A collection of intriguing competition level problems for secondary school students.
Wednesday, July 1, 2015
Optimization Contest Problem: Find the maximum of f(x) (Heuristic Solution)
Find the maximum of f(x)=x4−x2x6+2x3−1 where x>1.
If we want to maximize f(x)=x4−x2x6+2x3−1=1(x6+2x3−1x4−x2), this could be done if we are to find the minimum value for the expression x6+2x3−1x4−x2.
Note that
x6+2x3−1x4−x2=x2+1+2x3+x2−1x4−x2=x2+1+2x3x2(x2−1)+x2−1x2(x2−1)=x2+1+2xx2−1+1x2=(x−1x)2+2(x−1x)+3=(x−1x)2+1(x−1x)+1(x−1x)+3−−−(∗)
And since the term x−1x 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
(x−1x)2+1(x−1x)+1(x−1x)≥3
Equality occurs when (x−1x)2=1(x−1x), i.e.
(x−1x)3=1
x−1x=1
x2−x−1=0
(x−12)2−54=0
x=1+√52
Therefore, the minimum value for x6+2x3−1x4−x2 is 3+3=6 and that results in a maximum of
f(x)=x4−x2x6+2x3−1=1(x6+2x3−1x4−x2) as 16.
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