Minima - II
If ω , ω 1 , ω 2 are real variables such that tan ω + 2 tan ω 1 + 5 tan ω 2 = 6 0 , then find the minimum value of tan 2 ω + tan 2 ω 1 + tan 2 ω 2 .
The answer is 120.
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3 solutions
did same!!!
Think of a vector A= tan w i + tan w1 j+ tan w2 k And vector B= i + 2j + 5k Now A.B=ABcos(theta) .... so A.B =60 by putting the min value of cos(theta)=-1, u will get answer in a line.
Really a brilliant approach!!
Lagrange's multipliers makes the question a piece of cake.
Define a function, f ( x, σ )= a 2 + b 2 + c 2 + σ ( a + 2b + 5c - 60), where : a,b & c denote tan ( w 1 ), tan ( w 3 ) & tan ( w 2 ) respectively.
Now we partially differentiate the function with respect to a, b & c respectively to obtain an equation like b = 2 a = 2 5 c .
Find b and subsequently a & c.
Are there any other solutions for this problem?
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Not one I can think of immediately.
How about applying A.M. greater than or equal to G.M. first to tan(w), 2tan(w1) and 5tan(w2) and then applying the same to the squares?
The only restriction needed for that would be tan(w)* tan(w1)*tan(w2)>0.
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The lack of this given condition is the reason that motivated me to use Lagrange's multipliers :)
Just a application of Cauchy Shwarz inequality..