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5. Use the method of Lagrange multipliers to find the maxlmin of a. flx,y) e2ry subject to the constraint x +y = 16....

Question

5. Use the method of Lagrange multipliers to find the maxlmin of a. flx,y) e2ry subject to the constraint x +y = 16.

5. Use the method of Lagrange multipliers to find the maxlmin of a. flx,y) e2ry subject to the constraint x +y = 16.



Answers

Use the method of Lagrange multipliers to optimize the function subject to the given constraint. Maximize the function $f(x, y)=16-x^{2}-y^{2}$ subject to the constraint $x+y-6=0$.

We have F equals X times Y subject for the constraints that x squared plus y squared equals 16 or x squared plus y squared minus 16 is zero. So this is uh this means that x and y lie on a circle of radius four. Now our augmented equation is G equals X times Y plus lambert times X squared plus y squared minus 16. Yeah, take partial winters back to X. We get to lander X plus Y. Set that equal to zero. Take partial with respect to why we can X plus two lambda. Why set that equal to zero? Take partial with respect to the lambda. And we get C And that's equal zero. So now we have um you know, we can basically, so you know eliminate um say why and lambda from these uh get Y and lambda in terms of X from these two and then plug that into here and then get an equation for X. And what we see is we get we get to values um two solutions for X plus or minus square, two times the square to to and then we wind up with two values for each value of x. We get to values of why so plus or minus square to times square to to So in the end we wind up with four solutions and if we plug those in, you see that the solutions where we have the same signs in X and y are eight And we have different signs there -8 obviously. And so we wanted to maximize that. So we have to, we have to maximum and we have this is a maximum and this is a maximum and the and these are minimum.

For this problem we are asked to use the method of lagrange multipliers to optimize the function F. Of X. Y equals X. Y. Subject to the constraint. Uh X squared plus Y squared equals 16. Where particularly here we want to find both the maximum and minimum values. So we first set up Arlo Grandjean function. So that will be equal to X, Y minus X squared minus Y squared Plus. Yes it would be plus 16. Then we want to take the partial derivative of this with respect to X And set that equal to zero. So it'll be why -2, X equals zero. And then with respect to why? Ah actually need to be careful here. That should be negative lambda X squared and negative lambda Y squared plus lambda times 16 or plus 16 lambda. So we have y minus two lambda X equals zero. And then we would have x minus two lambda Y equals zero. So we can see then You have y equals 2 λ X. And then substituting that into our second equation. We'd have X Equals two times lamb or a to lambda times two lambda times X. So we'd have X equals four lambda squared X. Which means that we divide out the X. And we have that for lambda squared equals one. So lambda squared is going to equal 1/4 means then that we'd have lambda Equals one Over Root two. So we have why? In terms of lambda index and we have lambda. So that also means that we have, we can express why equal why as to over root two lamb debt or two over root two X. So that would be just route to X. Then we want to substitute that into our constraint equation. So we'd get that X squared plus two X squared equals 16. So we would have then that three X squared Must equal one moment here. Oh, I thought something looked funny. I need to correct myself here. That should be lambda equals plus or minus 1/2. So we'd get then that Y is going to equal plus or minus X. But since we want to have a well actually I'll just note that Y equals plus or minus X. Which then means that when we substitute this in we'd get that two X squared equals or uh yeah, it would be just X squared plus X squared equals 16. So X squared is going to equal eight which then means that X is going to be equal to plus or minus the square root of eight which is four times two. So that it can be X equals plus or minus two route to. So having that we then need to just evaluate our function at those points. So we would have clearly a maximum will occur at Positive two or a positive. Yeah, it would be at two route to to root to be where we have our maximum We have four times to giving a value of eight and then at the point um All right. It as plus or -2 route to then minus plus and actually that first one should be plus or minus two or two and then we'd have minima at plus or minus two to minus plus two route to. So they should be opposite signs, in which case we have a value of -8.

That we're going to solve proper number six have different thank right people about ex wife execute plus like you it could 16 the other Function off X Come All right Call to X cube plus y cube minus 16. So x cubed plus X cube equal to 16 If with respect to X jean with respect goods thought Lambda F flight equal to conceive thought Lambda execute. That's like you equal toe 16 Why do you know the products? Light Ricardo three x a square Don't land ex Hindu It'll products like or the three y squared Daughter Atlanta x y you doc, You don't put ex wife. You got three x cuba, nor to land. Now exploit Got into the products like you goto three like you would not lambda thing The ex cure don't lambda Equal toe three y cube will not lander x equal toe But then x cubed plus like you recall toe 16 two x cubicle toe 16 take sequel toe Been executed too. Light is equal toe so functioning to come out to is equal to Europe or to in due to which is equal to it about four. So there is extreme value, for example to do and where you could toe half off zero comma que brute off 16. It opposes zero in tow. Children off 16. You don't pose zero to go to what so maximum? It's next. So if you do about four for executive too and why you going toe? Thank you.

For this question, We're going to be considering the extreme points of F on the constraint G equals zero. So we're going to find the Grady INTs of each and our deport. Our ex grading is going to be, uh, one equals two x minus Y lambda are Why is when equals to one minus? Exclaimed uh, an er ze is one equals two, said Linda. We can set these equal to each other by dividing through and we wind up, uh, canceled. Her lamb does, and we'll get to one on his ex close to X minus y where X equals y if we want. Then we can plug our X equals y into our first lined equation on when you get a one equals X Linda, which is now directly comparable with R one equals two z lambda, which means that we can set X equal to two Z after dividing through Oh, are Z equals X over two. Uh, after that, we can plug those values into our constrain equation. So we get X squared plus X squared minus X squared plus x greater before equals five, which gives us X equals positive minus two. All of the others air relatively positive two x So we wind up with two solutions X equals And so we get to comment to come a one or negative to common negative to come A negative one. Uh, the values If if these give us our five and negative five, which means 2 to 1 is our maximum point and negative to negative to negative one is our minimum boy.


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