minimum in restricted area..
For 0 < x < π , find the minimum value of 2 x . s i n x 9 . x 2 . s i n 2 x + 4
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The answer is 6.
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3 solutions
Note: You should show that the equality case can indeed be achieved.
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The equality holds when 9 x s i n x = x s i n x 4 , ( x s i n x ) 2 = 9 4 , x s i n x = 3 2
sir are you asking about this thing
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You should state this can indeed be achieved. Technically, all that you have shown is that f ( x ) m i n ≥ 6 . You still need to show that f ( x ) m i n = 6 by finding a corresponding value of x .
Otherwise, as shown by this problem , if the equality case of AM-GM cannot be achieved, then the calculated value would not be the true minimum.
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@Calvin Lin – I think this is what you are asking for,
9 x 2 s i n 2 x − 1 2 x s i n x + 4 = 0 , ( 3 x s i n x − 2 ) 2 = 0 , x s i n x = 3 2
Sir now how do i find the value of x ?
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@U Z – You do not need to find the value of x . You just need to show that there exists a value of x which satisfies x sin x = 3 2 . This can be done, by using the Intermediate value theorem, since x sin x is a continuous function and 0 sin 0 = 0 , 2 5 π sin ( 2 5 π ) = 2 5 > 3 2 .
Exactly as it was expected...Good..Nice solution.
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Sir a good question on A.M - G.M inequality, asked in JOMO 10 , i can't post the question directly. I don't have the permission
My answer comes as 24 (hope our answer match)
JOMO 10 competition is finished (so i can discuss the question) , now JOMO 11 is the on-going competition
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Yeah, my answer is also 24. And that is achievable when x → 0 or y → 0 .. @megh choksi
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@Sandeep Bhardwaj – oh nice did'nt noticed, i did it this way
y [ 1 + x ( y 1 + y ) ] 2 + 3 y 2 + 3 x 2 y 2 ( y 1 + y ) 2 + 6 x y 2 ( y + y 1 )
= y [ 1 + x ( y 1 + y ) ] 3 y 2 ( x 2 ( y + y 1 ) 2 + 2 x ( y 1 + y ) + 1 ) + y [ 1 + x ( y 1 + y ) ] 2
= 3 y ( x ( y 1 + y ) + 1 ) + y [ 1 + x ( y 1 + y ) ] 2
A . M ≥ G . M
= 2 × 6 @Sandeep Bhardwaj thank you for being humble
I will have a look at this soon. By the way, can you tell me something about JOMO ? I don't have any idea about this ?
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Do you mean minimum value of the expression?
can be simplified as (9/2)xsinx + (2)/xsinx . ...........the range of xsinx= all real no.............let xsinx=a (any no.). .................so, we only need to find the min. of (9/2)a +(2)/a. ...................a=2/3 or min f(x)=6
0 < x < π therefore sinx positive
f ( x ) = 2 9 x s i n x + x s i n x 2
2 2 9 x s i n x + x s i n x 2 ≥ 2 9 . 2 x s i n x x s i n x
f ( x ) m i n = 6