Optimization Problem.

ffuh205

Junior Member
Joined
Mar 17, 2010
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Ok here is another problem. To get the best view of the Statue of Liberty you should be at the position where ? is a maximum. If the statue stands 92 meters high, including the pedestal, which is 46 meters high, how far from the base should you be? [Hint: Find a formula for ? in terms of your distance from the base. Use this function to maximize ?, noting that 0 ? ? ? ?/2]

Ok This is what I have.

? = Tan-196/d - Tan-146/d. Solving for zero graphically ? is maximized at 66.38 meters. Plugging that back into the original equation gives us 20.6 degrees.
 
Rewrite the correct problem verbatim from the book please, as I cant make heads or tails out\displaystyle Rewrite \ the \ correct \ problem \ verbatim \ from \ the \ book \ please, \ as \ I \ can't \ make \ heads \ or \ tails \ out

 of what you have written.\displaystyle \ of \ what \ you \ have \ written.
 
tan(θ) = 92d, tan(α) = 46d\displaystyle tan(\theta) \ = \ \frac{92}{d}, \ tan(\alpha) \ = \ \frac{46}{d}

β = θα, = arctan(92/d)arctan(46/d)\displaystyle \beta \ = \ \theta-\alpha, \ = \ arctan(92/d)-arctan(46/d)

dβdd = 46d2+211692d2+8464 = 0\displaystyle \frac{d\beta}{dd} \ = \ \frac{46}{d^{2}+2116}-\frac{92}{d^{2}+8464} \ = \ 0

Hence d =˙ 65 meters, the best view to see the lady without the pedestal.\displaystyle Hence \ d \ \dot= \ 65 \ meters, \ the \ best \ view \ to \ see \ the \ lady \ without \ the \ pedestal.

Note: α =˙ 35.280,β =˙ 19.470, and θ =˙ 54.750.\displaystyle Note: \ \alpha \ \dot= \ 35.28^{0}, \beta \ \dot= \ 19.47^{0}, \ and \ \theta \ \dot= \ 54.75^{0}.

Sorry ffuh205, what confused me was your notation after the problem.\displaystyle Sorry \ ffuh205, \ what \ confused \ me \ was \ your \ notation \ after \ the \ problem.
 
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