With an inequality constraint …, the multiplier at the minimum satisfies

These problems allow inequality constraints, optional but good.

With an inequality constraint $g\le k$, the multiplier at the minimum satisfies $\lambda \le 0$. If $k$ is increased, $f_{\min }$ goes down (since $\lambda =d{f}_{\min }/dk$). Explain the reasoning: By increasing $k$, (more) (fewer) points satisfy the constraints. Therefore (more) (fewer) points are available to minimize $f$. Therefore $f_{\min }$ goes (up) (down).

Gilbert Strang, Calculus, 3rd edition (Wellesley-Cambridge Press; MIT OpenCourseWare RES.18-001) · §13.7 Constraints and Lagrange Multipliers, #26