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anonymous
 5 years ago
Using ONLY calculus to prove that the minimum value of the function f(x)=xa1+xa2+xa3+...+xa100 is f(a50)
You are given a1<=a2<=a3<=.....<=a100.
anonymous
 5 years ago
Using ONLY calculus to prove that the minimum value of the function f(x)=xa1+xa2+xa3+...+xa100 is f(a50) You are given a1<=a2<=a3<=.....<=a100.

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watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Nice problem! We will prove in general that the minimum value of \[f(x)=\sum_{k=1}^{2n}xa_k\] where \(a_{k}<a_{k+1}\) for all \(k\) is \(f(a_n)\). Just for convenient, let \(a_0=\infty\) and \(a_{2n+1}=\infty\). Notice that for \(x\in(a_k,a_{k+1})\) we have \(xa_i=xa_i\) for all \(0< i\leq k\) and \(xa_i=(xa_i)\) for \(k< i<2n+1\). It follows that in \((a_k,a_{k+1})\) we have \[f(x)=\sum_{i=1}^{k}(xa_i)+\sum_{i=k+1}^{2n} (xa_i)=(2k2n)x+\sum_{i=k+1}^{2n}a_i\sum_{i=1}^k a_i\] Note that \(f'(x)=2k2n\). So \(f(x)\) is decreasing on \((a_k,a_{k+1})\) for all \(k<n\) and increasing for \(k> n\). Hence \(f(x)\) attain its's minimum value at \(a_n\).

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.1Hi, how about my answer above? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0very intuitive :) keep up. You are really good!
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