## 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)?

1. anonymous

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

2. Directrix

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

3. anonymous

Um just guessing, is $$\large \frac13$$ is the answer?

4. anonymous

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

5. anonymous

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

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

7. anonymous

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 $\tan 30 *\tan 30 = \frac{1}{3}$

Find more explanations on OpenStudy