As x ranges over all real numbers, what is the minimum value of
[ x 2 − 4 x + 7 − 2 2 + x 2 − 8 x + 2 7 − 6 2 ] 4 ?
This problem is posed by Subharthi C .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Great job!
I like the phrase "but we are not interested in the value of x ". That is not crucial to solving this problem, apart from knowing that A is on the line segment B C .
Log in to reply
I think this question donot deserve level 5 or 280 points.
Well, before reading this I would have said that the derivative method is the best way to do this problem; but now I'm thinking that your way is even better.
Good job. I worked out that x = 1 + 2 using calculus and then noticed that the solution makes ( x − 2 ) 2 = ( 2 − 1 ) 2 and ( x − 4 ) 2 = ( 2 − 3 ) 2 , but couldn't see the significance of that.
V good thought process :)
Nice ideas.
nice
Very nice.
man you're so good
Beauty
I did it the same way.....you presented it very nicely... :)
awesome work there..!
86 upvotes! Awesome
Average......................
We will use two inequalities to solve this problem:
Cauchy-Schwarz Inequality: x 2 + y 2 ≥ 2 1 ( x + y ) 2 (Equality occurs at x = y )
Absolute Value Inequality: ∣ a ∣ + ∣ b ∣ ≥ ∣ a + b ∣ (Equality occurs when a b ≥ 0 )
[ x 2 − 4 x + 7 − 2 2 + x 2 − 8 x + 2 7 − 6 2 ] 4
= [ ( x − 2 ) 2 + ( 2 − 1 ) 2 + ( x − 4 ) 2 + ( 2 − 3 ) 2 ] 4
≥ [ 2 1 ( ∣ x − 2 + 2 − 1 ∣ + ∣ x − 4 − 3 + 2 ∣ ) ] 4 (Cauchy-Schwarz)
= [ 2 1 ( ∣ x + 2 − 3 ∣ + ∣ − x − 2 + 7 ∣ ) ] 4 ( ∣ a ∣ = ∣ − a ∣ ∀ a)
≥ [ 2 1 ∣ x + 2 − 3 − x − 2 + 7 ∣ ] 4 (Absolute value inequalities)
= [ 2 1 ∣ 4 ∣ ] 4 = 6 4
We can easily check that equality occurs when x = 2 + 1 . Hence the minimum value is 64.
Nice usage of all these inequalities.
It is best to state the equality conditions each time you used an inequality, which would then clearly demonstrate that "equality occurs when x = 2 + 1 ".
This is an elegant elementary approach without using calculus. Very nice.
i like your way, when i tried to solve it, i didnt think about the inequalitie x^{2} + y^{2} \geq \frac {1}{2} (x+y)^{2}
Rearranging the given expression
\strut x 2 − 4 x + 7 − 2 2 + \strut x 2 − 8 x + 2 7 − 6 2
we get
\strut ( x − 2 ) 2 + ( 2 − 1 ) 2 + \strut ( x − 4 ) 2 + ( 2 − 3 ) 2
Hence, we set some points on the Cartesian Plane,
P = ( x , 2 )
A = ( 2 , 1 )
B = ( 4 , 3 )
To get
A P = \strut ( x − 2 ) 2 + ( 2 − 1 ) 2
P B = \strut ( x − 4 ) 2 + ( 2 − 3 ) 2
Subharti'sMinimum
By Triangle Inequality ,
A P + P B
≥ A B
≥ \strut ( 4 − 2 ) 2 + ( 3 − 1 ) 2
≥ 8
So the minimum value
\strut x 2 − 4 x + 7 − 2 2 + \strut x 2 − 8 x + 2 7 − 6 2 = 8
Hence, the minimum value
[ \strut x 2 − 4 x + 7 − 2 2 + \strut x 2 − 8 x + 2 7 − 6 2 ] 4 = 6 4
Can you all see the picture? I don't know what's wrong but it only has an icon...
but at equality case A , P , B must be collinear ! And For minimum P must be lie between the AB But it can't be Possible Since Y cordinate of P is less then both of These !
Please Indicate where I'am Going Wrong ??
@Christopher Boo
wow
Can somebody help me find solutions to this question?
Anagram Cracker!!
Anagrams are problems related to shuffled letters which are needed to be arranged and made into perfect meaningful sentences without repeating the letters (letters can be used only once).
