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Directrix

  • 4 years ago

If x and y are acute angles whose sum is 60 degrees, what is the largest possible value of the following: (tan x)(tan y)?

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  1. anonymous
    • 4 years ago
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    Are you upto solving this set: http://answers.yahoo.com/question/index?qid=20111214061522AA8yZ07 ?

  2. Directrix
    • 4 years ago
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    No. But, it appears that some yahoo may be guilty of _____. I wish I could see the date the question was posed.

  3. anonymous
    • 4 years ago
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    Um just guessing, is \( \large \frac13 \) is the answer?

  4. dumbcow
    • 4 years ago
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    yes it is:) i'm working on proving it though, taking derivative didn't get me anywhere

  5. anonymous
    • 4 years ago
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    Aha, I just used inequalities, the simple AM-GM one, in more simple terms the for constant sum the product of two variables is highest when they are equal.

  6. Directrix
    • 4 years ago
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    Yes, 1/3 is correct and can be concluded intuitively. The proof is not too bad.

  7. dumbcow
    • 4 years ago
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    i know im over complicating it, but here is a proof using trig tan(x+y) = tan(60) = sqrt3 \[\tan(x+y) = \frac{\tan x +\tan y}{1-\tan x \tan y}\] \[\rightarrow \frac{\tan x +\tan y}{1-\tan x \tan y} = \sqrt{3}\] solving for (tanx)(tany) \[\tan x \tan y = 1- \frac{\tan x +\tan y}{\sqrt{3}}\] Differentiating: \[\rightarrow - \frac{\sec^{2} x +y'\sec^{2} y}{\sqrt{3}}\] Now y = 60-x, so y' = -1 setting derivative equal to 0 \[ \frac{\sec^{2} y -\sec^{2} x}{\sqrt{3}} = 0 \rightarrow \sec^{2}y = \sec^{2}x\] \[\rightarrow y=x\] Thus proving (tanx)(tany) is maximized when x=y=30 \[\tan 30 *\tan 30 = \frac{1}{3}\]

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