bonanova Posted February 17, 2015 Report Share Posted February 17, 2015 Without resorting to differentiation (sorry, Y-San) find the minimum for (0< x< pi) of the function Quote Link to comment Share on other sites More sharing options...

0 gavinksong Posted February 25, 2015 Report Share Posted February 25, 2015 (edited) u = x sin(x) 9u + 4/u = m 9u^{2} - mu + 4 = 0 calculate the determinant: m^{2 }- 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 February 25, 2015 by gavinksong Quote Link to comment Share on other sites More sharing options...

0 phil1882 Posted February 17, 2015 Report Share Posted February 17, 2015 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. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted February 18, 2015 Author Report Share Posted February 18, 2015 The first step - making it the sum of two quantities - is correct. From there the minimum can be calculated directly. Quote Link to comment Share on other sites More sharing options...

0 dgreening Posted February 19, 2015 Report Share Posted February 19, 2015 (edited) 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 February 19, 2015 by dgreening Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted February 19, 2015 Report Share Posted February 19, 2015 There are two answers to this I believe Quote Link to comment Share on other sites More sharing options...

0 dgreening Posted February 19, 2015 Report Share Posted February 19, 2015 Yes. There are 2 answers 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 Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted February 19, 2015 Report Share Posted February 19, 2015 Yes. I found those using the Laurant series Quote Link to comment Share on other sites More sharing options...

0 k-man Posted February 19, 2015 Report Share Posted February 19, 2015 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. Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted February 20, 2015 Author Report Share Posted February 20, 2015 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 Yes. There are 2 answers 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? Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted February 25, 2015 Author Report Share Posted February 25, 2015 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? Quote Link to comment Share on other sites More sharing options...

0 Rainman Posted February 25, 2015 Report Share Posted February 25, 2015 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. 1 Quote Link to comment Share on other sites More sharing options...

0 bonanova Posted February 26, 2015 Author Report Share Posted February 26, 2015 That's it. 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. Quote Link to comment Share on other sites More sharing options...

0 gavinksong Posted February 26, 2015 Report Share Posted February 26, 2015 That's actually really interesting. Thank you. I learned something new today. Quote Link to comment Share on other sites More sharing options...

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