How to do this optimization? It invloves angles of view and maximizing the sight but I just do not understand...

Question

An art critic wishes to view a 6 foot tall painting which is hung so that the bottom of is 7ft above the ground. The critic knows that her best view of the painting will occur when the angle( theta) between her line of sight to the bottom of the painting and her line of sight to the top of the painting are at maximum. If her eye level is 5ft above the ground, how far from the wall should she stand in order to best view the painting.
It gave steps to take to get to the answer.

It first wants "express theta as beta-alpha"
I was given a picture where her eyelevel to the bottom of the painting is alpha, btottom to top of painting is theta and both together are beta.

It then says to express beta and alpha in terms of distance from the wall

I really don't see how these help me because I really do not know where to begin in terms of solving this problem.

Answer 1

Hi, Olivia.

 

I apologize that this explanation will be long.  I'm sorry that I can't draw a picture for you, but if you follow along, you should be able to draw the picture for yourself.

 

The solution to this problem requires you to take the first derivative of a function that you will create relating the distance from the wall to the angles beta and alpha.  To create that function, you need to remember your Trigonometry.

 

Assuming that the wall is perpendicular to the ground, draw a picture illustrating what you have been told in the problem.  You have a wall of unknown height (but that is not important).  A picture is hung on the wall so that the bottom of the picture is 7' above the ground and so that the top of the picture is 13' above the ground.  An art critic is standing some distance away from the wall.  If you draw a horizontal line at eye level of the art critic, that line will intersect the wall 5' above the ground and 2' below the bottom of the picture.  Now, draw lines from the art critic's eyes.  The first line goes to the bottom of the picture.  The second goes to the top of the picture.  You have just constructed two triangles formed by the wall, the eye level of the art critic, and the lines to the bottom and top of the picture.  Define alpha in the smaller of these triangles to be the angle from the eye level line to the line to the bottom of the picture.  Define beta in the larger triangle to be the angle from the eye level to the line to the top of the picture.  Finally, define the distance that the art critic's eyes are from the wall as "d".  Both the smaller triangle and the larger triangle share "d" as the base leg of each triangle.

 

Now, you should recall from Trigonometry that the tangent of an angle is the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle.  For the smaller triangle, "tan(alpha) = 2/d" or "alpha = tan^-1(2/d)".  For the larger triangle, "tan(beta) = 8/d" or "beta = tan^-1(8/d)".  Read "tan^-1" as the inverse tangent.  You can now define "theta" as the difference of the angles beta and alpha, or "theta = beta - alpha".  Substituting the Trigonometric expresses that you just created into this express, we see that

"theta = tan^-1(8/d) - tan^-1(2/d)", which defines "theta" as a function of "d".

 

To find the maximum angle of theta, take the first derivative of your function of "d", set it equal to zero, and solve for "d".  Once you have "d", go back to the Trigonometric functions and compute alpha and beta.

 

Dennis