Question

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)?

Answer 1

Are you upto solving this set: http://answers.yahoo.com/question/index?qid=20111214061522AA8yZ07 ?

Answer 2

No. But, it appears that some yahoo may be guilty of _____. I wish I could see the date the question was posed.

Answer 3

Um just guessing, is \( \large \frac13 \) is the answer?

Answer 4

yes it is:)
i'm working on proving it though, taking derivative didn't get me anywhere

Answer 5

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.

Answer 6

Yes, 1/3 is correct and can be concluded intuitively. The proof is not too bad.

Answer 7

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
\[\tan 30 *\tan 30 = \frac{1}{3}\]