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Where should the point $ P $ be chosen on the line segment $ AB $ so as to maximize the angle $ \theta $?

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03:59

Wen Zheng

01:07

Carson Merrill

10:50

Aparna Shakti

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

Beatrice J.

December 16, 2020

How can you do this same problem but as to minimize the angle theta?

Oregon State University

Baylor University

University of Michigan - Ann Arbor

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A review is a form of evaluation, analysis, and judgment of a body of work, such as a book, movie, album, play, software application, video game, or scientific research. Reviews may be used to assess the value of a resource, or to provide a summary of the content of the resource, or to judge the importance of the resource.

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Okay so here we have kind of a geometric problem and we have two triangles. Um at point P We can actually change where P is what our goal is. Where should we place P. So that data is maximized. Um we do have some values for length and um I'm gonna go ahead and define some extra value. So I'm gonna call this data one in this single data to I'm gonna call the distance from A. To P. X. And therefore From P to B will be 3 -1. Since the total vertical height is three. Okay. So I know that I have kind of a straight line where between PB and J. So I know that the angle there is 180. So I know that data one plus data plus data to mrs equal 1 80. I'm gonna use radiance. So I'm gonna just say equals pi. Um So data then is pi minus data one minus data to Okay, now we can come up with some data one. Um because we can take in first and we have the dimension. So we're gonna go minus In verse 10 of our opposite. Which is to over adjacent which is 3 -1. And data to then is inverse stand up opposite which is five over adjacent which is X. Alright, awesome. All right. We want to maximize status or data. That is our next function. We really want um uh data as um a function effects which we have now which is perfect. So now we're going to take the derivative to find critical points. Okay, so the derivative of data with respect to acts is equal to pi derivative is zero by power role. But now I've got to work all these inverse hands. Okay, so yet minus inverse. Tan in general is one over the argument squared. Um Plus one. But we still have chain role. So the derivative of 2/3 minus X is minus two over three minus X squared. But we get a double negative. So it's plus two because of chain role. And it helps if you rewrite it as twice, three minus x minus one, then you can see more easily that I will get minus two. But also another minus one because the chain role and then the three minus X goes to the minus two power. Which puts it on the bottom in a similar fashion. We will do minus In verse 10. So the derivative looks like the argument. Oops, that's a five. Let's make it look like a nicer five. Uh The argument quantity squared plus one and then times the derivative which is minus five over X squared. Okay, so this is kind of messy. And what we're actually going to do is um clean it up. And but before I do so it actually there's some natural cleaning up that happens because if you look on the first term I have three minus X squared. But that will cancel out the three minus exports on the bottom there. So let's clean this up. I get a -2, The first term denominator goes away. So I'll get four um Plus and then three minus X squared. So actually cleans up quite nicely. Then on the other one similar thing, I get that cancelation of this term in that term because Um when I distribute it will cancel so I get minus double negatives. So plus five over 25 plus X squared. So kind of scary. But not so bad. Okay, but what's scary now is we're actually gonna do a common denominator. Okay, so on the left I'm gonna multiply by this denominator top and bottom and on the right, I'm going to multiply by this denominator top and bottom. And so I'm going to do common denominators. So the bottom we'll have this whole denominator then multiplied together. Okay, So I'm gonna do the minus two times the right denominator and I'm gonna distribute. So I'm going to get minus 50 minus two X squared. And now five goes time system nominator. So I'll get and I'm going to distribute plus 20 Plus five times 3 -X. Books X squared. Is um yeah, three minus X squared. Ok. Next I'm gonna clean this up and foil. So I'm kind of starting to need some room. So I'm gonna race list on top and just keep going but I will keep the very bottom so that we know what we're working with and I'll remind ourselves what we're trying to do. But we need to just keep going here because there's lots of work to solve this and I need room on my board. Okay so let's keep going. We are now basically going to continue at the top. Okay so I am going to and this by the way was D. Theta D. X. So D. Theta D. X. Is let's see if I can clean this up. I'm going to have to foil out that term there. Um OK so I'm gonna go ahead and foil it out. I'll get 9 -6 x plus X squared. A lot of work for this problem. All right, all over the nice big um I think this is supposed to have a prophecy here. So that prophecy goes there, wow okay lots of work for this problem. Okay so now I can um distribute the five. I'll do that below. We need a lot of room here. So I get -50. Whoops. Um minus two X. Squared plus 20 Plus 45 because I can now distribute the five to each part minus 30 X Plus five X. Squared. And then I still have my really big denominator. Okay so finally we can clean this up to try to get a simple a simple quadratic. Let's see um Let's look at all the square terms, I have five X squared minus two X. Squared. So that gives me three X squared And then I have -30 apps and that's that one and then I got to clean up minus 50 plus 45 so minus five plus 26 plus 15. Whoa Getting there. Um Over the denominator. Okay we need to set this equal to zero for our critical point and I'm gonna now clean up the bottom so we can just keep going. Okay so this was quite a bit of work. Not something you could do really fast. Okay so to make the top equal zero I'm going to just focus on the new radar now so let's do that. So I can factor out of three. That will give me X squared minus 10 X Plus five equals 0. It'd be nice if we could factor but I'm not seeing a nice factoring. So we're going to use the quadratic formula to get our acts. This was a lot of work for this one. Let's do it. The extent is minus B. So 10 plus or minus square root of B Squared -4 times a. Times C. All over twice A. Alright so we get 10 plus or minus square root of, Let's see it looks like 100 -20. So that is 80 um over to. And so five plus or -80 I can do um 16 times five. So four root five divided by two. So two root five. Okay so that is our X. We can't choose the plus value because um that's too big. We can't have X bigger than three. So we're going to have to choose the minus value. So the one that makes sense for us is X -2 root five. So if I do Um five Let's make that a little bit cleaner. Let's go back. So if I choose 5 -2 root five and plug in the calculator I get 0.53. Whoa. Okay so we have found that are Um possible max data is at 0.53. We should do a quick sign chart uh to just to double check because we need data prime to be going from plus to minus. Let's check real quick. So at .53 that does make R. D. Theta. D. X. Go to zero. Let's look at our equation. Um The denominators positive. So really just looking at the numerator. If I make X really big like 100 you could see right away that I get a positive value here. If I make X. zero I get um um uh let's see something um is oh never mind I put in too big of a value because um we do have another critical point. And even though it's not physically possible um When I'm doing my plugging in I have to basically keep less than three we can't have three bigger than three. So let me pick something smaller. Like one if that's looking one for my numerator I will get a negative because I'll get one inside here I get one minus 10 to minus nine plus five. That's negative. That's like in zero. I do get a positive. So I do get a max. I do have a max because D. Theta D. X. Goes from plus to minus at Um X. equals 0.53 which is a critical point. So that Is where we can place our peace. Opie from a should be .53 and that's the distance that will maximize our theta. So anyway that was pretty cool. Good to know. So anyway hopefully that helped to have a wonderful day.

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