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Consider the following problem Ca Pobcinio Piv4d0 Maximize 2 = 5x1 5x2 13*3 supject t0 3*3 12*1 4*2 10x3 and (j=1,2,31. If we let *4 and X5 be the slack variables f...

Question

Consider the following problem Ca Pobcinio Piv4d0 Maximize 2 = 5x1 5x2 13*3 supject t0 3*3 12*1 4*2 10x3 and (j=1,2,31. If we let *4 and X5 be the slack variables for the espective constraints tne simplex method yields the fol Ilowing final set oi equations" 2x3 Sx4 = 100 3*3 16*1 2x3 4x4 X5 = 10. Now you are to conduct sensitivity analysis by independently investigating the following change in the origina mode" For tne following change use the sensitivity analysis procedure I0 revise

Consider the following problem Ca Pobcinio Piv4d0 Maximize 2 = 5x1 5x2 13*3 supject t0 3*3 12*1 4*2 10x3 and (j=1,2,31. If we let *4 and X5 be the slack variables for the espective constraints tne simplex method yields the fol Ilowing final set oi equations" 2x3 Sx4 = 100 3*3 16*1 2x3 4x4 X5 = 10. Now you are to conduct sensitivity analysis by independently investigating the following change in the origina mode" For tne following change use the sensitivity analysis procedure I0 revise Ihis sel of equations (in [ableau form) and convert to proper form from Gaussian elimination for identifying and evaluating Ihe current basic solution: Then test this solution for feasibility and for optimality: (Do not reoptimize Change the right-hand side of constraint to b1 Bas var EEq Coefficients of Right side Change the right-hand side of the constraint 2 to b2 BBass Coeficients of Right side Change the right-hand side t0 12 104 Bas var |Eq Coefiicients of Right side



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Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema:
a. Form the function $h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},$ where $f$ is the function to optimize subject to the constraints $g_{1}=0$ and $g_{2}=0 .$
b. Determine all the first partial derivatives of $h$ , including the partials with respect to $\lambda_{1}$ and $\lambda_{2},$ and set them equal to $0 .$
c. Solve the system of equations found in part (b) for all the unknowns, including $\lambda_{1}$ and $\lambda_{2} .$
d. Evaluate $f$ at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise.

Minimize $f(x, y, z, w)=x^{2}+y^{2}+z^{2}+w^{2}$ subject to the constraints $2 x-y+z-w-1=0$ and $x+y-z+w-1=0$

In this case we have um we're working in three dimensions so we can actually visualize this and I'll talk about that in a second. But what we're doing here is we're giving these two G one equals zero and G two equals zero. So those defined um surfaces in three dimensional space. This is just a cylinder and this is kind of like a felt wrote hyper hyper hyperbole of revolution or something. Um It's again you can see it down here in the yellow. Um And then what we're doing is we're we want both of these to be satisfied. So that means we're looking for points where these 22 functions or surfaces intersect and want to find points where those are minimum. Um the minimum distance to the origin by minimizing this function, this distance function or the distance squared. But if you minimize the distance squared you also minimizing the distance. It's just that when we plug in um I guess I never did plug in plug in these values that we got for. It looks like the actually the distance squared is one. So the distance is just one for the minimums. Yeah, I did program and actually so anyway in Mathematica again I defined F here G one and G two and then constructed H from those and the lagrange multipliers. And then I um took the gradient with respect to X. Y. Z. Lambda one, number two. So that gives us 55 components in our gradient. And then I said each component to zero. So that gives us five equations and five unknowns. And you can see here we just have you know, G one and G to produce when we take the grading, when we take the direct respect the lambda one in 92 we just get this equals zero. Is that this equals zero back. So we know that um oh the solutions must be on a point on points that were these two surfaces intersect. And we can find looking for all the real solutions. We see we get a bunch of them, we get 12345678 real solutions. And then so we can plug all those back into f and we can see that the first four solutions all have a distance squared of one, and the second floor solutions all have a distant square of square root of three halves. Um So we're looking for minimum, um It's the first four solutions that we're interested in, and so if we plot those, so here's our two surfaces and you can see that they basically kind of intersect, it looks like a circle, but it's, you know, obviously warped, so, you know, kind of a warp circle here. Um and this point here, this point here at this point here and the one that you can't see back here, those are all the ones that have 10 minus 100 zero minus 10 010 or 100 Those are those points here, and those are the ones that correspond to the distance from the origin being for one. And it's these points up here and here and here and then another one down here. Those actually are maximum, you can see that that that these are the maximum distance. Um Any point on this intersection curve is to the is from the origin. So those are those are the points where we actually have a maximum of the distance. So if we're looking for the minimum, it's just these first four here, the ones basically on the plus or minus X and y axes.

