245 followers problem
Given that x and y are positive reals satisfying x 1 + y 2 4 5 = 1 , the minimum value of x + y + x 2 + y 2 can be expressed as A + B C , where A , B and C are positive integers with C square-free. Find the value of A + B + C .
The answer is 516.
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4 solutions
Generalization : If x/a + y/b = 1 then minimum value of (a + b) + √(a² + b²) is 2(x + y + root(2xy))
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I'm just wondering, is there a geometry solution to this problem? My first thought was that x + y + x 2 + y 2 could represent the perimeter of a right angled triangle...
Agreed, but your generalisation is just a rescaling of the original result.
The value of g ( α ) in my above post is 2 [ α + 1 + 2 α ] .
If x a + y b = 1 then ξ 1 + η a − 1 b = 1 where ξ = a − 1 x and η = a − 1 y . Since x + y + x 2 + y 2 = a ( ξ + η + ξ 2 + η 2 ) we see that the minimum value of x + y + x 2 + y 2 , subject to x a + y b = 1 , is just a times the minimum value of ξ + η + ξ 2 + η 2 subject to ξ 1 + η a − 1 b = 1 , which is therefore a × 2 [ 1 + a − 1 b + 2 a − 1 b ] = 2 [ a + b + 2 a b ] .
How do we get to that?
Why can't we take x=36, y=252 to get the minimum value.
Won't an easier solution be using Lagrange Multipliers?
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It's do-able, but I am not sure (after having a brief go) if LM will get the answer any more easily. If you can do it, please show me.
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Okay, I will first do it, and surely put it up if it works!
Yeah, I tried it and it IS working. I will post it tomorrow.
First I put the restricting condition into the expression to minimize to get one equation with only one variable x. Then I found the minimum graphically and following numerically with XPLORE to value 536.271887242357. As last step I now had to determine which root ends with the same series of digits and I found 1 9 6 0 .
Using these results it was very simple to obtain the suitable shape after separating the 2 roots included in 1960.
Remark: If someone die-hard mathematician does not like my algorithm I can not change! As retired physicist I prefer such ways especially they possess a distinct scientific character!
The minimum value is actually 4 9 2 − 1 4 1 0
No it is not. Your value comes from using the negative root α − , which is not permitted.
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Without redoing this problem to check this, I'll go along with your explanation.
how is that?
We must have y > a = 2 4 5 . Putting y = a + z gives x = z z + a , and so we need to minimize g ( z ) = z z + a + z + a + ( z z + a ) 2 + ( z + a ) 2 = ( 1 + z a ) ( 1 + z + z 2 + 1 ) for z > 0 . Now g ′ ( z ) = z 2 z 2 + 1 ( z 2 − a ) z 2 + 1 + ( z 3 − a ) , so we need to solve the equation ( z 2 − a ) z 2 + 1 + ( z 3 − a ) = 0 . Solutions of this equation certainly satisfy (by squaring) ( z 2 − a ) 2 ( z 2 + 1 ) ( 2 a − 1 ) z 4 − 2 a z 3 − ( a 2 − 2 a ) z 2 ( 2 a − 1 ) z 2 − 2 a z − ( a 2 − 2 a ) z = = = = ( z 3 − a ) 2 0 0 α ± = 2 a − 1 a ± 2 a − 1 a − 1 2 a Since we must have z > 0 we deduce that z = α + = 2 a − 1 a + 2 a − 1 a − 1 2 a Indeed, α + is not only the positive option out of α + and α − ; it is the only solution of g ′ ( z ) = 0 . The other value α − is the solution of the equation ( z 2 − a ) z 2 + 1 − ( z 3 − a ) = 0 , and is therefore a solution of the rogue equation that was introduced when we squared the equation g ′ ( z ) = 0 to remove the square roots. It is easy to see that g ′ ( z ) < 0 for 0 < z < α + , with g ′ ( z ) > 0 for z > α + , and so z = α + is indeed a minimum for g ( z ) .
With a = 2 4 5 this gives z = 4 8 9 7 ( 3 5 + 2 4 4 1 0 ) and g ( z ) = 4 9 2 + 1 4 1 0 .