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| Hi, guys How to use FindInstance or others to make y2[[1]] maximum. Clear["`*"]; H = 6; L = 1; B = 2; \[Alpha] = 3/10; \[Eta]1a = \[Alpha]*(\!\( \*SubsuperscriptBox[\(\[Integral]\), \(Min[ \*FractionBox[\(p2 - \((p1 -... |
| Above question may be solved if we figure out the difference of the following two programmings : Programming (1) Clear["`*"]; L = 1; H = 2; \[Alpha] = 1/10; \[Eta]1q[p1_, p2_, g_] = \[Alpha]*\!\( ... |
| Thank you very much! |
| Hi, guys I use Table to calculate my problem on a discrete interval as follows: Clear["`*"];(*????????,??Table*) b = 2; y01 = N[Maximize[{-x*a - 6*x*b + 1, -6 {2, 5}] Y[[1, All, 2, All, 2]] ListPlot[Y[[1, All, 2, All,... |
| Thank you, Chaib, you are so warmhearted! |
| Good, Jõgeva, thank you for your answer! |
| Hello, friends Optimization 1: MaximalBy[{{4.9375, {x -> -(3/4)}}, {2.0792, {x -> 2/Sqrt[ 3]}}}, First] the result is: {{4.9375, {x -> -(3/4)}}} Optimization 2: MaximalBy[{{79/ 16, {x -> -(3/4)}}, {-1 + 16/( ... |
| Thank you, but I want to derive the result of a piecewise function, instead of Max or Min, just like the function in the picture. I just want to simplify the result, the first line of the result should not exist. |
| Thank you for your hint, Sander. After replacing = with == , it works and I think the syntax is right, but there is not even a point in the figure, it is empty. I do not know why it happens. |
| Thank you, but it seems doesn't work. |