The [url=https://en.wikipedia.org/wiki/Lagrange_multiplier]Lagrange multiplier[/url] theorem roughly states that at any [url=https://en.wikipedia.org/wiki/Stationary_point]stationary point[/url] of the function that also satisfies the equality constraints, the [url=https://en.wikipedia.org/wiki/Gradient]gradient[/url] of the function at that point can be expressed as a [url=https://en.wikipedia.org/wiki/Linear_combination]linear combination[/url] of the gradients of the constraints at that point, with the Lagrange multipliers acting as [url=https://en.wikipedia.org/wiki/Coefficient]coefficients[/url]. The relationship between the gradient of the function and gradients of the constraints leads to [b]Lagrangian function[/b].

[color=#0000ff][b]Example:[/b][/color][br]Minimize a function of two variables [color=#0000ff][i]z = xy[/i][/color] under a constraint qiven by equality [i][color=#0000ff]c:[/color] [math]x^2-3xy+y^2=20[/math][/i].

The great advantage of this method is that it allows the optimization to be solved without explicit [url=https://en.wikipedia.org/wiki/Parameterization]parameterization[/url] in terms of the constraints.[br]Let us denote function [color=#0000ff][i]z[/i] = f([i]x, y[/i])[/color] and the constraints given by equality as [i][color=#0000ff]g(x, y) = 0[/color][/i].[br]Lagrangian function[color=#0000ff] L([i]x, y)[/i] =[/color] [color=#0000ff]f([i]x, y[/i])[/color] - λ [i][color=#0000ff]g(x, y) [/color][/i][br]Stationary point of the Lagrangian function: [color=#0000ff]L'[sub]x[/sub] ([i]x, y) = [/i]0 [color=#000000]and[/color][/color][color=#0000ff]L'[sub]y[/sub] ([i]x, y) = [/i]0[/color] gives a [url=https://en.wikipedia.org/wiki/Necessary_condition]necessary condition[/url] for optimality in constrained problems.[br][br]For the hyperboloid [color=#0000ff][i]z[/i] = [i]x . y [/i][color=#000000]above the hyperbola [/color][/color][color=#0000ff]x[sup]2[/sup] [code]−[/code] 3xy + y[sup]2[/sup]= 20 [color=#000000]we have Langrangian function [br][/color][/color][color=#0000ff]L([i]x, y)[/i] =[/color] [color=#0000ff][i]x . y[/i][/color] - λ [i][color=#0000ff]([color=#0000ff]x[sup]2[/sup] [code]−[/code] 3xy + y[sup]2[/sup] [i][color=#0000ff][color=#0000ff][code]−[/code] [/color][/color][/i]20[/color])[br][/color][/i][color=#0000ff][color=#000000]Equations for stationary point [br][color=#0000ff]L'[sub]x[/sub] ([i]x, y) = y [/i][/color][/color][/color][color=#0000ff][color=#000000][color=#0000ff][i][i][color=#0000ff][color=#0000ff][code]−[/code] [/color][/color][/i] 2λx + 3[/i][/color][/color][/color][color=#0000ff][color=#000000][color=#0000ff][i][color=#0000ff][color=#000000][color=#0000ff][i]λ[/i][/color][/color][/color]y = [/i]0 [color=#000000]and[/color][/color][color=#0000ff]L'[sub]y[/sub] ([i]x, y) [color=#0000ff][color=#000000][color=#0000ff][i]= x + [/i][/color][/color][/color][color=#0000ff][color=#000000][color=#0000ff][i]3λx [i][color=#0000ff][color=#0000ff][i][color=#0000ff][color=#0000ff][code]−[/code][/color][/color][/i][/color][/color][/i] 2[color=#0000ff][color=#000000][color=#0000ff][i]λ[/i][/color][/color][/color]y[/i][/color][/color][/color] = [/i]0[/color][br]Add equations together and factor the member [color=#0000ff]([i]x+y[/i])[/color]. From the acquired reduction we have the relation for minimal points: [color=#0000ff]x = [i][i][i][i][code]−[/code][/i][/i][/i][/i] y[/color]. [br][br][b]Conclusion[/b][br]Function [i][color=#0000ff]z = x.y[/color][/i] has two minimal values above hyperbola: at points [color=#0000ff]Min1 = ([color=#0000ff][color=#000000][color=#0000ff][i][i][i][i][code]−[/code][/i][/i][/i][/i][/color][/color][/color]2, 2)[/color] and [color=#0000ff]Min2 = (2, [/color][/color][/color][color=#0000ff][color=#000000][color=#0000ff][color=#0000ff][color=#000000][color=#0000ff][i][i][i][i][code]−[/code][/i][/i][/i][/i][/color][/color][/color]2)[/color]. [/color][/color]

Find the extrema points of a function [i]z = xy[/i] subject to equality constraint [math]x^2+y^2=1[/math].

Parameterization of the circle gives us the function z(t) of the height over the circle. [br]z = xy = cos t . sin t[br]Stationary points: z'=0 [b]⇒[/b] [math]cos^2t-sin^2t=0[/math] [br]Maxima: [math]\left(\frac{\pm\sqrt{2}}{2},\frac{\pm\sqrt{2}}{2}\right)[/math] [br]Minima: [math]\left(\frac{\pm\sqrt{2}}{2},\frac{\mp\sqrt{2}}{2}\right)[/math]