Timeline for Maximizing a Definite Integral Subject to Constraints
Current License: CC BY-SA 2.5
8 events
when toggle format | what | by | license | comment | |
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Mar 22, 2011 at 14:35 | history | edited | Charles Matthews | CC BY-SA 2.5 |
update
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Mar 22, 2011 at 13:45 | comment | added | Jeff Morrow | This answer is great because it does even more than I asked for. It shows that the maximum value of the integral is attained only if f = g over [0, b]. I get the point that it is the general relationship of f to g and the equality of the integrals that drives the result, not the nature of g itself, but the specification of g grew out of the overall problem I had posed to myself. Anway thanks again to all. | |
Mar 22, 2011 at 7:33 | history | edited | Charles Matthews | CC BY-SA 2.5 |
notation
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Mar 21, 2011 at 23:43 | comment | added | Will Jagy | I typeset the answer, using the symbols of KConrad's edit, in a comment above. As Charles says, most of your conditions have no bearing on the answer. | |
Mar 21, 2011 at 23:22 | vote | accept | Jeff Morrow | ||
Mar 21, 2011 at 23:22 | comment | added | Jeff Morrow | Yes, that is exactly what I wanted. Thank you very much. I am also thankful to KConrad for the reformatting and to both him and Will Jagy for taking the time to look at a problem posed by the mathematically ignorant. I appreciate all the help. | |
Mar 21, 2011 at 22:45 | comment | added | Jeff Morrow | Yes I think you have given an elegantly simple answer. It is a little abbreviated so I am going to go work it out. If I have a question, I'll be back. If have none, I'll make sure to give you your reputation points. In any case, thank you for pointing out a direction. I suspected I was going to need a function that combined f and h so I reserved the letter g (and likewise L between j and k). Of course when we shifted to showing functions with lower-case letters L became unusable. BUT THANKS A LOT. | |
Mar 21, 2011 at 21:10 | history | answered | Charles Matthews | CC BY-SA 2.5 |