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Numerical optimization An important problem in numerieal computing is finding minimm * of [unction f (c). This is very mch related to the root - finding problem Ind...

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Numerical optimization An important problem in numerieal computing is finding minimm * of [unction f (c). This is very mch related to the root - finding problem Indeed. if f is diflerentiable. then minit 1* o f () is poiut at which the derivative f' (1) = 0 Unfortunately. approach based on applying the bisection method to f'(r) Mdy nol work since the derivalive values mnay nol be available in practice: Fortunately; there 15 algorithin similar to the bisection method can be used t0 find

Numerical optimization An important problem in numerieal computing is finding minimm * of [unction f (c). This is very mch related to the root - finding problem Indeed. if f is diflerentiable. then minit 1* o f () is poiut at which the derivative f' (1) = 0 Unfortunately. approach based on applying the bisection method to f'(r) Mdy nol work since the derivalive values mnay nol be available in practice: Fortunately; there 15 algorithin similar to the bisection method can be used t0 find the mini usiug values of f(1) ouly: Recall that the bisection method produces pairs of numbers [an.4 bn]: For finding Iinima will instead produce sequence o triples [Un.ba. Cn] that have the following bracketiug property f(an) > f (ba) and f (bn) < f(en) Hence ba can be used as an approximation t0 the minit the algorithm proceeds in the following war: aL step ". To compute sueh triples_ Choose HCI point using the formula: ~ (Ca bn) if (Cn bn) > (bn 7(4n ba) if (Cn bn) (bn Un) Update the triple using the formula: [a,, = ba] il _ < bn and f() < f (bn) [bn.x , Cn] i $ > bn and f(1) < f (bn) [r,bn, Cn] if < bn and f(r) > f(bn) bn.1] il I > bz and f(x) > f(bn) Jaa+" bn+1l, Cn+1] You may wish t0 verily that the update formula guarantees the bracketing property (1) holds at each step of the algorithm: Note that the choice o _ in (1) will allect the convergence rate. is possible t0 prove that the optital choice is ~ 3_W You should use this value throughout. Your task in this assignment is to study this algorith: First. wrile script t implement the algorithm: Once YOU have done this run it for N 100 iterations OH the fuction cosk) 1+12 f(c)



Answers

SOLVING OPTIMIZATION PROBLEMS USING CALCULUS

Okay, okay, so we are going to resolve this probability problem. And the first question says, let's see what the first person says here. It says, given that information, find the probability to report a probability that did not develop appears in a randomly selected photo. So we want the probability of an antelope. That's the first question. Now, if we go to our probabilities is listed, we are given only the probability of a complement of an antelope. So to get the probability of an antelope we say one minus its complement probability not an antelope. And this would be one minus several 0.70 Which then gives 1.30 And the second report says report the probability that the jackrabbits appear when a randomly selected photo, given that my daughter appears on the photos. So basically it's probability. But uh jack rabbits appear given that daughter appears. So this would be equal to probability of uh the intersection which is J. And D. Divided by the probability that it's a daughter, which is probability D. So this would be equal to what is J. N. D. Seven point several eight. Divided by probability that it's my daughter on the photo. It's zero point 64 And uh this would give set off 0.125 Okay, then the other one is. What is the probability that a bison api is given that the jack rabbit appears? So we want to find the probability all Beeson appearing given that the jackrabbit already appears. So this would be equal to probability almost B. In the section J divided by the probability of chair. And then this would be equal to. So what do we have here? We have the probability that the Czech rub it appears, which is 0.18 So we'll have P. B N. J divided by 7.18 Okay, So if we do have uh we have to work out B and J now to work that one out. Okay so I'll just leave it here and go and do a separate working to get B and J. So B and J. I can use J and B as an example. So probability of Mhm. Oh off sorry. The probability of G. B. Yeah or J given B is equal to the probability of J N. B. Which is something that we want here uh divided by the probability of a bison. And then what we want is this so we make it the subject. So probability of J. And be is able to probability of J. Given B. Applied by probability of the and then this would be equal to what is the probability of J. You can be 7.3 75 Uh multiplied by probability of being probability of B. We are given the compliment so one minus 0.8. It's 0.16 And if we put it in the cock litter that 0.375 multiplied by 0.16 Uh This give us several point several six. Set now this is what we want in order for us to calculate our probability of uh what is it be given? J. So let's take it down. We are taking this one down here. So the probability of B. Give and K. Is equals two probability of mhm P intersection G. Which is now worked out 0.60 divided by probability of che which is 7.18 And then this will give us 0.333 Then the last question relates to its unfortunately it was not copied. How if I can put it as A. D. Here it relates to uh if if three cards are selected a. Trundle what is the probability that it's A. D. D. D. Regardless of the fact that there is a wildlife or not? So what is the probability that will pick three daughters? So from the given information we can see that probability of picking your daughter is 3.64 So we want to find how many cards translate to that. If there are 50 cards. So how many cards are? 0.64 Yeah. Uh So it's celebrating 64 um settle play 64 Uh huh. Timeless 50. Yeah it's 32 cards. So there are 32 cards and they are being picked without replacement. So the probability of D. D. D. Is equals two 32 or 50 for our first card. Second card they are now that one cards uh with that one daughter cards. But now there are 49 cards left and then 30 but there are also 48 cards left. So this would be 32 times that you want, 2030 all divided by 50 times 49 times 48. So the probability will be zero point 253

