Picture Perfect

By:

Alex Moore

Suppose
we are in an art museum looking at magnificent works from the greatest artists
history has to offer. We walk up a
painting that is placed particularly high on the wall. What distance away should we stand to
get the best view of the painting?
That is, where should we stand to view the painting under the maximum
angle? First we picture the
scenario: The painting is a 4x4
picture, which is on the wall 2 feet higher than your eye level.

How
can we solve this problem? We
could do it completely geometrically using circles, but what is the first
method that comes to mind when you see the word ÒmaximumÓ in the problem? I think calculus, so here we approach
this problem using calculus!

Call
the angle we want to maximize t, and let T1 and T2 denote the two smaller
triangles formed and let T3 denote the union of the triangles, that is, the
overall triangle. Let x be the
distance from where you stand to the wall. Since we want to know where we should stand in order to
maximize t, we must create a function that measures t in terms of x, call it
t(x). Our approach will be to find
two different expressions for the area of T1 and set them equal to one another. The first expression is the more
familiar one. We know the area of
a triangle is (½)(base x height).
From the picture we see that T1 has a base of length 4 feet and a height
of length x feet. Therefore, Area(T1)
= (½)(4)(x) = 2x. The next
expression for the area will incorporate the angle t. Given a triangle with two of the sides having length A and B
and the angle between them having measure t, the area of the triangle is
(½)ABsin(t). How can we see
this?

Given
a general triangle, like the one above, let one side have length A, another
side have length B, and the angle between them be t. Let h denote the height of the triangle. After dropping the height we get two
new triangles, both of which are right triangles. Therefore, we have sin(t) = h/A, that is, h = Asin(t). Since the triangle has area (½)Bh
we simply substitute this expression in for h to get Area =
(½)ABsin(t).

Now
we use this expression on our original triangle. First we need to compute the lengths of the sides adjacent
to t. Since one length is the
hypotenuse of T2 and the other is the hypotenuse of T3, both of which are right
triangles, we can use the Pythagorean theorem to compute these lengths and we
get

where
A is the length of the hypotenuse of T3 and B is the length of the hypotenuse
of T2. Plugging these values into
our expression yields:

Setting
these two expressions for area equal to each other and solving for t we get:

Now
that we have our function we take the derivative with respect to x and set the
derivative equal to 0. After much
hard work we find the derivative of t(x) to be:

Now
we set this derivative equal to 0 and solve for x. First note that none of the terms in the denominator can be 0
since each term is strictly greater than 0. So we have:

To
be completely proper x^{2} is actually equal to 12 or -12, however, x
measures a distance so cannot have x be a complex number so we choose x^{2}
= 12. Also, when taking the square
root of 12, x could be the positive or negative root, but again, x measures a
distance so we choose the positive root.
After all of our hard work we conclude that you should stand approximately
3 ½ feet away from the wall when looking directly at the painting!