them equal to either x1 or x2 second I'll plug numbers for each constraint equation to form a line and get the area where the solution exists third I take the f(x) equation and solve it for different numbers under th area to find the minimum solution Is this correct?

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0I think you got it... the way you phrased your first step confuses me a bit, but if you're saying basically "graph the inequalities represented by the constraints", that's it.

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0don't let the g1 and g2 confuse you... think of those as labels or names... "constraint # 1" and "constraint # 2" The actual constraint is an inequality using 2 variables, x1, and x2... for convention, I would treat x2 like the y variable in a normal xy graph (i.e. make x2 the vertical axis)

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0Is this for an engineering class? Or straight math?

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0I found a PDF handout from a Purdue site online a bit ago... had to teach myself this stuff too :) It's attached if you want it for reference... it's more broad than just this one question, but it might be helpful.

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1Thanks that is helpful. Sorry if I'm asking too many questions but do you know how it can be programmed in Matlab?

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0I don't mind questions... love to help, plus I learned something also :) but unfortunately, I don't know Matlab at all... would love to help but that's not an area I can do. You might rephrase the question to focus on "how to program in matlab" and repost it as a new question... might get some fresh help from someone with Matlab skills. Also, I think there are some engineering forums too... might "cross post" it in one of those if you find one that looks promising.

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1Thank you so much for your help... I really appreciate it. This is my first time in this site, so I'm trying to learn the interface and features around here too :)

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0good luck with the site... btw, you can "join" multiple areas... look at the blue bar over "ask a question" and choose "find more subjects"... you can add physics or engineering or whatever without quitting the math page... you can jump back and forth quickly and get updates on both. I'm pretty new to this too, but it seems to work well. Lots of early math questions here though... I think there are a lot of home school and online school kids asking some really basic stuff, and worse, some are just wanting someone to hand them answers. I am glad you were willing to take the time to try to learn this optimization... it was refreshing to help with a problem that made me think pretty hard :)

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0also, for what it's worth, you can reward people who help you by clicking the little "best response' button beside the person's response in the thread above... that gives them a "medal" for helping, which adds to their score and helps identify them as someone who is generally trustworthy for help... i.e., better to get help from a "99" guy than a "6" guy... if the 6 guy is really good, he'll get medals and "move up the charts" in his score... score is like "math cred"

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0Except I have no idea how long it takes to get up to a 99!!! I have helped a bunch and my rate of change of score is slowing :) Fast progress at the start

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1I just reliazed I have another question, what if I happen to have two f(x) f (x)= (x12)^2 + (x25)^2 f(x) = (x14.5)^2 + (x28.5)^2 with the same constraints is it the same process?

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0but it's taken a week to go from like 60 to 64

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0What is the actual question on that one? Is it minimize f(x)?

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, it should be the same process, but I am confused a little by how you can name a function f(x), define it in terms of the inputs, and then name a separate function the same name f(x) but give it a different definition in terms of the inputs. Algebra would say f(x) = f(x), meaning the 2 expressions should be equivalent, but at a quick glance, they don't appear to be. I would just follow the same process of finding the min f(x) using the x1 and x2 constraint region... do it once for the first f(x), and then repeat for the 2nd f(x).... basically find min f(x) on the first expression, then min f(x) using the 2nd expression, as if it was 2 separate problems. But that's a weird way to ask it... I wonder if I'm missing something... (?)

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1yeah sorry the first is f1(x) and the other is f2(x)

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0Are you supposed to minimize each one individually? Or somehow deal with them both together?

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1it says in the book that they used the pareto optimal set but don't know what it means and they don't solve these equations mathmatically.

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1attached is the graph that is suppose to be for this problem and I need to know how they ended up drawing it like this

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1I think we need to deal with them together because it's mentioned under two objective optimization problem

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1This is the link for the book: http://caemm.zxq.net/em503/EM503%20%20Introduction%20to%20Optimum%20Design%20%20Jasbir%20Arora%20%202%20ed.pdf and it's in chapter 17

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0yep, I agree... unfortunately, you just moved past where I have much idea. I'll read about it, but at this point, I'm like someone walking into your class midterm and trying to catch up :) one last thing for now: notice that the plot in that picture is f2 vs. f1, so as you put in x1 and x2 values, you generate f2 and f1 results, which you could then plot against each other like they did in that drawing. Doing this by hand will start to be a serious pain though... a tool would be very useful (Matlab) Thanks for the textbook link... I will read up on it, but it will take me a bit and I have to log off here in a few minutes... I will check back later.

fa272
 3 years ago
Best ResponseYou've already chosen the best response.1ok.. thanks for your help... appreciate your time :)

JakeV8
 3 years ago
Best ResponseYou've already chosen the best response.0It was interesting... makes me wish I had taken this class :) Good luck!
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