Calculus maximization Help?

Please solve the following

To get the best view of Abe Lincoln at Mount Rushmore, you should be at a position to maximize your viewing angle. if the the top of his face is 92 meters above the ground, with the bottom of his beard 46 meters above the ground, how far from the base should you be?

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  • 1 decade ago
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    This for me was a very difficult problem, for me anyway. There must be a better way to do it, but the way I set it up was the following.

    Think of TH (big theta) the angle between your line of sight to the top of Abe's head and the horizontal.

    Think of th (little theta) as the angle between your line of sight to the bottom of Abe's beard and the horizontal.

    From basic trig. relationships you know that sin(TH) = 92/r where r is the distance from your eyes to the top of Abe's head.

    From the Pythagorian Theorem, you also know that d^2 + 92^2 = r^2 where d is the horizontal distance from the base. Therefore you can rewrite sin(TH) = 92/r in terms of d only, and get rid of r.

    sin(TH) = 92/sqrt(d^2 + 92^2)

    and the inverse

    TH = arcsin(92/sqrt(d^2 + 92^2))

    You can do the same thing for your little theta (th) too, using the value of 46 instead of 92.

    sin(th) = 46/sqrt(d^2 + 46^2)

    and the inverse

    th = arcsin(46/sqrt(d^2 + 46^2))

    Now, your angle of interest (i.e., the view angle) is TH - th. Let's call this angle A, which is what you want to maximize. For a visual, see http://i411.photobucket.com/albums/pp194/watari_pa...

    A = arcsin(92/sqrt(d^2 + 92^2)) - arcsin(46/sqrt(d^2 + 46^2))

    The only thing that sucks, is you have to maximize A by finding A' = 0, and it's a murderous derivative. I used an online solver for it.

    A' = 46(sqrt((d^2 + 92^2)/(d^2 + 46^2)) - 2*sqrt((d^2 + 46^2)/(d^2 + 92^2)))/sqrt((d^2+46^2)(d^2+92^2))

    so you need to set A' to 0 and solve for d

    It's not a simple matter to do that either. I used a graphing program, and got

    d = 65.05382 meters

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