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# A mean, mean minimization problem

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u = x sin(x)

9u + 4/u = m

9u2 - mu + 4 = 0

calculate the determinant:

m- 4(9)(4) ≥ 0

m ≥ 12

Thus, 12 is the minimum. We ignored m ≤ -12 because m ≥ 0 for 0 ≤ x ≤ pi.

Edit - corrected a sign error.

Edited by gavinksong
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9*x*(sin x) +4/(x*sin x)

if x = pi/2:

16.68

if x = pi/4:

12.2

if x = pi/8:

27.97

if x = 3*pi/8:

13.47

if x = 3*pi/16:

15.17

so it looks like its pi/4.

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The first step - making it the sum of two quantities - is correct.

From there the minimum can be calculated directly.

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Following the earlier post, let's look at both parts of the equation

9*x*(sin x) +4/(x*sin x)

over the range of 0 to pi

9*x*(sin x) starts at 0, ends at 0 and would expect it to peak  somewhere past 0.5 pi

4/(x*sin x) starts very large, ends relatively large and you would expect it to reach a minimum somewhere before 0.5 pi

Since the peak of one function and the minimum of the 2nd function do not coincide, You would expect 2 relative minima.

Rearranging 2 pieces, we get to

9 * (xsinx) = 4/(xsinx)

or

(xsinx)^2 = 4/9

therefore

xsinx = sqrt (4/9) = 2/3

I am sure that there is an elegant way to solve this, but I cheated

This occurs at about 0.27 pi

It also occurs at about 0.93 pi

Edited by dgreening
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This occurs at about x= 0.27 pi

It also occurs at about x= 0.93 pi

f(0.27 pi) = f(0.93 pi) = about 12.0

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1. Function f(x) = 9x*sin(x) + 4/(x*sin(x)) can be represented as f(x) = 9*g(x) + 4/g(x), where g(x) = x*sin(x)

2. For functions in the form f(x) = a*g(x) + b/g(x) (when a>0, b>0) the g(x) is irrelevant in finding the value of f(x) when f'(x)=0. [spoiler=I cheated here and used differentiation to confirm that]f'(x) =  a*g'(x) - b*g'(x)/g(x)2

f'(x)=0 when a=b/g(x)2 or g(x)=sqrt(b/a)

3. Any local extremums for f(x) will occur when the value of g(x)=sqrt(b/a) and plugging sqrt(b/a) into the formula for f(x) yields f(x) = 2*sqrt(ab)

4. Inserting 9 and 4 in place of a and b finds the minumum of 12.

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Following the earlier post, let's look at both parts of the equation

9*x*(sin x) +4/(x*sin x)

over the range of 0 to pi

9*x*(sin x) starts at 0, ends at 0 and would expect it to peak  somewhere past 0.5 pi

4/(x*sin x) starts very large, ends relatively large and you would expect it to reach a minimum somewhere before 0.5 pi

Since the peak of one function and the minimum of the 2nd function do not coincide, You would expect 2 relative minima.

Rearranging 2 pieces, we get to

9 * (xsinx) = 4/(xsinx)  I don't understand how you got this.

or

(xsinx)^2 = 4/9

therefore

xsinx = sqrt (4/9) = 2/3

I am sure that there is an elegant way to solve this, but I cheated

This occurs at about 0.27 pi

It also occurs at about 0.93 pi

There are two answers to this I believe  Well, the single answer is 12, but it occurs for two values of x

This occurs at about x= 0.27 pi

It also occurs at about x= 0.93 pi

f(0.27 pi) = f(0.93 pi) = about exactly 12.0

1. Function f(x) = 9x*sin(x) + 4/(x*sin(x)) can be represented as f(x) = 9*g(x) + 4/g(x), where g(x) = x*sin(x)

2. For functions in the form f(x) = a*g(x) + b/g(x) (when a>0, b>0) the g(x) is irrelevant in finding the value of f(x) when f'(x)=0. [spoiler=I cheated here and used differentiation to confirm that]f'(x) =  a*g'(x) - b*g'(x)/g(x)2

f'(x)=0 when a=b/g(x)2 or g(x)=sqrt(b/a)

3. Any local extremums for f(x) will occur when the value of g(x)=sqrt(b/a) and plugging sqrt(b/a) into the formula for f(x) yields f(x) = 2*sqrt(ab) That is the key finding.

4. Inserting 9 and 4 in place of a and b finds the minumum of 12.

You're flirting with derivatives. Suppose this were an algebra problem.

What if the title were not chosen arbitrarily?

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Your solution fits the OP so I'll mark it solved. Persistence paid off.

Anyone see another approach that fits the title? What means are available here, with a little manipulation? What do we know about various means? Know of any inequalities that have to do with means? Can I think of even more clues? ##### Share on other sites

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The AM-GM inequality gives us (9t+4/t)/2 ≥ sqrt(9t4/t), which yields 9t+4/t ≥ 12, with equality iff 9t = 4/t.

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Someone pointed out in an article I read recently that

AM>GM solves a whole slew of optimization problems.

An application easily but not normally made.

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That's actually really interesting. Thank you. I learned something new today.

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