We're going to find the point P of Xy on the line. Y equals two. X plus three. That is closest to the origin. We will talk about some points here. Where is the objective function? Dfx? What is available? We are going to be controlling. We'll call it X. What function modules? Security. This problem. What are the coordinates of the point satisfying the statement of the problem. And how far is that point from the origin? We are going to sketch the line also and market the point P mm hmm. On that line. And also the origin. So the spread line is why it was to explore three. Yeah. Mhm. Okay. Okay. So we know that when we have X equal um zero. Why equal three? So we know at this point is on the line at this point here and when X equals one for example, mhm. Why goes five? So it's gonna be upward? And we can say that for example negative one to interview one. Negative one is negative, two plus three is one. So at this point here get to be on the line. So the line get to be something like this or less. So we can say that the point is somewhere around this on this line and we had the origin here. So we want to find the point there is closest to the urgent on this blue line that is closest to the origin. So it's going to be around. Remember that the minimum distance corresponds to the uh or total distance. That is the point that is closest the point on the line that is closest to the original form uh 90 degrees between the light, the blue line and the line joining the point that is closest to the origin and the origin. Yeah. Mhm. Okay, that is the geometric situations. So mhm. Uh we are going to write the distance between a point that is on this line and the origin. Remember that? The distance between the points X one, Y one and X two, Y two is giving us the square root of X one minus x two square plus white one minus way to square. Yeah. And we want to find the minimum of that function. But uh we can say that the minimum of dysfunction is a team where the expression that is inside this group is minimum. Mhm. So we can say that the problem corresponds to minimize the expression side the square root. First of all, we are going to write the specific function we have here specific function is the distance from a point X, Y Z are defined this equation and 00 the origin. And this distance is equal to x minus serious square. There is x squared plus y minus serious square. There is white square. Mhm. And okay, we can say then that the function D of x Y equals square root of x square plus y square is our objective function and Y equals two X plus three. That is a condition that the point get to satisfy. So we got to minimize this function. Yeah, subject to these conditions. That is a problem we have and we can say that this problem is equivalent to the problem of minimizing this. Yeah, sorry, here, I forget the square minimizing the is subject to these conditions. That is the function square root of X squared plus y squared is going to be minimum when x squared plus y squared is minimum. So this is the optimization problem. We have this one, minimize this function subject to these conditions. And we can reduce this problem to the problem of one variable by using the fact that why I got to be equal to two X plus three. So why it was two X plus three implies that the can be viewed as a function of one variable eggs which is available. We are controlling the X coordinate of the point on the line and D. Is going to be equal to X square plus to explore three which is expression for Y square. So this is X square plus four X square plus 12 X plus nine, which is equal to five X square plus 12 eggs plus nine. So the problem has been reduced to minimize D of X equal to five X square plus 12 fakes plus nine subject two eggs on the line. Which is implicit in the fact that we use these condition for the second coordinated the point. That is we have in fact no restriction tina minimize this function and in this case X can be positive or negative. There is no restriction on that because the point had to be on the line. The important thing is that the point is on the line which is implicit in this substitution we had made here right, mm hmm. This one. So now to solve this, we get to find the derivative of tea which is 10 X plus 12. And distributive zero. If and only if 10 X is equal to negative 12, which is equivalent to X equal to negative 12/10. That is X equal, yeah, negative 6/5 mm or negative 1.2 if we want. So we can see that X is negative value. And that is consistent with the fact that in this graph, which is not perfect. Give us an idea. We saw that the point where we must have the perpendicular distance to the origin is a point with a negative coordinate X coordinate, so why must be positive instead? And that we will see later. But for now we're going to just to justify that. This is in fact the minimum value of the And we can say easily here because we can recognize here Parabola opens upwards because the coefficient, the coefficient of X square is positive. So this corresponds certainly to a minimum value of T. But we can see another way is to find the second derivative which is 10, which is always positive independently of the value of X. So this is a concave of function like this. And there is a minimum value for this function which is attained at X equals negative 6/5. Yeah. So we can say, yeah, he is the value where the of X has uh minimum and this is a global minimum. That is an absolute minimum because of these properties. So here function is always concave up. Yeah, that means there is only one minimum in the whole domain on the real numbers. So uh now we find why using the equation of the line, that is two, X plus three. Okay, Why equals uh two X plus three, That is two times negative. 6/5 plus three. That is negative 12 +45 or three. That is negative 12, 15/5. That is three or five is a positive value. So on the graph, yeah. Then the point Yeah. P Yes, negative 6/5, 3/5. That is 1/5 times upon eight of 63 Which is um where the distance to the origin or we're going to see better is the closest point to the origin on the line. Mhm To pds the closest point to the origin which is on the line, y equals two x 43 And now we want to find how far this point is from the origin. Okay. In that we find simply by calculating the distance from this point of the origin. No mhm 96 of the five 3/5 to the origin. And we saw that this square root. Now we got to take this because the square root, because we are calculating the actual distance. So we get to use the formula to distance that is the square root of negative six of five minus zero square plus three of the five minus serious square. That is square root off six square over five square plus three square over five square. That is one or five square root of six square three square. Uh That is 1/5 squared of 36 plus nine. One of the five squares of 45 mm and that is equal to 1/5 square root of nine times five square root of 93. So it's 3/5 square to five pes three over 500 to 5 apart from the original. So this is the minimum distance possible to the origin from points lying on the line Y equals to explore three. So that's the point place around here. And uh as we can see this problem was originally originally an optimization problem with two variable foot, we can reduce it to an optimization problem problem of one variable using the condition from which we can solve for one of the variables and stay with only one variable. And besides that, we have made the simplification of taking the radical expression instead of the whole expression with the square root. Because we know that the function with the square root or the distance is going to be minimum if the expression inside the square root of minimum. So we have done this to step. So originally disease the optimization problem. This is the first simplification, making the observation that these two functions has the minimum value at the same point. And then we have reduced the problem to one variable. Using the condition where we can solve for one of those virus. And finally it was solving the optimization problem in one variable, we find that the point that is closest to the origin and which lies from the line way to to explore. Three is the point uh negative 6/5 and 3/5. That is x is negative 6/5 and y is three over five. And the actual distance from that point to the Aborigines, three square to 5/5

