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Wednesday, July 1, 2015

Optimization Contest Problem: Find the maximum of f(x) (Heuristic Solution)


Find the maximum of f(x)=x4x2x6+2x31 where x>1.

If we want to maximize f(x)=x4x2x6+2x31=1(x6+2x31x4x2), this could be done if we are to find the minimum value for the expression x6+2x31x4x2.

Note that

x6+2x31x4x2=x2+1+2x3+x21x4x2=x2+1+2x3x2(x21)+x21x2(x21)=x2+1+2xx21+1x2=(x1x)2+2(x1x)+3=(x1x)2+1(x1x)+1(x1x)+3()

And since the term x1x is a positive in the given domain, we then are allowed to apply the AM-GM to the first three terms of the expression (*) and that gives

(x1x)2+1(x1x)+1(x1x)3

Equality occurs when (x1x)2=1(x1x), i.e.

(x1x)3=1

x1x=1

x2x1=0

(x12)254=0

x=1+52

Therefore, the minimum value for x6+2x31x4x2 is 3+3=6 and that results in a maximum of

f(x)=x4x2x6+2x31=1(x6+2x31x4x2) as 16.


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