Here are some anagrams which you need to crack:
1) tuteauaewribeifslh
2) geaperioitrdspawsagnhabineod
3) enaednenetorfyimrw
Remember to arrange and make a meaningful sentence (one sentence from each group of letters), not single word. If you are able to solve this anagrams please inform me the answers as well as how you found the solutions to the anagrams.
Details and assumptions:
Example:
"My name is Anil" can be written in the form of group of letters as:
meailaysmnni
We have f (x) = [√(x²-4x+7-2√2) + √(x²-8x+27-6√2)]⁴
We note : f (x) attains minimum value when √(x²-4x+7-2√2) + √(x²-8x+27-6√2) attains minimum value
Re-write it as √[(x-2)²+(√2-1)²] + √[(x-4)²+(√2-3)²]
= Sum of the distances of the point (x,√2) from the points (2,1) and (4,3) which is minimum only when the three points are collinear
=> x = √2 + 1
=> min [√(x²-4x+7-2√2) + √(x²-8x+27-6√2)] = 2√2
=> f_min (x) = (2√2)⁴ = 64
Can somebody help me find solutions to this question?
Anagram Cracker!!
Anagrams are problems related to shuffled letters which are needed to be arranged and made into perfect meaningful sentences without repeating the letters (letters can be used only once).
Here are some anagrams which you need to crack:
1) tuteauaewribeifslh
2) geaperioitrdspawsagnhabineod
3) enaednenetorfyimrw
Remember to arrange and make a meaningful sentence (one sentence from each group of letters), not single word. If you are able to solve this anagrams please inform me the answers as well as how you found the solutions to the anagrams.
Details and assumptions:
Example:
"My name is Anil" can be written in the form of group of letters as:
meailaysmnni
to get minima of the function, we use derivative of the function g ( x ) by using:
let it f ( x ) = ( g ( x ) ) 4 , g ( x ) = x 2 − 4 x + 7 − 2 2 + x 2 − 8 x + 2 7 − 6 2
so d x d g ( x ) = 0 is 1 + 2 and by replacing in g ( x ) we've g ( 1 + 2 ) = 2 2 ,
finally f ( 2 2 ) = 6 4
I think a few dozen lines of working were omitted ..:)
The last line seems wrong to me.
Log in to reply
The last line seems wrong to me.
Indeed. I believe it was suppose to say " f ( 1 + 2 ) = 6 4 ".
denote g = x 2 − 4 x + 7 − 2 2 and h = x 2 − 8 x + 2 7 − 6 2 and f = g + h . observe that f > 0 for all x
so I will find minimum of f replace f 4 (by calculus)
we get d x d f = g x − 2 + h x − 4
and have to solve the equation
g x − 2 = h 4 − x (observe that x ∈ ( 2 , 4 ) ) ( x − 2 ) 2 ( 4 − x ) 2 − 3 = g h − 3 → x 2 − 4 x + 4 − 2 x 2 + 4 x + 4 = g − 2 x 2 + 4 x + 6 → − 2 x 2 + 4 x + 4 − 2 x 2 + 4 x + 6 − 1 = x 2 − 4 x + 4 g − 1 → − x 2 + 2 x + 2 1 = x 2 − 4 x + 4 3 − 2
to use your brute force to solve the last equation (easy to know the method) you will get x = 2 − 2 1 , 2 + 1 but x ∈ ( 2 , 4 ) so x = 2 + 1 that f ( 2 + 1 ) = 2 2 and f 4 = 6 4
Problem Loading...
Note Loading...
Set Loading...
Note that , x 2 − 4 x + 7 − 2 2 = ( x − 2 ) 2 + ( 2 − 1 ) 2 ,
x 2 − 8 x + 2 7 − 6 2 = ( x − 4 ) 2 + ( 2 − 3 ) 2 .
Hence, x 2 − 4 x + 7 − 2 2 + x 2 − 8 x + 2 7 − 6 2 is sum of distance from ( x , 2 ) to ( 2 , 1 ) , ( 4 , 3 ) in co-ordinate plane .
Assign name A to ( x , 2 ) , B to ( 2 , 1 ) , C to ( 4 , 3 ) .
_From Triangular inequality , _
A B + A C ≥ B C ,* Equality holds when A,B,C are co-linear *.
As , 1 < 2 < 3 , we must can find x for which A is on line segment BC .
(of course x can be found by taking slope of AB = slope of AC, but we are not interested in *value of x ) * .
Hence, m i n ( A B + A C ) = B C = 8
& ( m i n ( A B + A C ) ) 4 = 6 4