Lagrange Multipliers

Method of Lagrange Multipliers

Earlier, as a part of a problem were we were finding absolute maxima and minima of a continuous function on a closed disk, we had to find maxima and minima of the function \(f(x,y) = 2x^3 - y^4\) on the boundary of the unit disk \(x^2 + y^2 \le 1\). The boundary has the equation \(x^2 + y^2 = 1\). We did it by splitting the boundary into the upper and lower semicircles, parametrizing each of them by \(x\), and solving the two resulting single variable optimization problems. We ended up with critical points at \((0, \pm 1)\), and \((-\frac{1}{2}, \pm\frac{\sqrt{3}}{2})\), and with end-points \((\pm 1, 0)\). The absolute maximum was attained at \((1,0)\), the absolute minimum at \((-1,0)\).

The graph of the function looks like this:

Now we are going to learn a different way of finding the maximum and minimum of \(f\) on the boundary. This kind of problem is called constrained optimization: find the maximum and minimum of a function of several variables \(f(x_1, x_2, \dots, x_n)\) subject to a constraint given by an equation in the form \(g(x_1, x_2, \dots, x_n) = c\) for some function \(g\) and a constant \(c\).

The contour plot of the function \(f(x,y) = 2x^3 - y^4\) together with the unit circle \(x^2 + y^2 = 1\) is shown below. The points where a contour is tangent to the unit circle are marked and labeled.

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More examples

Ellipse constraint

Find the absolute maxima and minima of \(f(x,y) = 3x^2 + 2xy\) on the ellipse \(4x^2 + y^2 = 5\).

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Unbounded constraint

Find the absolute maxima and minima of \(f(x,y) = 3x^2 + 2y^2\) on the hyperbola \(xy = 6\).

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Three variables

Find the absolute maxima and minima of \(f(x,y,z) = e^{xyz}\) subject to \(2x^2 + y^2 + z^2 = 24\).