How Can You Maximize the Product of Integers Given a Fixed Sum?

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SUMMARY

The discussion focuses on maximizing the product of integers given a fixed sum of 73. The optimal strategy involves using the integer 3 as much as possible, as it provides a higher product compared to using 2 or 4. Specifically, the solution suggests utilizing twenty-three 3s and one 4 to achieve the maximum product. This approach is based on the mathematical principle that 3 raised to the power of one-third is greater than 2 raised to the power of one-half and 4 raised to the power of one-fourth.

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mhill
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we must obtain 'N' and a(i) i=1,2,3,....,N on condition that

the product [tex]a(1)a(2)a(3)...a(N)[/tex] is the highest possible

[tex]a(1)+a(2)+a(3)+...+a(N)=73[/tex]

every a(i) is positive

here N (this is the hardest part) is not known and must be calculated
 
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Well, since [itex]3^{1/3}>2^{1/2}=4^{1/4}[/itex], I'd suggest using as many 3s as will fit, making the rest a 2 or 4. In this case twenty three 3s and one 4 would seem to be the best. Now just check my work and make sure I didn't miss something silly.
 

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