In one line
To optimise \(f(x,y,z)\) subject to \(\phi(x,y,z)=0\), form \(F=f+\lambda\phi\) and solve \(F_x=F_y=F_z=0\) together with \(\phi=0\).
Explained simply
Suppose you want the point on a road (the constraint) that is closest to a temple (the function to minimise). You are not free to go anywhere; you must stay on the road. At the best point, the direction in which the distance changes fastest is perpendicular to the road. Lagrange's method turns this geometric fact into equations with an extra unknown \(\lambda\), the multiplier.
Technically, at a constrained extreme point the gradients satisfy \(\nabla f=-\lambda\nabla\phi\) for some constant \(\lambda\). Writing \(F=f+\lambda\phi\) and setting its partial derivatives to zero gives exactly these conditions. This is a self-learning topic; the method gives stationary points, and the nature is usually decided from the physical problem.
How it works
- Write the function \(f\) to optimise and the constraint as \(\phi=0\).
- Form \(F=f+\lambda\phi\).
- Write \(F_x=0\), \(F_y=0\), \(F_z=0\).
- Eliminate \(\lambda\) (often by expressing each equation as "\(\lambda=\dots\)") and use \(\phi=0\) to find \(x,y,z\).
- Substitute to get the extreme value.
Key formula
\(F(x,y,z)=f(x,y,z)+\lambda\,\phi(x,y,z)\); conditions \(F_x=F_y=F_z=0\), \(\phi=0\). \(\lambda\) is a constant called the Lagrange multiplier.
Quick example
Find the minimum of \(x^2+y^2+z^2\) subject to \(x+y+z=3\).
\(F=x^2+y^2+z^2+\lambda(x+y+z-3)\).
\(F_x=2x+\lambda=0\), \(F_y=2y+\lambda=0\), \(F_z=2z+\lambda=0\). So \(x=y=z=-\lambda/2\).
Constraint: \(3x=3\Rightarrow x=y=z=1\). Minimum value \(=1+1+1=3\).
Geometric meaning: the point of the plane \(x+y+z=3\) nearest the origin is \((1,1,1)\), at distance \(\sqrt3\).
Common mistakes
- Forgetting to use the constraint equation at the end.
- Dividing by a variable that could be zero and losing solutions.
- Writing \(F=f\cdot\lambda\phi\) instead of \(f+\lambda\phi\).
How MU asks it
As a self-learning topic it may appear as a 5/6-mark application: "Using Lagrange's method, find the maximum value of \(xyz\) subject to \(x+y+z=a\)" (answer \(a^3/27\) at \(x=y=z=a/3\)). Write \(F\), the three equations, the elimination of \(\lambda\), the point and the value.