Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 20 / a

For a differentiable objective $f$ and regular constraint $g=r$, a constrained stationary point satisfieswhere $\lambda$ is a Lagrange multiplier. Solve these equations for both the coordinates and $\lambda$.

Geometrically, the gradient $\nabla g$ is normal to the constraint level set when $\nabla g\ne 0$. Every feasible infinitesimal displacement $\mathbf{v}$ is tangent to that level set, so $\nabla g\cdot \mathbf{v}=0$. At a constrained stationary point, the directional change $\nabla f\cdot \mathbf{v}$ must also vanish for every such tangent displacement. Thus $\nabla f$ is normal to the same tangent plane and is a scalar multiple of $\nabla g$. This is the geometric basis of the Lagrange multiplier method.

For a second constraint $h=s$, with linearly independent vectors $\nabla g$ and $\nabla h$, the feasible tangent directions are perpendicular to both. Their normal space is the linear span of these two gradients, so introduce a second Lagrange multiplier:These equations find candidates; deciding whether they are maxima, minima or saddle points requires further analysis. Constraint points where the required gradient independence fails must be examined separately.