# Plot multivariable function with constraints [closed]

Is it possible to plot some function similar to $f(x,y)=x+2y$ with a constraint like this? $x2+y2=5$

We want to optimize (i.e. find the minimum and maximum value of) a function, $f(x, y)$, subject to the constraint $g(x,y)=0$. And I would like to know if it is possible to plot function f with the restriction g

Thanks

• This question does not make sense to me. What do you mean "print" a function ? Do you mean define and plot ? – Lotus Oct 10 '17 at 10:57
• Use ContourPlot, but fix your variable names first. – Alan Oct 10 '17 at 11:06
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• The problem is this: We want to optimize (i.e. find the minimum and maximum value of) a function, f(x, y), subject to the constraint g(x,y)=0. And I would like to know if it is possible to plot function f with the restriction g – Ana Oct 10 '17 at 12:25
• Do you mean x^2 + y^2 == 5? – m_goldberg Oct 25 '17 at 3:52

Something like?

Plot3D[x + 2 y, {x, -Sqrt, Sqrt}, {y, -Sqrt, Sqrt},
RegionFunction -> Function[{x, y}, x^2 + y^2 < 5]]

• Plot3D[4000*x*y^3, {x, 0, 1}, {y, 0, 2}, RegionFunction -> Function[{x, y}, x + 2/3*y - 1.3 = 0]] works in the same way for me with or without constraint – Ana Oct 10 '17 at 12:36
• Try use less/greater in Function[{x, y}, x + 2/3*y - 1.3 <> 0. – Eduard Oct 10 '17 at 12:50
• that is what I was looking for :) – Ana Oct 10 '17 at 13:11

Using ConditionalExpression:

Plot3D[ConditionalExpression[x + 2 y, x^2 + y^2 <= 5],
{x, -Sqrt, Sqrt}, {y, -Sqrt, Sqrt},
BoundaryStyle -> Directive[Thick, Red], Mesh -> None] Add the option PlotStyle->None to display only the boundary: Using MeshFunctions:

Plot3D[x + 2 y, {x, -Sqrt, Sqrt}, {y, -Sqrt, Sqrt},
MeshFunctions -> Function[{x, y, z}, x^2 + y^2], Mesh -> {{5}},
MeshStyle -> Directive[Thick, Red],
MeshShading -> {Automatic, None}, BoundaryStyle -> None] Use MeshShading -> {None, None} to get the boundary only: • that is what I was looking for :) – Ana Oct 10 '17 at 13:10