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Picture Perfect


Alex Moore

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         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. 

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         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 “maximum” in the problem?  I think calculus, so here we approach this problem using calculus! 

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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? 

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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

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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:

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

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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:

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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:

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To be completely proper x2 is actually equal to 12 or -12, however, x measures a distance so cannot have x be a complex number so we choose x2 = 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!