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Problem of the Week #68 - July 15th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Let $\alpha$, $\beta$, and $\gamma$ be angles of a triangle. Use Lagrange multipliers to find the maximum value of the function $f(\alpha,\beta,\gamma) = \cos\alpha\cos\beta\cos\gamma$, and determine the angles for which the maximum occurs.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
This week's problem was correctly answered by MarkFL. You can find his answer below.

We are given the objective function:

\(\displaystyle f(\alpha,\beta,\gamma)=\cos(\alpha)\cos(\beta)\cos(\gamma)\)

subject to the constraint:

\(\displaystyle g(\alpha,\beta,\gamma)=\alpha+\beta+\gamma-\pi=0\)

Because of the cyclic symmetry of the 3 variables, we know the critical value for the objective function must come from:

\(\displaystyle \alpha=\beta=\gamma=\frac{\pi}{3}\)

Observing that decreasing one of the variables by a small amount, and increasing another by the same amount results in a smaller value of the objective function, we can conclude this extremum is a maximum.

Hence:

\(\displaystyle f_{\max}=\cos^3\left(\frac{\pi}{3} \right)=\frac{1}{8}\)
 
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