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Suppose f(r) = 2r? + 14r +1 Solve 2r 14c + 1 = 0 for I.-0.07217,-6.92782PreviewWhat are the roots of f?-0.07217,-6.92782PreviewWhat T-value produces maximum (Or min...

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Suppose f(r) = 2r? + 14r +1 Solve 2r 14c + 1 = 0 for I.-0.07217,-6.92782PreviewWhat are the roots of f?-0.07217,-6.92782PreviewWhat T-value produces maximum (Or minimum) value for f(z)?2 = L72.-47/2Previewd. What is the maximum (Or minimum) value of f(z)?47E2-4712PreviewSubmitQuestion 4. Points possible: _ Unlimited attempts_ Score on last attempt: 2 Score in gradebook: 2 Message instructor about this questioniCcnsSuppose the function f is defined as f(z) ar +bx + for parameters a, b and c Match

Suppose f(r) = 2r? + 14r +1 Solve 2r 14c + 1 = 0 for I. -0.07217,-6.92782 Preview What are the roots of f? -0.07217,-6.92782 Preview What T-value produces maximum (Or minimum) value for f(z)? 2 = L72.-47/2 Preview d. What is the maximum (Or minimum) value of f(z)? 47E2-4712 Preview Submit Question 4. Points possible: _ Unlimited attempts_ Score on last attempt: 2 Score in gradebook: 2 Message instructor about this question iCcns Suppose the function f is defined as f(z) ar +bx + for parameters a, b and c Match each of the following expressions with what it rcprcscnts in the context of this function: The roots of f. b. The distance from the T-value of the vertex t0 each rooL: The €-value of the maxImin value of f(z): The maxImin value of f(z). 4ac 2a 2a 4ac 2a Submit Question 5. Points possible: Unlimited attempts_ Score on last attempt: 05_ Score in gradebook: 05 iCens



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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.

Determine the distance from the line $y=x+1$ to the parabola $y^{2}=x .$ (Hint: Let $(x, y)$ be a point on the line and $(w, z)$ a point on the parabola. You want to minimize $(x-w)^{2}+(y-z)^{2} . )$

In this case we have F being a function of W. X. Y. And Z. G. And G. One. G. Two are also functions of W. X. Y. And Z. And then h his forum from F. G. One multiplied by lambda one and D. Two multiplied by land to So if we sent G1 equals zero, this is some kind of hyper plane and a four dimensional space. And to equal to zero. That's also a hyper plane four dimensional space. And we can see that F. Is the distance squared In four dimensional space. Now. Um What what this problem is is it's not looking for um the minimum distance between these two hyper planes because they're going to intersect actually. So the minimum distance between them would be zero. But what we're looking for is the minimum distance from the origin to a point where they intersect. So this has to be satisfied. This has to be satisfied. That leads to points where these hyper planes intersect. And then to minimize that, we're looking for the one the point on that intersection. Um I guess it's kind of a hyper line you think about it that is closest to the origin. Now unfortunately we can't visualize this because it's in four dimensions, but we can work through the math. And again I use Mathematica to do this. So we have F. G one G two and then we form H. Out of those. So here's our H function here and then we can take the gradient. I'm using the Mathematica is grad command and then we take the gradient with respect to X. Y Z. W. Lambda one lambda too. And so we get six equations here for six and 6 A six component vector And we sell each of those components to zero. And that gives us six equations for six unknowns. Then we asked Mathematica to solve it. And it comes up with a nice clean solution, relatively clean solution, And it only comes up with one solution. Um And that is X. W. Let's see here, that would be excess. Two thirds Y is one, nine, Z is -1 9th W is one night. And if we plug that back into their the distance or the higher dimensional distance from the origin to this particular point here in this four dimensional space Is 13/27. So that's the optimal um basically the point on the intersection of these two hyper planes that is closest to the origin.

Of earthly shouldn't be noisy. Axe equals x Y you think creating I e green GTO? This was one I minus. Okay, you hear? No, Y c Q X ecstasy. Why? Why make it is inserted into the equation here in the why cze x Explain why? Why? He's why see here x x cute. Before that we deal. We do y squared, right extremes y squared times x x cubed because X equals zero Well, acts until y squared equal to x Q After that were rearranged these pieces here we insert escorting toe Why score it into the equation here? So three squared get going one x were quoted to third X squared to seize also equal to plus where hey, and why you want to plus all right, the minimum of his supposed todo three. Wait, wait, wait

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.

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.


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