Math Is Fun Forum

Au101

Member

Registered: 2010-12-01

Posts: 353

Final "Higher Derivatives" Question

Okay, this is the last question on this exercise and it's got me completely confused in so many ways. This is the question:

Now, I've got all of the pieces of the puzzle. Obviously:

Next, I said:

Now, if we sub that value of x back into y, we get this:

So we've got the two parts of our inequality, but I'm just stuck "considering the minimum value of the function". I tried finding the second derivative:

Now, if I put a/2 into that, I get:

which I reckon is surely positive if a > 0, which suggests that a/2 is our minimum. The bad news is that Wolfram|Alpha thinks that the function has no global minimum. The worse news is that I actually found y to be a maximum at x = a/2 when I tried this approach:

Again Wolfram|Alpha thinks that the function has no global maximum, either.

So I'm a little bit at sea here - I'm clearly doing something wrong, I just don't know what!

Olinguito

Member

Registered: 2014-08-12

Posts: 649

Re: Final "Higher Derivatives" Question

Why are you considering

? Doesn't the question specifically say

?

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Bassaricyon neblina

Au101

Member

Registered: 2010-12-01

Posts: 353

Re: Final "Higher Derivatives" Question

That's a fair point actually, it does, I just wanted a value of x that would make my algebra nice and easy, so let's try something else:

I believe it comes to the same thing, doesn't it?

Last edited by Au101 (2015-02-23 07:25:36)

Olinguito

Member

Registered: 2014-08-12

Posts: 649

Re: Final "Higher Derivatives" Question

Well, I think you've worked out the problem correctly using calculus, so never mind Wolfram|Alpha.

This problem is typical of olympiad inequality problems requiring quick manipulations using algebra rather than calculus. Thus, applying AM–GM, we have

It is easily seen that in the interval

the expression

attains a maximum at

, the maximum being

. Hence the minimum of

in that interval is

and so:

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Bassaricyon neblina