14.2 Extremal Values on a Curve in Two Dimensions

Suppose we have a curve, C that is defined by an equation, G(x, y) = 0, and we seek an extremal value of F(x, y) among points restricted to lie on this curve.

Imagine, for example that G represents an ellipse, ax2 + by2 = 1, and we want the maximum of xy on C.

At any point q on C, we are free to move while staying on C only in the direction of the tangent line to the curve. Our condition above for an extremum then tells us that for q to be an extremum of F, F must have 0 derivative in the direction of the tangent, t, to the curve defined by G.

This means that the gradient of F must be perpendicular to t. But the gradient of G is perpendicular to t as well, so that in two dimensions the gradient of F and the gradient of G must be parallel, for F to have an extremum on G.

There are two standard ways to express this condition.

One is to notice that it means that the parallelogram formed by F and G has no area, so that the determinant with these vectors as columns must be 0.

The other is to notice that it means that F = cG for some constant c.

Either observation allows us to find the extrema.


The second method is called that of "Lagrange Multipliers", and the constant c is called a Lagrange Multiplier.



In the example you can click on above, if you write out the three equations defined by G = ax2+by2 -1 =0, i(F - cG) = 0 and j(F - cG) = 0, you may solve them for x, y and c, and arrive at the same solutions obtained.

Again, computing second derivatives (or examining values of F) must be used to determine the local and/or global maxima and minima.

When a curve is defined parametrically with parameter t you can write F = F(x(t), y(t)) and apply the single variable condition that


14.1 Work out the details of the Lagrange Multipliers approach to the example above.

14.2 Suppose we want to maximize the volume of a vertically oriented cylinder given a fixed value q for the surface area of its sides and its top (but not its bottom). What radius and height should it have?