In this time we have an area of 625 and we want to find this rectangle such that the perimeter is a minimum. And so we can we know that the area of a rectangle of the straw picture. Uh Here we have our hyped and here we have a wet uh so our area is pitch times W. So we know that each times W is going to have to be 6 25. Now our perimeter we know is two times the height, plus two times the whip. So we want to try to get we need the area uh to be the 6 25 and we want this preliminary such that the area is at a minimum. And so now what we can do um since we want to find a minimum, we need to take a derivative but we have two different variables last year and it's easiest if we have just one variable. So what we can do, we can solve this equation to be, It is a great to 6 25. Oh they w and then we can just plug this into this equation. So here we get our preliminary this is equal to two times 6 25 divide by W plus two W. So now we have an equation with just one variable. And sure now we can find the minimum by taking the derivative. So D. P. With a D. W. Mhm. So we can take the derivative of this first term. So that is negative two times 6 25. Which we can actually write the value of that. Yes. So two times 6 25 is 12 50 have W to the negative too. And then we can take the debate of this time which is just plus two. Yeah. And so if we want the minimum we set the secret at zero. And so now we can just talk for W here. So if we do that we can add this term to both sides. And shall we get please 50? And then we can multiply by W. Square as you quit to W square. So this is the same thing is W is equal to 12 50 divided by two. And then we take the square root, so that's the square root of 6 25. Mhm. Yeah. Which is equal to 25. Okay so we got that W is equal to 25. And so now we can use that to solve for height. So we know that uh eight times I went is equal 6 25. So that means we have 6 25 times um sorry equal to 25 times eight. Which is the same thing as same H as you're 25. Yeah. Yeah. So we know that our height is 25 are with this 25. So that means we actually get a square here. And so now the last thing we can do is we can find what our perimeter is equal to. Um We will have 25 on every side. So that means I have a perimeter is equal to 100. So that means if we have an area of 6 25 the minimum our perimeter can be this 100.

You want to find the point on the proble why you push back squared? That is closest to the 0.0.0 comma zero. Oh. And so this is actually somewhat easy because we know that when X zero, Y. Is zero. So the 00.0 come zero will be exactly at the point coming zero comma zero. So we actually already know what the answers. But we want to try solving it with tuberculosis to practice doing optimization. So to find the distance of a certain point 20 comma zero, we can use the distance one which is X squared plus rice spread square root. And so this formula comes from just applying the pythagorean theorem. And so now here we know what why is though? So we can plug that in here. So we get X squared and when you plug in for why we have X squared and we square that. So then to get extra before, so now we have an equation just in terms of X. And now to um find the optimum value, we can take the derivative. But now one kind of cool trick we can use uh is that whatever the minimum value of D. S. That will also be the minimum value of D squared. So if we square both sides of the equation, we get this squared is you called two X squared plus X to the fourth. And so the reason why we squared both sides. And there's this little trick is just to realize that um if we don't have a square root, taking derivatives will be a little easier. So now we have the derivative of distance with respect to acts as you called to axe plus X to the fourth. So the derivative X to the fourth, which was flexed to the third. And we want to find the minimum value so we can set that equal to zero. And so if we sat this liquid is zero um we find that um xia can be zero or we can see if there's any other solutions. So we know that extra zero. But also if we saw that we can get or X squared is equal to negative two, which is the same thing as X squared is equal to negative one half. And so for swearing a value and we get a negative, this would give us a imaginary values and so we don't care about imaginary value. Still we just care about the one real value which is X equals you. And we can use that in combination of work with our original equation to get that Y you got to zero squared which is just zero and show the point that we're looking for. It's just zero comma zero. Which is what we said at the beginning. Just by looking at it from a geometrical point of view.


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