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How to get the following ContourPlot

ContourPlot[4 (s + i) -  Log[s] +  Log[i], {s, 0, 1}, {i, 0, 1}]

over a non rectangular region? First try

ContourPlot[4 (s + i) - Log[s] + Log[i], 
 Element[{s, i}  ,  {s > 0 && i > 0 && s + i < 1}]]

does not work.

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    $\begingroup$ For your first try, wrap the region condition with ImplicitRegion: ContourPlot[4 (s + i) - Log[s] + Log[i], Element[{s, i}, ImplicitRegion[{s > 0 && i > 0 && s + i < 1}, {s, i}]]], but this is considerably slower and disables the refinement option PlotPoints. $\endgroup$ Commented May 16, 2021 at 16:38

2 Answers 2

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For a region of the form $a \le x \le b$, $f(x) \le y \le g(x)$, one can use

ContourPlot[eqn, {x, a, b}, {y, f[x], g[x]}]

In the OP's case:

ContourPlot[4 (s + i) - Log[s] + Log[i], {s, 0, 1}, {i, 0, 1 - s}]

enter image description here

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Try the RegionFunction option:

ContourPlot[4 (s + i) - Log[s] + Log[i], {s, 0, 1}, {i, 0, 1}, 
 RegionFunction -> (#1 + #2 < 1 &)]

enter image description here

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