So we were asked to use a computer algebra system. I use Mathematica here um to find to minimize dysfunction here, F equals X times Y plus Y times Z. Given these two constraints here. And what these constraints are are basically uh Well, they're cylinders, it has two cylinders perpendicular to one, another one with an axis in the Z direction, along with an axis in the Y. Direction. So um let's see here. Well we construct each, so here's our age function. And let's see here we have, you know, it's just a function of X. Y. Z. Lambda one, # two. So we use Mathematica is gradient function and take the gradient with partial with respect to X. Y. Z. We love to one lamb to two. And we get these five equations setting those equal to zero, that we get our two constraint equations back. And then we said these 20 And when we solved that, so I said to make with a zero using this little um kind of notation here, but I'd basically mapped Equals zero over all of these things. So I got an array of equations here and then I can solve them. And we see even if we assume we have real numbers, we can see we get 12345678 solutions. So we have eight solutions and we can see that they're kind of plus or minus. Um they come in kind of plus or minus pairs, Right? So plus plus minus minus, plus minus plus plus, you know? And then minus minus minus plus plus. So and then so some of these are all, there's two at minus Z equals minus 12 as equals one And four z equals zero. So you can kind of see that there's some symmetries here. Right, So all the values are the same or close to the same. Um what did I say minus? No, They get -1. Zeke, was this value and plus or minus minus minus plus plus, um Plus plus minus minus and then our Lagrange multipliers which we really don't care about. Um So we can again here's all of our solutions um and we can we can plug in pulling all these solutions into our function and see what we get and so we get values that either it looks like square to to know that's not square did too. Anyway. Um one plus, quarter to maybe. Um so we get -2.4 Plus 2.4 -2 plus and then we get these .4 ones. Um Again basically too much less than this. Uh So we can see that that our maximum and minimum are here and here, but they actually occur at at two different points. So the first one is this is a maximum here, This X, Y. Z. Here, um that's a minimum. This is a maximum, this is another minimum and another maximum here and then the rest are not. So what's going on here? Well, I tried to make a plot of this, but it gets really ugly. But here we can see kind of what's going on. These are all level sets of X, um X times Y plus Y, times Z. So these are various level sets of those, and I and I plugged them in, I plugged in the solutions. Um and, you know, I plugged in, let's see the level sets I made Yeah, I made level sets of the solutions. And so here's our one cylinder, here's the other cylinder. And then these balls here are where we have solutions um back here. And so what we can see is that obviously these occur on the intersection of the two cylinders and then the, you know, the level set of the function f you know what this value I think is is uhh You know the largest one solution Here, I think that's what that one is. And then we have some other ones like back here. So there's lots of symmetries here and that's what you can see here, that here, the other levels, that's kind of in the interior. Um These are where the other um extremely we got work, but they weren't maximum or minimum. So again, basically they, they all occur on the intersection of these two cylinders and then whenever, you know, the level sets of Hongshan, our largest, and so that we get there and there. And so again, it's kind of hard to see, but that's what's going on here. Um I'm not sure any other physical explanation or geometric explanation for what these are. Level sets are, but what we're looking for is the maximum of these level sets On the intersection of these two cylinders.

Everyone directions that we have in this example, Or if thinks I'm going, you quit X squared Those two words square with the mainstream functional G off X roi equal four x minus six Oil because 24th together around immigration forms. Responsive. If my most longer G it was zero, this will mean partial differentiation. Off X will be equipped with a bunch of foundation off X, and the first concert is that for excellent boy will be to Ace Mine's London abortion, but by the person from Station Off fixes the second function, which is four. This will give us two X equal for long, full of the same concept. Departure from fishing before I will be able. Full boy brainless longer nearly six. This should give us four Loy because negative six lot is our first equals for our seven requirement. If we take X, people presume this will mean that longer people zero. Which means that worrying before you embrace verse. That's that's mean. If if X is equal to zero or before this court to resume their long live will be because you ends Expo on the corresponding expert work with the old speak one is our second for ourselves permanent if you took If we take the two functions which is two X full longer and for warring equal negative six long in front of different The first equation by to worry This will give us for X roi equal eight. War longed up and we want to. But second equation Boy X four x warning The equal negative six x long by taking unequal for the two equations in boy Lunda were with *** to six x longer and this will give us for one equal new to ST X Way Will takes tourism and implement is ah Jeanne fix on droid. We would have four X minus six. Deploy negative street over four X Peak 1 25 This will give us an X equals two 25 over a On was there so we would have you, Lloyd equal. Negative. 74th over 34. This is our critical point to see if, uh, this critical boy is corresponding to a minimum or a maximum value of f. We will apply the slow. Really? Which leaves that. Are this club off the? Is there slow? 00 G fix genetics. Enjoy function one over X, which is the equivalent negative three or four you see over full. This means if we moved a distance off X, we will move in distance before new. Expect four on by going Negative street. This way loses dysfunction tohave another point on the line running away from our critical point which would be our X to x two equal 649 and going to equal negative voice. I want to also can have an excess tree equal then for mine. And Boy Street equalled military in Windows by implementing these two values off X and boy into the function f x If x ongoing, we we have and if one equals two 18.4 if two equals to do if she because Because I'm sorry, if two will be equal 102 want some dissonance f street with the bull 254 foreign Dismiss. This means that by running away from the court from the point of the critical point that we got from the ground integration, a differentiation we way is the value of his F function increase which require that they be is corresponding to a minimum value off. If and this is our answers, thank you


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