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Problem 78 Hard Difficulty

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?)

Name: a painting in an art gallery has height h and is hung so that its lower edge is a distance d above t
Uploaded: 2021-08-26
Channel: Numerade
Description: 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?)

Answer

$x=\sqrt{h d+d^{2}}$

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Video Transcript

in this problem we're told that we've got to painting, I've drawn the painting here in blue and it is mounted a distance D. Above the line of sight of an observer. So persons I is here and the painting is mounted a distance a D. On the wall above the level of the observer's eye. And the height of the painting itself is this um Hi h and we are being asked to figure out what is the ideal distance X. For the observer to stand away from the wall to have the best view of the painting. By which we mean to make this angle theta maximum. So this is a maximization problem. We want to find the derivative of that angle to so that we can determine When that derivative equals zero and look for the maximum there. Well, I know that this is a right angle. And I'm going to use tangent to try and start finding relationships between these distances. So that angle created at the bottom of the painting here. I'm going to call Alpha And I'm going to say that the tangent of that angle, Alpha. Actually I'm going to try and just type this instead of writing it because it will be easier for everybody to see a tangent of the angle helped him. It is going to equal the ratio of the distance above the observer D divided by the distance the observers standing away from the wall X. Mhm. And at the same time, yeah. The tangent of angle also. Oh plus a single fade up. Yeah. Right. Who is going to equal the sum of the distances. D plus age. Yes, divided by that distance X. So I'm starting to get a relationship between all of the measurements. I really want to find a relationship between D. And H. And that um uh D. And H. And the rate that the angles are changing. So I am going to and I want to find the rate that the angles are changing. So I'm going to rewrite this. So that angle alpha is equal to inverse tangent. Mhm Yeah. Of that ratio D divided by X. And also I'm going to rewrite this so that angle alpha. Okay. Because while staying gold fade huh is equal to the universe tangent. Mhm of this ratio. Okay. D plus H divided by X. Mhm. Now I've got something that I can differentiate. Find the rate that the angles are changing. So yeah. The derivative of angle health. A as a function of the distance is going to equal the derivative of arc tangent. The dragon derivative of art tangent is the reciprocal of one plus the argument squared. So that is A B over X squared. Okay. And and I also have to include the derivative arguments. So that will be the derivative with respect to X. Of D over X. And doing a little bit of rewriting Yeah. Multiplying everything by X squared is going to give me explain aired. Yeah. Divided by X squared plus B squared And the derivative of one over X is negative one over X squared D is a constant. So it's just multiplied. So all this is being multiplied by clark negative D over X squared. And then I can rewrite a little bit more X squared divided by X squared is just one. So and I am left with negative ratio of D two x squared plus. Yeah. Mhm Yeah. So that gives me the derivative of angle alpha. Then I'm going to do the same thing for angle alpha plus data and I will get when I do all of the same arithmetic. I will get that the derivative of angle alpha with respect to X plus the derivative of you can make that and I'll find out. Okay. Plus the derivative of angle fada with respect to X is going to be equal to a negative ratio of the close H all over X squared plus the U plus H squared. I'm going to go ahead and resist the temptation to expand that. Plus a general is going to keep it as its own quantity. And just to save room. I'm going to go ahead and delete all of the math that teams up to this since you've already seen it. Okay. Yeah. Uh huh I can yeah. Mhm It's fundamental from the tangent of boat. The sum of the angles and angle theta I have derivatives of just helpful. I have derivatives of alpha plus data. What I want is just the derivative of data. So the derivative of theater is going to be this entire derivative of alpha plus data minus the derivative of alpha. So uh is going to equal Yeah. Yeah. Okay. Mhm After a bit of Alpha plus data which is a negative ratio. People's H mhm Yeah over X squared plus B plus each squared minus. Did you really get it with cal fire? Okay. Mhm. Mhm. But I know that the derivative of Outfront equals this negative D over X squared plus city so I am going to replace mhm But yeah. Uh huh. Okay. Okay. Okay. Mhm. Mhm Yeah. Okay. Yeah. So I have I think derivative of a cluster a bit of a beta here and I am subtracting just the derivative of the alfa here, negative being subtracted becomes positive. So now I have my derivative succeed I'm going to do a little bit of rewriting. Okay. But to find the maximum value of data, what I want to do is find out when the derivative of data equals zero. So I'm going to set all of that equal to zero. Yeah and zero equals I'm going to multiply both pieces by the denominator. So I get a common denominator so I will have the ratio. Uh huh. The negative B plus Yeah. A speech times X squared plus t square. Okay. Plus the times the other denominator is X squared yes. Keep us h squared and the mhm And that is all divided by X squared Class D. Plus H. I swear this prophecies. Okay. Yeah. Mhm. Okay Times X squared plus B squared. It's yeah. Mhm. Now for this to equal zero if I multiply both sides by this denominator I am going to um I'm multiply both sides by this denominator zero times anything is still zero. So really I don't need to worry about the denominator denominator is not going to influence zeros. You only the numerator. Well I think that yeah. Mhm. Uh huh. Mhm. Mhm. Multiplying everything by X squared plus B plus H squared and X squared plus T squared leaves me with this. Zero equals yeah D. Plus H. Times X squared times. I'm sorry D plus H times X squared plus B squared plus D. Times X squared plus de plus H squared. And then solving this for X takes a little bit of manipulation. Right? If I multiply this all through zero equals a negative d. x squared minus H. X. Square. Yeah. Okay. Okay. Pine nuts be cute. Well actually I'm gonna do my SD times. It's okay. Yeah. Yeah. And then plus D. X squared lunch the yeah times D. Plus H squared. Mhm. Yeah. Trade and then continuing to rewrite no it didn't like that. Why didn't like that? Yeah. The X squared minus D. X squared statistic zero. I can cancel those out. Yeah. Oh. Mhm. And I'm left with minus H. X squared minus D. Squared times D. H. Plus D. Times D. H squared equals zero. Rewriting that, wow. I'm going to put the H X squared on one side. Yeah and that equals I'm going to factor out A. D. I'm going to factor out D. H. So this becomes the times B plus H times 80 plus H minus B. So complete. And then if I divide both sides by H I get X squared equals. Okay listen D. Times D plus H divided by H. D minus D. Here cancels out. So D plus H minus is just a church. Okay. Mhm. And each divided by H. is one. Yeah. Yeah. If I am left with X squared equals B. Times D plus H. Or. Okay. The ideal distance for an observer to stand. Okay. Yes. Is at a distance equal to the square root of the height above the eye line times the total height painting about the high line. Uh huh.

Michael C.

University of Southern California

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems