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A painting in an art gallery has height $ h $ and is hung so that its lower edge is a distance $ d $ above the eye of an observer (as in the figure). How far from the wall should the observer stand to get the best view? (In other words, where should the observed stand so as to maximize the angle $ \theta $ subtended at his eye by the painting?)

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Calculus 1 / AB

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Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

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October 28, 2020

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A painting in an art galle…

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The lower edge of a painti…

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Optimal Viewing Angle A pi…

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The lower edge of a wall h…

Alright. So here we have a scenario where we have a picture and um basically it's placed such that it's a little bit above what would be the eye view, distance D above the eye view of the person. So they're looking up And our goal is where do we have the person stand back to maximize their data? Data is shown here in the red. I'm going to go ahead and also define data one right there. Um Okay so our goal is to basically um find um our maximum uh Theta where should we stand as a function of ants. So we need an equation for data. Then we can take a derivative with respect to X. And show that we have a max. So this is our goal. Okay, so if I drew the extra theater one because we have this really nice big triangle and we can do a couple. Um basically use our trig. We know that Data one should be in verse 10 of our opposite over our adjacent so D. Over X. Um we also know that data plus data one. That's this full angle here should be in verse 10 uh are full opposite length which is H plus D. All over X. Once again. So that gives us a couple of relationships. That will be helpful. Um And so basically then since I want basically I want to find, I want to basically find um D Theta. D. X. And I'm gonna set it equal to zero and then prove I have a map. So we need to get data by itself. So data then will be equal to inverse tan of eight plus D over x minus data one. But we know what data one is. Theta one is inverse. Candy over X. I'll go ahead and write that in for theta one. Uh huh. So minus in verse 10 H plus D over eps. Okay, so you can see these problems have quite a bit of a scary looking equations and a lot of steps. So we'll go through it carefully. Okay so now we want to find D. Theta D. X. Set equal to zero and find our critical points which means we're gonna take these lovely derivatives. Ok So in verse 10 normally looks like for a derivative one over the argument Squared plus one times the derivative of the argument. Um And so we will get for the this is like just to help you see it, H plus D over X is like HP plus D. X to minus one. So if I do power role I get a minus H plus D and then expand minus two and I can write that on the bottom. So I get um I'll get minus one over x minus one times age plus de All over X square. Okay so that's the first derivative let's do the other one. Let's say in verse 10 again. So one over argument Squared Plus one. They're going to take the derivative. Um and similar to before it's going to have the 4-. Um I feel like I plugged in the wrong thing I did. Let's fix it real quick. This should just be actually plugged in the combo again. This is just the over X. Okay so that I need to fix real quick so this just becomes one over the argument square. So that's just D. Over X. And then the derivative of this is minus D. Over X. Squared. Similar form. Okay. There is some nice cleaning up. If I distribute on the left fraction the expert across and it cancels in the bottom. So we do get a little bit of cleanup here. So I get minus H. Plus D. Over H. Plus the quantity squared plus X. Squared. Um And then double negative. So plus D. Over X squared cancels at bottom X. So I get D squared plus X. Squared. Okay so we want to set this equal to zero but we only want to do that with we have a common denominator so we have a single fraction. So let's go ahead and do that. We'll keep going down here below and we change color. Okay so that gives us D. Theta D. S. Equals to seek common denominators. So I'm going to multiply the left numerator by the right denominator because I'm gonna get common denominators. So I'm going to get minus H. Plus D. Times D. Squared plus X. Squared. And then I'm going to do D. Times the other denominator. So then plus D. Times H. Plus D. Squared. Plus X. Squared all over the big bottom. Um And that's a big denominator. Okay let's uh clean up what we can we have a bunch of oils to do. So let's see if we can do this carefully. I'm going to distribute the H. First of both and then the deed of both. So minus H. Um D. Squared minus H. X. Squared minus D. Q. Minus D. X. Square. That's all just the distribution there. And then I'm going to distribute on the right so I'll get plus D. Times. I'm gonna have to foil this out. Eight squared plus two HD. Plus T. Squared plus X. Squared. Um All over the really scary big denominator. Okay I'm gonna need room so I'm going to clear off spot and be right back. Okay so let's um focus on that numerator because we want D. Day to D. X. To go to zero. So I'm gonna just do um set the numerator equal to zero. But I still have to finish distributing distributing. So this is quite an extensive numerator. Um Okay now I'm going to distribute the D. So I'm working with this numerator and I'm finally distributing the D. So I'll get plus um D. H. Squared plus two H. D. Squared, lusty cute Plus D. X. Squared. Okay so we look for anything nice to cancel and if it does we're going to be really happy and let's see we do ye we're gonna get this D. X. Squared. Subtracting that one. We get A. D. Q. Subtracting this one. Oh this is good news. Anything else canceling? Let's see um I think mhm. Is that everything? Okay? So we end up with, so this is our numerator equals look at the export term because we have to solve for X. So we have only the one with X. And then the rest all just slumped together. They all have so then I have plus I have a minus H. D. Squared. I have a plus D. H. Squared and a two H. D. Squared. So it does clean up a little bit to be minus H. X squared. I can combine the HD squares to just be H. D. Squared. Um Philosophy H squared. All right cool. Okay we're going to set that equal to zero because we want that numerator to go to zero and so therefore we can solve for X. We're almost there. This is great. Okay so X squared then is minus H. D. Squared plus D. Eight squared subtracting that parentheses from both sides divided by minus H. And so expert is actually I can divide through and I'll get D squared plus D. H. That was a lot of work D squared plus th so X band is the square root D squared plus D. H. And um that is the X. Value. This is how far back you should stand to get the best best view of the picture. We want to prove it's a max real quick so let's just do that real quick here. So we're looking at for X versus basically d. Theta dx we want to know and we could have a plus or minus here but the plus only makes sense for our distance and once we're on the other side of the picture um So let's see what happens to our function. If we look at our numerator and we plug in something um really big for X. That will make this term super negative. So that would be a minus one. X. Is really big When it's a small like zero. Um Then we'll get a positive. So we definitely have a max because our derivative defeated dx goes from plus to minus at our critical point. And so we solved it. This was huge work but we did it. So you should stand back square root B squared plus D. H. Alright. Hopefully that helps to have a wonderful day.

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